FW physics 10192011 843 PM page 1 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

1

F12

F21

m1

m2

r2

r1

d

The torque by a force applied to a point with mass m is defined as

(11)

with the magnitude =rFsin

perpendicular component of r (with respect to F) times F

perpendicular component of F (with respect to r ) times r

r F r F rF

r F

rF

Consider an object which is free to rotate around a fixed axis A If we apply

a single exterior force to this object every single point of it will experience

the same angular acceleration α We must calculate (sum) the terms

corresponding to the rotational acceleration of all the individual masses mi

To get the total torque of a system of particles we must sum over all the

torques acting on the individual masses mi

(12)i i ir F

However for an extended

object of point masses all

interior forces come in

equal and opposite pairs

and therefore cancel each

other out The same is true

for the internal torques

corresponding to pairs of

equal and opposite forces

Therefore the total torque on an extended object is equal to the vector

sum of the external torques only

θ

Fi

ri

FW physics 10192011 843 PM page 2 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

2

(13) external external

i i

= i i ir F

For a single external force and torque this reduces itself to the formula

(14)

only those components of the two vectors contribute which are

perpendicular to each other

sin

r F

rF rF r F

The torque on an extended object or a system of particles is caused by the

external torque only

Kinetic Energy of Rotation for a cylinder rotating around a fixed axis (21)

2

1 1 1

2 21 The work done by a torque is equal to the change in

2

rotational kinetic energy

N N N

ext i i i i i i i i

i i ia to b a to b a to b a to b a to b

I

b a rot

a to b

dW F dr m a dr m rrd m r d I d

dt

W I d I K

of the rotating object

Different approach

(22)

1 2 1 2 1 2 1 2 1 2

2 21

2

The work done by a torque rotating a set of masses is equal to the change in

rotational kine

ext ext

a to b to to to to to

b a rot

dW F dr F r d d I d I dt I d

dt

W I K

tic energy of the rotating set of masses

FW physics 10192011 843 PM page 3 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

3

We can also see that work done by a torque is equal to the integral of the torque over the

angle of rotation similarly to the work done by a linear force which is equal to the

integral of the force over the distance travelled

(23)

For a constant torque we get

dWPower = (Compare to P=Fv for linear motion)

dt

(24)

2

1

For a constant torque we get

W=

a b

N

i i

ia tob b to a to b

I

b a

W F dr m r d d

In the case of a constant torque we also get an easy expression for the power delivered by

the torque

(25)

v

path

F ma F dr W K

dW drF r W Power F F

dt dt

The torque on an extended object causes an accelerated rotation around an axis This is

the extension of Newtons second law for linear motion Acceleration means increase in

kinetic energy Therefore we get 2 21

2b a rotW I K

If there is no net exterior torque acting an object rotating at a constant rate will keep

rotating at this rate (Newtons first law applied to rotations)

The direction of the torque around a fixed axis of rotation is always perpendicular to the

plane of rotation ie parallel to the axis of rotation

Angular Momentum L Let us now consider the angular momentum of the system of particles

The angular momentum L for one particle in motion is given by

(11) l r p

The derivative of the angular momentum is

(12)

0 torque

dl d dr dpr p p r r F

dt dt dt dt

FW physics 10192011 843 PM page 4 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

4

For a system of particle we must sum over all the angular momenta and then take the

derivative with respect to t

(13)

where k is the summation index for

the exterior torques only the interior torques cancelling each other in pairs

i i i

i i

ii i k

i i k

L l r p

dpdL dl r

dt dt dt

(14) exterior

k

k

dL

dt

The direction of the torque around a fixed axis of rotation is always perpendicular to the

plane of rotation ie parallel to the axis of rotation

The angular momentum of an extended object in a rotation about a point is defined by the

vector sum of all the angular momenta li All radius vectors start at the same point Thus

it is evident that the angular momentum vector depends on the chosen origin

(15) ivi i i ii

i i

L l r p r m

We consider here a constant rotation around a fixed axis say the z-axis while we use the

x-y plane as the plane of rotation In the case of a counter clock wise rotation the

direction of the angular momentum points in the positive z direction We get

(16)

i

i

v

v

In a rotation around a fixed axis the vector is always perpendicular to the

tangential velocity

i i i iii i

i

i

i

L l r p r m

r

r

ru

(17)

2

i

2 2

object

v

I is called the moment of inertia=

We see that

i i i i i z i i

i i i

i i

i

L r m r m ru u m r

L I m r r dm

dL dI I

dt dt

The change in total momentum of a system of particles (or an extended object) is

equal to the vector sum of the exterior torques

FW physics 10192011 843 PM page 5 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

5

This is also valid for the relationship between the total torque of an extended object and

its angular momentum We write the relationship only for the magnitude using (14)

(18)

exterior

k

k

dL

dt

dL d dI I I

dt dt dt

Torque and angular momentum We just need to prove that torque is equal to the derivative of angular momentum

just like force is the derivative of linear momentum

(19) 2=mr just like

where =

z

z r

dL dpu F ma

dt dt

u u u

Example If we apply an exterior force to a disk capable of rotating around an axis each

individual point mass im of the disk is subjected to a clockwise (or ccw) tangential time

change of momentum vi i

i

dp dm

dt dt

or i i i im a m r In reference to the axis of rotation

this becomes the change of angular momentum for each individual mass im

2i ii i i i i i i i i i i i

dL dpdr p r rm a rm r m r

dt dt dt

(110) 2 2

total i i i i i i i

i i i i

R F m a m r m r I

(111) dp dL d dr

r F r r p pdt dt dt dt

torque

dpr

dt

As velocity and momentum are parallel vectors only the last term remains and we have

(112)

The torque created by the exterior force is equal to the change

of angular momentum of the rotating object Or the exterior torque changes the

angular momentum of a rotating object

dLr F

dt

Next we need to prove that Newtons second law becomes for the rotation of a solid

object around a fixed axis AI

FW physics 10192011 843 PM page 6 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

6

The angular momentum of an extended object in a rotation about a point is defined by the

vector sum of all the angular momenta li All radius vectors start at the same point Thus

it is evident that the angular momentum vector depends on the chosen origin

(113) ivi i i ii

i i

L l r p r m

We consider here a constant rotation around a fixed axis say the z-axis while we use the

x-y plane as the plane of rotation in which we define the radial and tangential unit

vectors In the case of a counter clock wise rotation the direction of the angular

momentum points in the positive z direction We get

(114)

i

i

v

v

In a rotation around a fixed axis the vector is always perpendicular to the

tangential velocity The vector is perpendicular to the plane of rotatio

i i i iii i

i

i

i

L l r p r m

r

r

ru

i

n We call the

unit vector in that direction such that

v

z r z

i z i r i

u u u u

r u ru ru

(115)

2

i

2 2

object

v

I is called the moment of inertia=

i i i i i z i i

i i i

z

i i

i

L r m r m ru u m r

L I

m r r dm

This is also valid for the relationship between the total torque of an extended object and

its angular momentum We write the relationship only for the magnitude

(116) dL d d

I I Idt dt dt

Example Consider a solid disk of mass 25 kg and radius 10 cm The disk is free to rotate

without friction around a fixed axis through its

center of mass A long massless cord is wound

around the perimeter of the disk and a mass of 18

kg is suspended from it

torque =

Tr=Iα=05Mr2α

mg

T-mg=ma=mrα

M=25kg

FW physics 10192011 843 PM page 7 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

7

(117)

2

2

1 1 1

2 2 2

1

2

18123 T=154N

1 143

2

Tr Mr Mra T Ma

T mg ma Ma mg ma

mga g m s

m M

The exterior torque is Tr=154N01m=154Nm

Rolling motion We assume that a sphere rolls without slipping

(118) i ir R

We take the time derivative of this equation to get the velocities

(119) v vi icm i i

dr dRdR

dt dt dt

The velocity of a point of a rolling object is therefore the velocity of the center of mass

plus the velocity of this point around the center of mass

For a point on the rim we have

CM

mi

O

vector from origin

to mi

vector from origin

to center of mass

vector from

center of mass to

FW physics 10192011 843 PM page 8 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

8

A C

mi

(120)

cm

v v

v =

cm

dr d dRR

dt dt dt

R

Kinetic energy of the rolling object We sum over all the kinetic energies of the points making up the object

(121)

2

2

2 2

2 2

1 1v v

2 2

1 1v ( ) v ( )

2 2

1 1v v

2 2

i i i i cm i

i cm i i i cm i

cm cm cm i i

K K m m R

m m R m R

M I m R

The last term is 0 (why) and therefore the total kinetic energy of a rolling object equals

the kinetic energy of its center of mass plus the kinetic energy of rotation of the object

around the axis through its center of mass

(122) 2 21 1

v2 2

cm cmK M I

One can convince oneself easily with the aid of the parallel axis theorem that this kinetic

energy is equal to the kinetic energy of a pure rotation around a point A at the surface of

the rotating object

Proof of the parallel axis theorem the axes of rotation are at A and C(center of mass)

perpendicular to the plane of writing (see also angular momentum)

(123)

2

2 2 2

2 2

2

2

2

0 because 0 why

A i i i i i i i

i i i i i

cm i i

I m R m D r m D r D r

D m m r D m r

MD I m r

Thus

(124)2 where D is the distance between the two axesA cmI MD I

FW physics 10192011 843 PM page 2 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

2

(13) external external

i i

= i i ir F

For a single external force and torque this reduces itself to the formula

(14)

only those components of the two vectors contribute which are

perpendicular to each other

sin

r F

rF rF r F

The torque on an extended object or a system of particles is caused by the

external torque only

Kinetic Energy of Rotation for a cylinder rotating around a fixed axis (21)

2

1 1 1

2 21 The work done by a torque is equal to the change in

2

rotational kinetic energy

N N N

ext i i i i i i i i

i i ia to b a to b a to b a to b a to b

I

b a rot

a to b

dW F dr m a dr m rrd m r d I d

dt

W I d I K

of the rotating object

Different approach

(22)

1 2 1 2 1 2 1 2 1 2

2 21

2

The work done by a torque rotating a set of masses is equal to the change in

rotational kine

ext ext

a to b to to to to to

b a rot

dW F dr F r d d I d I dt I d

dt

W I K

tic energy of the rotating set of masses

FW physics 10192011 843 PM page 3 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

3

We can also see that work done by a torque is equal to the integral of the torque over the

angle of rotation similarly to the work done by a linear force which is equal to the

integral of the force over the distance travelled

(23)

For a constant torque we get

dWPower = (Compare to P=Fv for linear motion)

dt

(24)

2

1

For a constant torque we get

W=

a b

N

i i

ia tob b to a to b

I

b a

W F dr m r d d

In the case of a constant torque we also get an easy expression for the power delivered by

the torque

(25)

v

path

F ma F dr W K

dW drF r W Power F F

dt dt

The torque on an extended object causes an accelerated rotation around an axis This is

the extension of Newtons second law for linear motion Acceleration means increase in

kinetic energy Therefore we get 2 21

2b a rotW I K

If there is no net exterior torque acting an object rotating at a constant rate will keep

rotating at this rate (Newtons first law applied to rotations)

The direction of the torque around a fixed axis of rotation is always perpendicular to the

plane of rotation ie parallel to the axis of rotation

Angular Momentum L Let us now consider the angular momentum of the system of particles

The angular momentum L for one particle in motion is given by

(11) l r p

The derivative of the angular momentum is

(12)

0 torque

dl d dr dpr p p r r F

dt dt dt dt

FW physics 10192011 843 PM page 4 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

4

For a system of particle we must sum over all the angular momenta and then take the

derivative with respect to t

(13)

where k is the summation index for

the exterior torques only the interior torques cancelling each other in pairs

i i i

i i

ii i k

i i k

L l r p

dpdL dl r

dt dt dt

(14) exterior

k

k

dL

dt

The direction of the torque around a fixed axis of rotation is always perpendicular to the

plane of rotation ie parallel to the axis of rotation

The angular momentum of an extended object in a rotation about a point is defined by the

vector sum of all the angular momenta li All radius vectors start at the same point Thus

it is evident that the angular momentum vector depends on the chosen origin

(15) ivi i i ii

i i

L l r p r m

We consider here a constant rotation around a fixed axis say the z-axis while we use the

x-y plane as the plane of rotation In the case of a counter clock wise rotation the

direction of the angular momentum points in the positive z direction We get

(16)

i

i

v

v

In a rotation around a fixed axis the vector is always perpendicular to the

tangential velocity

i i i iii i

i

i

i

L l r p r m

r

r

ru

(17)

2

i

2 2

object

v

I is called the moment of inertia=

We see that

i i i i i z i i

i i i

i i

i

L r m r m ru u m r

L I m r r dm

dL dI I

dt dt

The change in total momentum of a system of particles (or an extended object) is

equal to the vector sum of the exterior torques

FW physics 10192011 843 PM page 5 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

5

This is also valid for the relationship between the total torque of an extended object and

its angular momentum We write the relationship only for the magnitude using (14)

(18)

exterior

k

k

dL

dt

dL d dI I I

dt dt dt

Torque and angular momentum We just need to prove that torque is equal to the derivative of angular momentum

just like force is the derivative of linear momentum

(19) 2=mr just like

where =

z

z r

dL dpu F ma

dt dt

u u u

Example If we apply an exterior force to a disk capable of rotating around an axis each

individual point mass im of the disk is subjected to a clockwise (or ccw) tangential time

change of momentum vi i

i

dp dm

dt dt

or i i i im a m r In reference to the axis of rotation

this becomes the change of angular momentum for each individual mass im

2i ii i i i i i i i i i i i

dL dpdr p r rm a rm r m r

dt dt dt

(110) 2 2

total i i i i i i i

i i i i

R F m a m r m r I

(111) dp dL d dr

r F r r p pdt dt dt dt

torque

dpr

dt

As velocity and momentum are parallel vectors only the last term remains and we have

(112)

The torque created by the exterior force is equal to the change

of angular momentum of the rotating object Or the exterior torque changes the

angular momentum of a rotating object

dLr F

dt

Next we need to prove that Newtons second law becomes for the rotation of a solid

object around a fixed axis AI

FW physics 10192011 843 PM page 6 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

6

The angular momentum of an extended object in a rotation about a point is defined by the

vector sum of all the angular momenta li All radius vectors start at the same point Thus

it is evident that the angular momentum vector depends on the chosen origin

(113) ivi i i ii

i i

L l r p r m

We consider here a constant rotation around a fixed axis say the z-axis while we use the

x-y plane as the plane of rotation in which we define the radial and tangential unit

vectors In the case of a counter clock wise rotation the direction of the angular

momentum points in the positive z direction We get

(114)

i

i

v

v

In a rotation around a fixed axis the vector is always perpendicular to the

tangential velocity The vector is perpendicular to the plane of rotatio

i i i iii i

i

i

i

L l r p r m

r

r

ru

i

n We call the

unit vector in that direction such that

v

z r z

i z i r i

u u u u

r u ru ru

(115)

2

i

2 2

object

v

I is called the moment of inertia=

i i i i i z i i

i i i

z

i i

i

L r m r m ru u m r

L I

m r r dm

This is also valid for the relationship between the total torque of an extended object and

its angular momentum We write the relationship only for the magnitude

(116) dL d d

I I Idt dt dt

Example Consider a solid disk of mass 25 kg and radius 10 cm The disk is free to rotate

without friction around a fixed axis through its

center of mass A long massless cord is wound

around the perimeter of the disk and a mass of 18

kg is suspended from it

torque =

Tr=Iα=05Mr2α

mg

T-mg=ma=mrα

M=25kg

FW physics 10192011 843 PM page 7 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

7

(117)

2

2

1 1 1

2 2 2

1

2

18123 T=154N

1 143

2

Tr Mr Mra T Ma

T mg ma Ma mg ma

mga g m s

m M

The exterior torque is Tr=154N01m=154Nm

Rolling motion We assume that a sphere rolls without slipping

(118) i ir R

We take the time derivative of this equation to get the velocities

(119) v vi icm i i

dr dRdR

dt dt dt

The velocity of a point of a rolling object is therefore the velocity of the center of mass

plus the velocity of this point around the center of mass

For a point on the rim we have

CM

mi

O

vector from origin

to mi

vector from origin

to center of mass

vector from

center of mass to

FW physics 10192011 843 PM page 8 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

8

A C

mi

(120)

cm

v v

v =

cm

dr d dRR

dt dt dt

R

Kinetic energy of the rolling object We sum over all the kinetic energies of the points making up the object

(121)

2

2

2 2

2 2

1 1v v

2 2

1 1v ( ) v ( )

2 2

1 1v v

2 2

i i i i cm i

i cm i i i cm i

cm cm cm i i

K K m m R

m m R m R

M I m R

The last term is 0 (why) and therefore the total kinetic energy of a rolling object equals

the kinetic energy of its center of mass plus the kinetic energy of rotation of the object

around the axis through its center of mass

(122) 2 21 1

v2 2

cm cmK M I

One can convince oneself easily with the aid of the parallel axis theorem that this kinetic

energy is equal to the kinetic energy of a pure rotation around a point A at the surface of

the rotating object

Proof of the parallel axis theorem the axes of rotation are at A and C(center of mass)

perpendicular to the plane of writing (see also angular momentum)

(123)

2

2 2 2

2 2

2

2

2

0 because 0 why

A i i i i i i i

i i i i i

cm i i

I m R m D r m D r D r

D m m r D m r

MD I m r

Thus

(124)2 where D is the distance between the two axesA cmI MD I

FW physics 10192011 843 PM page 3 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

3

We can also see that work done by a torque is equal to the integral of the torque over the

angle of rotation similarly to the work done by a linear force which is equal to the

integral of the force over the distance travelled

(23)

For a constant torque we get

dWPower = (Compare to P=Fv for linear motion)

dt

(24)

2

1

For a constant torque we get

W=

a b

N

i i

ia tob b to a to b

I

b a

W F dr m r d d

In the case of a constant torque we also get an easy expression for the power delivered by

the torque

(25)

v

path

F ma F dr W K

dW drF r W Power F F

dt dt

The torque on an extended object causes an accelerated rotation around an axis This is

the extension of Newtons second law for linear motion Acceleration means increase in

kinetic energy Therefore we get 2 21

2b a rotW I K

If there is no net exterior torque acting an object rotating at a constant rate will keep

rotating at this rate (Newtons first law applied to rotations)

The direction of the torque around a fixed axis of rotation is always perpendicular to the

plane of rotation ie parallel to the axis of rotation

Angular Momentum L Let us now consider the angular momentum of the system of particles

The angular momentum L for one particle in motion is given by

(11) l r p

The derivative of the angular momentum is

(12)

0 torque

dl d dr dpr p p r r F

dt dt dt dt

FW physics 10192011 843 PM page 4 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

4

For a system of particle we must sum over all the angular momenta and then take the

derivative with respect to t

(13)

where k is the summation index for

the exterior torques only the interior torques cancelling each other in pairs

i i i

i i

ii i k

i i k

L l r p

dpdL dl r

dt dt dt

(14) exterior

k

k

dL

dt

The direction of the torque around a fixed axis of rotation is always perpendicular to the

plane of rotation ie parallel to the axis of rotation

The angular momentum of an extended object in a rotation about a point is defined by the

vector sum of all the angular momenta li All radius vectors start at the same point Thus

it is evident that the angular momentum vector depends on the chosen origin

(15) ivi i i ii

i i

L l r p r m

We consider here a constant rotation around a fixed axis say the z-axis while we use the

x-y plane as the plane of rotation In the case of a counter clock wise rotation the

direction of the angular momentum points in the positive z direction We get

(16)

i

i

v

v

In a rotation around a fixed axis the vector is always perpendicular to the

tangential velocity

i i i iii i

i

i

i

L l r p r m

r

r

ru

(17)

2

i

2 2

object

v

I is called the moment of inertia=

We see that

i i i i i z i i

i i i

i i

i

L r m r m ru u m r

L I m r r dm

dL dI I

dt dt

The change in total momentum of a system of particles (or an extended object) is

equal to the vector sum of the exterior torques

FW physics 10192011 843 PM page 5 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

5

This is also valid for the relationship between the total torque of an extended object and

its angular momentum We write the relationship only for the magnitude using (14)

(18)

exterior

k

k

dL

dt

dL d dI I I

dt dt dt

Torque and angular momentum We just need to prove that torque is equal to the derivative of angular momentum

just like force is the derivative of linear momentum

(19) 2=mr just like

where =

z

z r

dL dpu F ma

dt dt

u u u

Example If we apply an exterior force to a disk capable of rotating around an axis each

individual point mass im of the disk is subjected to a clockwise (or ccw) tangential time

change of momentum vi i

i

dp dm

dt dt

or i i i im a m r In reference to the axis of rotation

this becomes the change of angular momentum for each individual mass im

2i ii i i i i i i i i i i i

dL dpdr p r rm a rm r m r

dt dt dt

(110) 2 2

total i i i i i i i

i i i i

R F m a m r m r I

(111) dp dL d dr

r F r r p pdt dt dt dt

torque

dpr

dt

As velocity and momentum are parallel vectors only the last term remains and we have

(112)

The torque created by the exterior force is equal to the change

of angular momentum of the rotating object Or the exterior torque changes the

angular momentum of a rotating object

dLr F

dt

Next we need to prove that Newtons second law becomes for the rotation of a solid

object around a fixed axis AI

FW physics 10192011 843 PM page 6 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

6

The angular momentum of an extended object in a rotation about a point is defined by the

vector sum of all the angular momenta li All radius vectors start at the same point Thus

it is evident that the angular momentum vector depends on the chosen origin

(113) ivi i i ii

i i

L l r p r m

We consider here a constant rotation around a fixed axis say the z-axis while we use the

x-y plane as the plane of rotation in which we define the radial and tangential unit

vectors In the case of a counter clock wise rotation the direction of the angular

momentum points in the positive z direction We get

(114)

i

i

v

v

In a rotation around a fixed axis the vector is always perpendicular to the

tangential velocity The vector is perpendicular to the plane of rotatio

i i i iii i

i

i

i

L l r p r m

r

r

ru

i

n We call the

unit vector in that direction such that

v

z r z

i z i r i

u u u u

r u ru ru

(115)

2

i

2 2

object

v

I is called the moment of inertia=

i i i i i z i i

i i i

z

i i

i

L r m r m ru u m r

L I

m r r dm

This is also valid for the relationship between the total torque of an extended object and

its angular momentum We write the relationship only for the magnitude

(116) dL d d

I I Idt dt dt

Example Consider a solid disk of mass 25 kg and radius 10 cm The disk is free to rotate

without friction around a fixed axis through its

center of mass A long massless cord is wound

around the perimeter of the disk and a mass of 18

kg is suspended from it

torque =

Tr=Iα=05Mr2α

mg

T-mg=ma=mrα

M=25kg

FW physics 10192011 843 PM page 7 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

7

(117)

2

2

1 1 1

2 2 2

1

2

18123 T=154N

1 143

2

Tr Mr Mra T Ma

T mg ma Ma mg ma

mga g m s

m M

The exterior torque is Tr=154N01m=154Nm

Rolling motion We assume that a sphere rolls without slipping

(118) i ir R

We take the time derivative of this equation to get the velocities

(119) v vi icm i i

dr dRdR

dt dt dt

The velocity of a point of a rolling object is therefore the velocity of the center of mass

plus the velocity of this point around the center of mass

For a point on the rim we have

CM

mi

O

vector from origin

to mi

vector from origin

to center of mass

vector from

center of mass to

FW physics 10192011 843 PM page 8 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

8

A C

mi

(120)

cm

v v

v =

cm

dr d dRR

dt dt dt

R

Kinetic energy of the rolling object We sum over all the kinetic energies of the points making up the object

(121)

2

2

2 2

2 2

1 1v v

2 2

1 1v ( ) v ( )

2 2

1 1v v

2 2

i i i i cm i

i cm i i i cm i

cm cm cm i i

K K m m R

m m R m R

M I m R

The last term is 0 (why) and therefore the total kinetic energy of a rolling object equals

the kinetic energy of its center of mass plus the kinetic energy of rotation of the object

around the axis through its center of mass

(122) 2 21 1

v2 2

cm cmK M I

One can convince oneself easily with the aid of the parallel axis theorem that this kinetic

energy is equal to the kinetic energy of a pure rotation around a point A at the surface of

the rotating object

Proof of the parallel axis theorem the axes of rotation are at A and C(center of mass)

perpendicular to the plane of writing (see also angular momentum)

(123)

2

2 2 2

2 2

2

2

2

0 because 0 why

A i i i i i i i

i i i i i

cm i i

I m R m D r m D r D r

D m m r D m r

MD I m r

Thus

(124)2 where D is the distance between the two axesA cmI MD I

FW physics 10192011 843 PM page 4 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

4

For a system of particle we must sum over all the angular momenta and then take the

derivative with respect to t

(13)

where k is the summation index for

the exterior torques only the interior torques cancelling each other in pairs

i i i

i i

ii i k

i i k

L l r p

dpdL dl r

dt dt dt

(14) exterior

k

k

dL

dt

The direction of the torque around a fixed axis of rotation is always perpendicular to the

plane of rotation ie parallel to the axis of rotation

The angular momentum of an extended object in a rotation about a point is defined by the

vector sum of all the angular momenta li All radius vectors start at the same point Thus

it is evident that the angular momentum vector depends on the chosen origin

(15) ivi i i ii

i i

L l r p r m

We consider here a constant rotation around a fixed axis say the z-axis while we use the

x-y plane as the plane of rotation In the case of a counter clock wise rotation the

direction of the angular momentum points in the positive z direction We get

(16)

i

i

v

v

In a rotation around a fixed axis the vector is always perpendicular to the

tangential velocity

i i i iii i

i

i

i

L l r p r m

r

r

ru

(17)

2

i

2 2

object

v

I is called the moment of inertia=

We see that

i i i i i z i i

i i i

i i

i

L r m r m ru u m r

L I m r r dm

dL dI I

dt dt

The change in total momentum of a system of particles (or an extended object) is

equal to the vector sum of the exterior torques

FW physics 10192011 843 PM page 5 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

5

This is also valid for the relationship between the total torque of an extended object and

its angular momentum We write the relationship only for the magnitude using (14)

(18)

exterior

k

k

dL

dt

dL d dI I I

dt dt dt

Torque and angular momentum We just need to prove that torque is equal to the derivative of angular momentum

just like force is the derivative of linear momentum

(19) 2=mr just like

where =

z

z r

dL dpu F ma

dt dt

u u u

Example If we apply an exterior force to a disk capable of rotating around an axis each

individual point mass im of the disk is subjected to a clockwise (or ccw) tangential time

change of momentum vi i

i

dp dm

dt dt

or i i i im a m r In reference to the axis of rotation

this becomes the change of angular momentum for each individual mass im

2i ii i i i i i i i i i i i

dL dpdr p r rm a rm r m r

dt dt dt

(110) 2 2

total i i i i i i i

i i i i

R F m a m r m r I

(111) dp dL d dr

r F r r p pdt dt dt dt

torque

dpr

dt

As velocity and momentum are parallel vectors only the last term remains and we have

(112)

The torque created by the exterior force is equal to the change

of angular momentum of the rotating object Or the exterior torque changes the

angular momentum of a rotating object

dLr F

dt

Next we need to prove that Newtons second law becomes for the rotation of a solid

object around a fixed axis AI

FW physics 10192011 843 PM page 6 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

6

The angular momentum of an extended object in a rotation about a point is defined by the

vector sum of all the angular momenta li All radius vectors start at the same point Thus

it is evident that the angular momentum vector depends on the chosen origin

(113) ivi i i ii

i i

L l r p r m

We consider here a constant rotation around a fixed axis say the z-axis while we use the

x-y plane as the plane of rotation in which we define the radial and tangential unit

vectors In the case of a counter clock wise rotation the direction of the angular

momentum points in the positive z direction We get

(114)

i

i

v

v

In a rotation around a fixed axis the vector is always perpendicular to the

tangential velocity The vector is perpendicular to the plane of rotatio

i i i iii i

i

i

i

L l r p r m

r

r

ru

i

n We call the

unit vector in that direction such that

v

z r z

i z i r i

u u u u

r u ru ru

(115)

2

i

2 2

object

v

I is called the moment of inertia=

i i i i i z i i

i i i

z

i i

i

L r m r m ru u m r

L I

m r r dm

This is also valid for the relationship between the total torque of an extended object and

its angular momentum We write the relationship only for the magnitude

(116) dL d d

I I Idt dt dt

Example Consider a solid disk of mass 25 kg and radius 10 cm The disk is free to rotate

without friction around a fixed axis through its

center of mass A long massless cord is wound

around the perimeter of the disk and a mass of 18

kg is suspended from it

torque =

Tr=Iα=05Mr2α

mg

T-mg=ma=mrα

M=25kg

FW physics 10192011 843 PM page 7 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

7

(117)

2

2

1 1 1

2 2 2

1

2

18123 T=154N

1 143

2

Tr Mr Mra T Ma

T mg ma Ma mg ma

mga g m s

m M

The exterior torque is Tr=154N01m=154Nm

Rolling motion We assume that a sphere rolls without slipping

(118) i ir R

We take the time derivative of this equation to get the velocities

(119) v vi icm i i

dr dRdR

dt dt dt

The velocity of a point of a rolling object is therefore the velocity of the center of mass

plus the velocity of this point around the center of mass

For a point on the rim we have

CM

mi

O

vector from origin

to mi

vector from origin

to center of mass

vector from

center of mass to

FW physics 10192011 843 PM page 8 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

8

A C

mi

(120)

cm

v v

v =

cm

dr d dRR

dt dt dt

R

Kinetic energy of the rolling object We sum over all the kinetic energies of the points making up the object

(121)

2

2

2 2

2 2

1 1v v

2 2

1 1v ( ) v ( )

2 2

1 1v v

2 2

i i i i cm i

i cm i i i cm i

cm cm cm i i

K K m m R

m m R m R

M I m R

The last term is 0 (why) and therefore the total kinetic energy of a rolling object equals

the kinetic energy of its center of mass plus the kinetic energy of rotation of the object

around the axis through its center of mass

(122) 2 21 1

v2 2

cm cmK M I

One can convince oneself easily with the aid of the parallel axis theorem that this kinetic

energy is equal to the kinetic energy of a pure rotation around a point A at the surface of

the rotating object

Proof of the parallel axis theorem the axes of rotation are at A and C(center of mass)

perpendicular to the plane of writing (see also angular momentum)

(123)

2

2 2 2

2 2

2

2

2

0 because 0 why

A i i i i i i i

i i i i i

cm i i

I m R m D r m D r D r

D m m r D m r

MD I m r

Thus

(124)2 where D is the distance between the two axesA cmI MD I

FW physics 10192011 843 PM page 5 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

5

This is also valid for the relationship between the total torque of an extended object and

its angular momentum We write the relationship only for the magnitude using (14)

(18)

exterior

k

k

dL

dt

dL d dI I I

dt dt dt

Torque and angular momentum We just need to prove that torque is equal to the derivative of angular momentum

just like force is the derivative of linear momentum

(19) 2=mr just like

where =

z

z r

dL dpu F ma

dt dt

u u u

Example If we apply an exterior force to a disk capable of rotating around an axis each

individual point mass im of the disk is subjected to a clockwise (or ccw) tangential time

change of momentum vi i

i

dp dm

dt dt

or i i i im a m r In reference to the axis of rotation

this becomes the change of angular momentum for each individual mass im

2i ii i i i i i i i i i i i

dL dpdr p r rm a rm r m r

dt dt dt

(110) 2 2

total i i i i i i i

i i i i

R F m a m r m r I

(111) dp dL d dr

r F r r p pdt dt dt dt

torque

dpr

dt

As velocity and momentum are parallel vectors only the last term remains and we have

(112)

The torque created by the exterior force is equal to the change

of angular momentum of the rotating object Or the exterior torque changes the

angular momentum of a rotating object

dLr F

dt

Next we need to prove that Newtons second law becomes for the rotation of a solid

object around a fixed axis AI

FW physics 10192011 843 PM page 6 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

6

The angular momentum of an extended object in a rotation about a point is defined by the

vector sum of all the angular momenta li All radius vectors start at the same point Thus

it is evident that the angular momentum vector depends on the chosen origin

(113) ivi i i ii

i i

L l r p r m

We consider here a constant rotation around a fixed axis say the z-axis while we use the

x-y plane as the plane of rotation in which we define the radial and tangential unit

vectors In the case of a counter clock wise rotation the direction of the angular

momentum points in the positive z direction We get

(114)

i

i

v

v

In a rotation around a fixed axis the vector is always perpendicular to the

tangential velocity The vector is perpendicular to the plane of rotatio

i i i iii i

i

i

i

L l r p r m

r

r

ru

i

n We call the

unit vector in that direction such that

v

z r z

i z i r i

u u u u

r u ru ru

(115)

2

i

2 2

object

v

I is called the moment of inertia=

i i i i i z i i

i i i

z

i i

i

L r m r m ru u m r

L I

m r r dm

This is also valid for the relationship between the total torque of an extended object and

its angular momentum We write the relationship only for the magnitude

(116) dL d d

I I Idt dt dt

Example Consider a solid disk of mass 25 kg and radius 10 cm The disk is free to rotate

without friction around a fixed axis through its

center of mass A long massless cord is wound

around the perimeter of the disk and a mass of 18

kg is suspended from it

torque =

Tr=Iα=05Mr2α

mg

T-mg=ma=mrα

M=25kg

FW physics 10192011 843 PM page 7 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

7

(117)

2

2

1 1 1

2 2 2

1

2

18123 T=154N

1 143

2

Tr Mr Mra T Ma

T mg ma Ma mg ma

mga g m s

m M

The exterior torque is Tr=154N01m=154Nm

Rolling motion We assume that a sphere rolls without slipping

(118) i ir R

We take the time derivative of this equation to get the velocities

(119) v vi icm i i

dr dRdR

dt dt dt

The velocity of a point of a rolling object is therefore the velocity of the center of mass

plus the velocity of this point around the center of mass

For a point on the rim we have

CM

mi

O

vector from origin

to mi

vector from origin

to center of mass

vector from

center of mass to

FW physics 10192011 843 PM page 8 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

8

A C

mi

(120)

cm

v v

v =

cm

dr d dRR

dt dt dt

R

Kinetic energy of the rolling object We sum over all the kinetic energies of the points making up the object

(121)

2

2

2 2

2 2

1 1v v

2 2

1 1v ( ) v ( )

2 2

1 1v v

2 2

i i i i cm i

i cm i i i cm i

cm cm cm i i

K K m m R

m m R m R

M I m R

The last term is 0 (why) and therefore the total kinetic energy of a rolling object equals

the kinetic energy of its center of mass plus the kinetic energy of rotation of the object

around the axis through its center of mass

(122) 2 21 1

v2 2

cm cmK M I

One can convince oneself easily with the aid of the parallel axis theorem that this kinetic

energy is equal to the kinetic energy of a pure rotation around a point A at the surface of

the rotating object

Proof of the parallel axis theorem the axes of rotation are at A and C(center of mass)

perpendicular to the plane of writing (see also angular momentum)

(123)

2

2 2 2

2 2

2

2

2

0 because 0 why

A i i i i i i i

i i i i i

cm i i

I m R m D r m D r D r

D m m r D m r

MD I m r

Thus

(124)2 where D is the distance between the two axesA cmI MD I

FW physics 10192011 843 PM page 6 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

6

The angular momentum of an extended object in a rotation about a point is defined by the

vector sum of all the angular momenta li All radius vectors start at the same point Thus

it is evident that the angular momentum vector depends on the chosen origin

(113) ivi i i ii

i i

L l r p r m

We consider here a constant rotation around a fixed axis say the z-axis while we use the

x-y plane as the plane of rotation in which we define the radial and tangential unit

vectors In the case of a counter clock wise rotation the direction of the angular

momentum points in the positive z direction We get

(114)

i

i

v

v

In a rotation around a fixed axis the vector is always perpendicular to the

tangential velocity The vector is perpendicular to the plane of rotatio

i i i iii i

i

i

i

L l r p r m

r

r

ru

i

n We call the

unit vector in that direction such that

v

z r z

i z i r i

u u u u

r u ru ru

(115)

2

i

2 2

object

v

I is called the moment of inertia=

i i i i i z i i

i i i

z

i i

i

L r m r m ru u m r

L I

m r r dm

This is also valid for the relationship between the total torque of an extended object and

its angular momentum We write the relationship only for the magnitude

(116) dL d d

I I Idt dt dt

Example Consider a solid disk of mass 25 kg and radius 10 cm The disk is free to rotate

without friction around a fixed axis through its

center of mass A long massless cord is wound

around the perimeter of the disk and a mass of 18

kg is suspended from it

torque =

Tr=Iα=05Mr2α

mg

T-mg=ma=mrα

M=25kg

FW physics 10192011 843 PM page 7 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

7

(117)

2

2

1 1 1

2 2 2

1

2

18123 T=154N

1 143

2

Tr Mr Mra T Ma

T mg ma Ma mg ma

mga g m s

m M

The exterior torque is Tr=154N01m=154Nm

Rolling motion We assume that a sphere rolls without slipping

(118) i ir R

We take the time derivative of this equation to get the velocities

(119) v vi icm i i

dr dRdR

dt dt dt

The velocity of a point of a rolling object is therefore the velocity of the center of mass

plus the velocity of this point around the center of mass

For a point on the rim we have

CM

mi

O

vector from origin

to mi

vector from origin

to center of mass

vector from

center of mass to

FW physics 10192011 843 PM page 8 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

8

A C

mi

(120)

cm

v v

v =

cm

dr d dRR

dt dt dt

R

Kinetic energy of the rolling object We sum over all the kinetic energies of the points making up the object

(121)

2

2

2 2

2 2

1 1v v

2 2

1 1v ( ) v ( )

2 2

1 1v v

2 2

i i i i cm i

i cm i i i cm i

cm cm cm i i

K K m m R

m m R m R

M I m R

The last term is 0 (why) and therefore the total kinetic energy of a rolling object equals

the kinetic energy of its center of mass plus the kinetic energy of rotation of the object

around the axis through its center of mass

(122) 2 21 1

v2 2

cm cmK M I

One can convince oneself easily with the aid of the parallel axis theorem that this kinetic

energy is equal to the kinetic energy of a pure rotation around a point A at the surface of

the rotating object

Proof of the parallel axis theorem the axes of rotation are at A and C(center of mass)

perpendicular to the plane of writing (see also angular momentum)

(123)

2

2 2 2

2 2

2

2

2

0 because 0 why

A i i i i i i i

i i i i i

cm i i

I m R m D r m D r D r

D m m r D m r

MD I m r

Thus

(124)2 where D is the distance between the two axesA cmI MD I

FW physics 10192011 843 PM page 7 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

7

(117)

2

2

1 1 1

2 2 2

1

2

18123 T=154N

1 143

2

Tr Mr Mra T Ma

T mg ma Ma mg ma

mga g m s

m M

The exterior torque is Tr=154N01m=154Nm

Rolling motion We assume that a sphere rolls without slipping

(118) i ir R

We take the time derivative of this equation to get the velocities

(119) v vi icm i i

dr dRdR

dt dt dt

The velocity of a point of a rolling object is therefore the velocity of the center of mass

plus the velocity of this point around the center of mass

For a point on the rim we have

CM

mi

O

vector from origin

to mi

vector from origin

to center of mass

vector from

center of mass to

FW physics 10192011 843 PM page 8 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

8

A C

mi

(120)

cm

v v

v =

cm

dr d dRR

dt dt dt

R

Kinetic energy of the rolling object We sum over all the kinetic energies of the points making up the object

(121)

2

2

2 2

2 2

1 1v v

2 2

1 1v ( ) v ( )

2 2

1 1v v

2 2

i i i i cm i

i cm i i i cm i

cm cm cm i i

K K m m R

m m R m R

M I m R

The last term is 0 (why) and therefore the total kinetic energy of a rolling object equals

the kinetic energy of its center of mass plus the kinetic energy of rotation of the object

around the axis through its center of mass

(122) 2 21 1

v2 2

cm cmK M I

One can convince oneself easily with the aid of the parallel axis theorem that this kinetic

energy is equal to the kinetic energy of a pure rotation around a point A at the surface of

the rotating object

Proof of the parallel axis theorem the axes of rotation are at A and C(center of mass)

perpendicular to the plane of writing (see also angular momentum)

(123)

2

2 2 2

2 2

2

2

2

0 because 0 why

A i i i i i i i

i i i i i

cm i i

I m R m D r m D r D r

D m m r D m r

MD I m r

Thus

(124)2 where D is the distance between the two axesA cmI MD I

FW physics 10192011 843 PM page 8 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

8

A C

mi

(120)

cm

v v

v =

cm

dr d dRR

dt dt dt

R

Kinetic energy of the rolling object We sum over all the kinetic energies of the points making up the object

(121)

2

2

2 2

2 2

1 1v v

2 2

1 1v ( ) v ( )

2 2

1 1v v

2 2

i i i i cm i

i cm i i i cm i

cm cm cm i i

K K m m R

m m R m R

M I m R

The last term is 0 (why) and therefore the total kinetic energy of a rolling object equals

the kinetic energy of its center of mass plus the kinetic energy of rotation of the object

around the axis through its center of mass

(122) 2 21 1

v2 2

cm cmK M I

One can convince oneself easily with the aid of the parallel axis theorem that this kinetic

energy is equal to the kinetic energy of a pure rotation around a point A at the surface of

the rotating object

Proof of the parallel axis theorem the axes of rotation are at A and C(center of mass)

perpendicular to the plane of writing (see also angular momentum)

(123)

2

2 2 2

2 2

2

2

2

0 because 0 why

A i i i i i i i

i i i i i

cm i i

I m R m D r m D r D r

D m m r D m r

MD I m r

Thus

(124)2 where D is the distance between the two axesA cmI MD I

FW physics 10192011 843 PM page 9 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

9

If we want to calculate the work done due to a rotation we proceed similarly

(125)

2

1

i i i i i i

i i i

i

dW F dr m r rd

dW m r rd I d W d

dW dP I

dt dt

We also see again easily that the total work done by all exterior torques is equal to the

change in rotational kinetic energy

(126)

2

1

2 21

2f i rot

dW dW d I d I dt I d

dt

I K

The work done by all exterior forces causing an object to rotate is equal to the change of

rotational kinetic energy of the object

For a rolling object the kinetic energy is given by equation (122)

Rolling down an incline For example a sphere rolling down an incline of 30 degrees for 5 m has a potential

energy U=mgh = mmiddotgmiddotxmiddotsinθ At the bottom of the incline it will have a kinetic energy of

(127)

2 2

2 2 2 2

2

1 1v

2 2

1 2 1 7v v

2 5 2 10

10sin or v sin

7

rot cm cmK K I M

MR M M

Mgx gx

For the acceleration this means

(128)

2 10 5v 2 sin 2 or a= sin

7 7ax gx ax g

We can consider the rolling motion as a

motion around the point A created by the

torque MgRsinθ=IAα where IA is the moment

of inertia around point A The rotation is

clockwise

As rolling motion is characterized by the fact that at the contact point there is no

relative motion The two objects are fixed and static in this point It is therefore

Mg

R

A

CM

θ

fs

Macm

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Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

10

not trivial to determine the direction of the static friction In general f opposes

the motion that would occur in its absence and f is le μN

We can consider the motion as being caused by a friction torque which creates the

rotation around the center of mass plus the downward acceleration of the center of mass

caused by mg The torque which creates the rotation around the center equals

(129)

2

2

and setting

cos

s cm cm

s cm s s

R f I I MR

Rf I MR f Mg Ma

In addition to the rotational motion around the center of mass we must consider the linear

motion of the center of mass

(130) sins sf Mg Ma f Mg Ma

If we combine (130) with (129) we get for the acceleration

(131)

2

2

2 2

5cos sin sin

7

sin ( )

sinsin (1 )

1

cm

s

Ia

R

cm cmcm

Mg Mg Ma a g

I IMg Ma a a M I MR

R R

gg a a

We must just ensure that all the equations are consistent In the case of a solid sphere we

have that β=25

(132)22 2

5 5

s cm

s cm s

R f I

Rf I MR f Ma

(133) sins sf Mg Ma f Mg Ma

Now we can also calculate a value for the coefficient of static friction

2 2 5) cos sin

5 5 7

2) tan

7

s s

s

a f Mg Ma M g

b

FW physics 10192011 843 PM page 11 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

11

Spinning top A spinning top provides a nice example in which the vector nature of angular momentum

and torque become indispensable for the understanding of the physical situation

The cone shaped top with mass m rotates around its own axis with the angular speed

ω The exterior force mg creates a torque around the point at which the top rests on the

ground

This torque is equal to the vector change dL

dt

The change in the direction of the angular momentum creates a precession rotation with

the angular precession frequency Ωp

2 1L L L is a vector in the tangential direction perpendicular to L It describes a

circular arc of the magnitude of L L where is the angle between the vectors

1 2L and L pt

is the angular

velocity of precession of the vector L

(134)

) sin sin sin

sin)

sin

p

p

A

da mgR L L

dt

mgR mgRb

L I

The horizontal circle described by the

angular momentum L has a radius of

Lsinβ Therefore ΔL=LsinβΔθ

The precession frequency is then given by

(134) b)

Another approach is to look at the fact that we are dealing with a vector L I which

rotates in a circle For any such vector an equation similar to

dr

rdt

is valid

In our case this means that

(135)

p

dLL

dt

This is of course equal to the torque created by the force of gravity on the center of mass

of the rotating top which is given by

A

CM

FW physics 10192011 843 PM page 12 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

12

(136)

R mg

p

dLL R mg

dt

Both cross-products have the same direction which is to say that the angles in both

formulas are the same This is why we can write the equality for the magnitudes and

cancel sinβ

(137)

P P

Rmg RmgL Rmg

L I

FW physics 10192011 843 PM page 13 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

13

The rest is optional

Rolling of a spool

Consider the rolling spool above It consists of a set of two disks of radius R=1m with

mass M=30kg The interior disk serves as an axle and has a mass m=05kg and radius

r1=02 m It has a cord tightly wound around it A person pulls this cord to the right with

a constant force F=3N causing the disk to roll without friction to the right The person

pulls the string for x=15 m Find the velocity of the center of the disk at the end point of

15m

The interesting thing is that the distance x by which the cord is pulled to the right is not

equal to the distance covered by the center of the disk This can be understood easily

when one considers that the point C moves at a speed of 1vC R r with respect to

the ground In the time t the cord unfolds to a distance 1 1( ) ( )x R r t R r which

means that the angle in question is 1

x

R r

To find the speed of the rolling cylinder we need to use the work energy theorem

(138)

2 2 2 2 2

1

2

2 2 2 2

2

1 1

2

1 1 1 1 05( )v 15 004 15

2 2 2 2 2

v 1 1 v 1545 ( ) v 1125 v 123v

2 2 12

45v v 0605

123

W Fx M m I I MR mr kgm

Fx M m IR r R r

m

s

Spools of wire A spool of wire of mass M and radius R is unwound under a constant force F applied

horizontally to the right at the top of the spool Assuming that the spool is a solid cylinder

that does not slip show that (a) the acceleration of the center of mass is 4F3M and b)

D

C

A

B r1

R

FW physics 10192011 843 PM page 14 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

14

that the force of friction is to the right and equal in magnitude to F3 c) If the cylinder

starts from rest and rolls without slipping what is the speed of its center of mass after it

has rolled through a distance d 4

v3

Fd

M

We take the torques around point A

(139) 2 21 3 42 2

2 2 3

FRF MR MR F Ma a

M

Take the torques which provide the rolling around the center

(140) 21 1 1 1 4

2 2 2 2 3 3s s s

F FRF f R MR F f Ma f F Ma F M

M

c) work energy theorem

(141) 2 2 23 3 42 v v

2 2 3

FdW F d K MR M

M

The force must be applied for the distance 2d while the center moves to the

right by distance d

D

A

B

R

FW physics 10192011 843 PM page 15 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

15

The falling spool A string is wound around a uniform disk of radius R and mass M The disk is suspended

from a vertical string at its rim The string is fixed at a ceiling

a) Show that the tension in the string is 13 of the weight b) that the acceleration of the

center of the disk is 23 g and c) that the speed of the center of the disk after falling a

distance d is equal to 4

v=3

gd Verify this by

using energy considerations

a)

(142)

2

torques around the center

1 1 1RT=I

2 2 2

1

2

2

3

1 2 1

2 3 3

cm

mg T ma

mR maR T ma

mg ma ma

a g

T m g mg

To obtain the speed of the center of the disk we use the kinematic formulas

(143) 2 2 4v 2 2 v

3 3ad d g gd

To obtain the same result using energy conservation we need to realize that we have

conservation of energy The tension does not do any work

(144)

2 2 2 21 1 3 3v

2 2 2 4

4v

3

Amgd I mR m

gd

The dragged spool A spool of thread consists of a cylinder of radius R1with end-caps of radius R2 The mass

of the spool including the thread is m and its moment of inertia around the axis through

its center is I The spool is placed on a rough horizontal surface so it rolls without

slipping when a force T acting to the left is applied to the free end of the thread around

R1 Show that the magnitude of the friction force exerted by the surface on the spool is

given by 1 2

2

2

I mR Rf T

I mR

Determine the direction of the force of friction Calculate the

acceleration of the spool

The spool will roll to the right

f Tf T ma a

m m

R2

R1

FW physics 10192011 843 PM page 16 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

16

We take the torques around the center to get a clockwise rotation of the spool with an

acceleration of the center to the right

(145)

2

1 2 1 2 2

2

IaTR fR TR R fR Ia

R

Eliminate a from the two equations

(146)

2 2

1 2 2 1 2 2

1 2

2

2

1 2 1 2

2 2

2 2

1

mTR R mfR If IT mTR R IT If mfR

I mR Rf T

I mR

I mR R I mR RT T Ta

I mR m m m I mR

One can see that there is an acceleration as long as R1ltR2

Why do rolling objects slow down If rolling objects had perfect circular shapes they would never slow down while rolling

on a flat horizontal surface However a sphere resting or rolling on a surface is slightly

flattened at the contact area This results in a net torque due to the normal force of the

surface on the sphere opposing the direction of rolling

direction of rolling

creates a torque around the center point

opposing the rolling and bringing it to a

stop

FW physics 10192011 843 PM page 17 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

17

Moving plank on rollers To illustrate the point about the difficulty of discerning the correct direction of fs in the

context of rolling motion let us look at the following problem

A plank with a mass of M=600 kg rides on top of two identical solid cylindrical rollers

that have radii of R=500 cm and masses m=200 kg The plank is pulled by a constant

horizontal force of magnitude F=600 N (see picture) The cylinders roll without slipping

on a flat surface and no slipping occurs between the rollers and the plank Find the

acceleration of the plank and the rollers find the frictional forces

The reason why the plank moves twice the distance as the center of mass of the rollers is

that with respect to the ground the top of each roller moves with twice the speed

v 2top R and 2 2top tops R a R

The forces on the rollers ft-fb=mar and (ft+ fb )R=Iα=05 mR2 α=05Rmar

add the last two equations to get

2 ft=15mar=3ar ft=0752ar=15ar

fb=-025 mar=

from the plank equation we get -3ar+6=6a=12 ar ar=0400 a=0800

fb=-025 204=-0200N

ft=0600N

The interesting point here is that we have chosen fb in the wrong direction The friction

force fb points to the right

forces on the plank -2ft +F= Ma

2Nt-2Mg=0 ft is the friction force at the top both on the plank and on the rollers just in

opposite directions

F=600 N M=6kg

As each roller moves forward by the distance s the plank moves twice as far This

means that the acceleration of each roller is frac12 the acceleration of the plank

fb

fb

ft

ft

mg

mg

N

t

N

t

FW physics 10192011 843 PM page 18 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

18

Moment of inertia tensor

For your personal enjoyment only it will not be required in a test

For a more general derivation of the moment of inertia let us proceed without any

assumptions as to the kind of rotation We assume only that we are dealing with a rigid

body

(147)

i iv v

i i i

i

i i i i i i i i

i i

x y z

L r m r

L r m r m r r r r

r x y z

We need to calculate the three components for the vector L

(148)

2

Omitting the running index i in the x and components

x i x

i

L m x

2 2

xy z x x 2 2

2 2

y z i x y z

i

y i y

i

y z m y z x y z

L m x y

2

x yz y x y 2 2

2 2 2

z i y x z

i

z i z

i

z m x z y x z

L m x y z

x y zz x y z 2 2

i z x y

i

m x y z x y

This 3 by 3 ldquomatrixrdquo is called the moment of inertia tensor It reveals among other things

that the angular momentum L is not necessarily parallel to the angular velocity vector ω

(149)

2 2

2 2

2 2

2 2 2 2

1 etc

etc

x x

y i y

i

z z

xx i i i

i body

xy i i i

i body

L y z xy xz

L m yx x z yz

L zx zy x y

I I m y z y z dm

I m x y xydm

If the symmetry axes of a uniform symmetric body coincide with the coordinate axes the

mixed terms xy zx and so on vanish and the tensor has a simple diagonal form These

axes are then called principal axes

Remarkably it is always possible for a body of any shape and mass distribution to find a

set of three orthogonal axes such that the off-diagonal elements (the so-called products of

inertia) vanish

FW physics 10192011 843 PM page 19 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

19

(150)

0 0

0 0

0 0

x xx x

y yy y

z zz z

L I

L I

L I

For a uniform solid sphere or a shell any three orthogonal axes are principle axes and

we have

(151)

2

2

1

2 2 2

2 2 2 2

4 5 5 3 3

2 1 2 1

4

3 2 4

2 4 2 44

3 3 5 3

xx yy zz

zz

sphere

xx yy zz

spherer

R

R

I I I

I x y dm dm dV r dr

I I I I x y z dm dm r dr

I r dr R R M R R

For the inner radius equal 0 we get 22

5I MR which is the moment of inertia of a solid

sphere rotating about any diameter

For a thin spherical shell of mass m we get (no integration necessary)

(152) 2

2 2 2 2

2

2 2 2

3 2 4 4

2 24

3 3

xx yy zz

sphereR

mI I I I x y z dm m R

R

I R R mR

To calculate the moment of inertia of a thin hoop for a rotation around its center

The hoop has a line density ρ every point of the hoop has the same distance R from the

center

2

2 2 2 3 2

0

2I r dm r Rd R Rd R MR

since M=2πRρ

Calculating various moments of inertia

If we want to find the moment of inertia I for a rotating solid cylinder we have to choose

a cylindrical volume element which has a variable distance r (from 0 to R) from the axis

of rotation

(21) with 0 r R 0 2 and 0 z hdm rdrd dz

We can integrate over θ and z to give a volume element of

(22) 2dm rhdr

to arrive at the simple integral

FW physics 10192011 843 PM page 20 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

20

Ri ri

D A C

mi

(23)

43

2

0

2

M2 2 and with =

4 V

1

2

RR M

I h r dr hR h

I MR

The same can be achieved more readily with one simple integral if one chooses as

volume element a thin cylindrical disk of height h and thickness dr

(24)

42 2 3 2

0

2

12 2 2

4 2

with M= R

RR

I r dm r rhdr h r dr h MR

h

If we integrate r from an inner radius A to an outer radius B we get the moment of

inertia for a cylindrical shell with thickness B-A

(25)

4 43

2 2

2 2

( ) M2 2 and with =

4 V

1( )

2

B

A

B A MI h r dr h

B A h

I M B A

To calculate the moment of inertia of a rectangular stick of width A height H and

length L we put the axis of rotation at one end of the stick at the edge of the width A

The volume element

(26) dm= Hmiddotdxmiddotdy 0 A and 0y x L

(27) 3 3

2 2

0 0

( )3 3

L A

x y

L AI H x y dxdy H A L

The volume of the stick is H∙L∙A and its density is MH∙L∙A and therefore

(28) 2 2( )

3

M L AI

which is the moment of inertia for a rotation around one end (corner) To get

the moment of inertia around the center of mass we use the parallel axis

theorem

(29) 2 21( )

12cmI M L A

Proof of the parallel axis theorem the axes of rotation

are at A and C(center of mass) perpendicular to the

plane of writing

FW physics 10192011 843 PM page 21 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

21

(210)

2

2 2 2

2 2

2

2

2

0 because 0 why

A i i i i i i i

i i i i i

cm i i

I m R m D r m D r r D

D m m r D m r

MD I m r

Thus

(211) 2 where D is the distance between the two axesA cmI MD I

For a circular cone rotating around the z

axis we choose the volume element in

cylindrical coordinates

(212)

2 2 2

dV=rmiddotdrmiddotd middotdz 0 r 0

hr h z= and dz= and x +y =r

R R

R z h

R hdr

r z

(213)2

2 2

0 0 0

R h

cm

r z

I r dV r rdrdz d

As z depends on r we must first change the

boundaries of z which varies from hrR to h

and integrate it before we integrate over r

The integration over θ is simply 2π

(214)

542 3

0

0 0

4 4 4 42

2

2 2 24 5

32 2

14 5 20 10 10

3

R h R

R

cm

z hr R

hr hr hrI dzr rdr h r dr

R R

hR hR R R Mh h MR

R h

r

h

R

x

z

y

FW physics 10192011 843 PM page 22 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

22

Cycloids optional The speed of the center of mass is equal to the speed of a point on the rim (rolling

without slipping at constant velocity) with respect to the center For the x and y

coordinates of a point on the rim we get therefore

(215) r R

(216) 0

sin( ) and cos( )x t x R t R t y t R R t

If we start our motion with the point on the rim at x=0 and y=0 we get

(217) sin and cosx t R t R t y t R R t

For a point inside of the circle (RrsquoltR) the coordinates are given by

(218) sin and cosx t R t R t y t R R t

We see that the horizontal distance traveled remains equal to 2πR (not 2πRrsquo)

Assume for example that Rrsquo=05R

x0

In the time t the point on the rim of the circle (starting at the contact point

of the circle with the x-axis moves through the angle ωt to the left while

at the same time the center of the circle moves to the right by ωtR A point

inside the circle travels a shorter distance

x

FW physics 10192011 843 PM page 23 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

23

(219)

2 22 2 2 2

2 2

2

2 2

0

2 2

2

0 0

cos sin

(1 cos 2cos ) sin

2 1 cos 2 1 cos

1 12 1 cos sin 1 cos 2 sin 1 cos

2 2 2

2 2sin 2 sin 2 2 sin2 2

dx R dt R tdt dy R tdt

dx R dt t t dy R dt t

ds dx dy R t dt Rd

xS R d x x x

S R d R d R zdz

2

04 cos 82

R R

For a point inside the circle we get

(220)

222 2 2 2 2 2 2

2 2 2 2

22 2

cos sin

( cos 2 cos ) sin

2 cos

1 5for example R= we get ds= cos cos

2 4 4

dx R dt R tdt dy R tdt

dx dt R R t RR t dy R dt t

ds dx dy R R RR tdt

RR R R tdt R tdt

End of cycloids

FW physics 10192011 843 PM page 10 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

10

not trivial to determine the direction of the static friction In general f opposes

the motion that would occur in its absence and f is le μN

We can consider the motion as being caused by a friction torque which creates the

rotation around the center of mass plus the downward acceleration of the center of mass

caused by mg The torque which creates the rotation around the center equals

(129)

2

2

and setting

cos

s cm cm

s cm s s

R f I I MR

Rf I MR f Mg Ma

In addition to the rotational motion around the center of mass we must consider the linear

motion of the center of mass

(130) sins sf Mg Ma f Mg Ma

If we combine (130) with (129) we get for the acceleration

(131)

2

2

2 2

5cos sin sin

7

sin ( )

sinsin (1 )

1

cm

s

Ia

R

cm cmcm

Mg Mg Ma a g

I IMg Ma a a M I MR

R R

gg a a

We must just ensure that all the equations are consistent In the case of a solid sphere we

have that β=25

(132)22 2

5 5

s cm

s cm s

R f I

Rf I MR f Ma

(133) sins sf Mg Ma f Mg Ma

Now we can also calculate a value for the coefficient of static friction

2 2 5) cos sin

5 5 7

2) tan

7

s s

s

a f Mg Ma M g

b

FW physics 10192011 843 PM page 11 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

11

Spinning top A spinning top provides a nice example in which the vector nature of angular momentum

and torque become indispensable for the understanding of the physical situation

The cone shaped top with mass m rotates around its own axis with the angular speed

ω The exterior force mg creates a torque around the point at which the top rests on the

ground

This torque is equal to the vector change dL

dt

The change in the direction of the angular momentum creates a precession rotation with

the angular precession frequency Ωp

2 1L L L is a vector in the tangential direction perpendicular to L It describes a

circular arc of the magnitude of L L where is the angle between the vectors

1 2L and L pt

is the angular

velocity of precession of the vector L

(134)

) sin sin sin

sin)

sin

p

p

A

da mgR L L

dt

mgR mgRb

L I

The horizontal circle described by the

angular momentum L has a radius of

Lsinβ Therefore ΔL=LsinβΔθ

The precession frequency is then given by

(134) b)

Another approach is to look at the fact that we are dealing with a vector L I which

rotates in a circle For any such vector an equation similar to

dr

rdt

is valid

In our case this means that

(135)

p

dLL

dt

This is of course equal to the torque created by the force of gravity on the center of mass

of the rotating top which is given by

A

CM

FW physics 10192011 843 PM page 12 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

12

(136)

R mg

p

dLL R mg

dt

Both cross-products have the same direction which is to say that the angles in both

formulas are the same This is why we can write the equality for the magnitudes and

cancel sinβ

(137)

P P

Rmg RmgL Rmg

L I

FW physics 10192011 843 PM page 13 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

13

The rest is optional

Rolling of a spool

Consider the rolling spool above It consists of a set of two disks of radius R=1m with

mass M=30kg The interior disk serves as an axle and has a mass m=05kg and radius

r1=02 m It has a cord tightly wound around it A person pulls this cord to the right with

a constant force F=3N causing the disk to roll without friction to the right The person

pulls the string for x=15 m Find the velocity of the center of the disk at the end point of

15m

The interesting thing is that the distance x by which the cord is pulled to the right is not

equal to the distance covered by the center of the disk This can be understood easily

when one considers that the point C moves at a speed of 1vC R r with respect to

the ground In the time t the cord unfolds to a distance 1 1( ) ( )x R r t R r which

means that the angle in question is 1

x

R r

To find the speed of the rolling cylinder we need to use the work energy theorem

(138)

2 2 2 2 2

1

2

2 2 2 2

2

1 1

2

1 1 1 1 05( )v 15 004 15

2 2 2 2 2

v 1 1 v 1545 ( ) v 1125 v 123v

2 2 12

45v v 0605

123

W Fx M m I I MR mr kgm

Fx M m IR r R r

m

s

Spools of wire A spool of wire of mass M and radius R is unwound under a constant force F applied

horizontally to the right at the top of the spool Assuming that the spool is a solid cylinder

that does not slip show that (a) the acceleration of the center of mass is 4F3M and b)

D

C

A

B r1

R

FW physics 10192011 843 PM page 14 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

14

that the force of friction is to the right and equal in magnitude to F3 c) If the cylinder

starts from rest and rolls without slipping what is the speed of its center of mass after it

has rolled through a distance d 4

v3

Fd

M

We take the torques around point A

(139) 2 21 3 42 2

2 2 3

FRF MR MR F Ma a

M

Take the torques which provide the rolling around the center

(140) 21 1 1 1 4

2 2 2 2 3 3s s s

F FRF f R MR F f Ma f F Ma F M

M

c) work energy theorem

(141) 2 2 23 3 42 v v

2 2 3

FdW F d K MR M

M

The force must be applied for the distance 2d while the center moves to the

right by distance d

D

A

B

R

FW physics 10192011 843 PM page 15 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

15

The falling spool A string is wound around a uniform disk of radius R and mass M The disk is suspended

from a vertical string at its rim The string is fixed at a ceiling

a) Show that the tension in the string is 13 of the weight b) that the acceleration of the

center of the disk is 23 g and c) that the speed of the center of the disk after falling a

distance d is equal to 4

v=3

gd Verify this by

using energy considerations

a)

(142)

2

torques around the center

1 1 1RT=I

2 2 2

1

2

2

3

1 2 1

2 3 3

cm

mg T ma

mR maR T ma

mg ma ma

a g

T m g mg

To obtain the speed of the center of the disk we use the kinematic formulas

(143) 2 2 4v 2 2 v

3 3ad d g gd

To obtain the same result using energy conservation we need to realize that we have

conservation of energy The tension does not do any work

(144)

2 2 2 21 1 3 3v

2 2 2 4

4v

3

Amgd I mR m

gd

The dragged spool A spool of thread consists of a cylinder of radius R1with end-caps of radius R2 The mass

of the spool including the thread is m and its moment of inertia around the axis through

its center is I The spool is placed on a rough horizontal surface so it rolls without

slipping when a force T acting to the left is applied to the free end of the thread around

R1 Show that the magnitude of the friction force exerted by the surface on the spool is

given by 1 2

2

2

I mR Rf T

I mR

Determine the direction of the force of friction Calculate the

acceleration of the spool

The spool will roll to the right

f Tf T ma a

m m

R2

R1

FW physics 10192011 843 PM page 16 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

16

We take the torques around the center to get a clockwise rotation of the spool with an

acceleration of the center to the right

(145)

2

1 2 1 2 2

2

IaTR fR TR R fR Ia

R

Eliminate a from the two equations

(146)

2 2

1 2 2 1 2 2

1 2

2

2

1 2 1 2

2 2

2 2

1

mTR R mfR If IT mTR R IT If mfR

I mR Rf T

I mR

I mR R I mR RT T Ta

I mR m m m I mR

One can see that there is an acceleration as long as R1ltR2

Why do rolling objects slow down If rolling objects had perfect circular shapes they would never slow down while rolling

on a flat horizontal surface However a sphere resting or rolling on a surface is slightly

flattened at the contact area This results in a net torque due to the normal force of the

surface on the sphere opposing the direction of rolling

direction of rolling

creates a torque around the center point

opposing the rolling and bringing it to a

stop

FW physics 10192011 843 PM page 17 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

17

Moving plank on rollers To illustrate the point about the difficulty of discerning the correct direction of fs in the

context of rolling motion let us look at the following problem

A plank with a mass of M=600 kg rides on top of two identical solid cylindrical rollers

that have radii of R=500 cm and masses m=200 kg The plank is pulled by a constant

horizontal force of magnitude F=600 N (see picture) The cylinders roll without slipping

on a flat surface and no slipping occurs between the rollers and the plank Find the

acceleration of the plank and the rollers find the frictional forces

The reason why the plank moves twice the distance as the center of mass of the rollers is

that with respect to the ground the top of each roller moves with twice the speed

v 2top R and 2 2top tops R a R

The forces on the rollers ft-fb=mar and (ft+ fb )R=Iα=05 mR2 α=05Rmar

add the last two equations to get

2 ft=15mar=3ar ft=0752ar=15ar

fb=-025 mar=

from the plank equation we get -3ar+6=6a=12 ar ar=0400 a=0800

fb=-025 204=-0200N

ft=0600N

The interesting point here is that we have chosen fb in the wrong direction The friction

force fb points to the right

forces on the plank -2ft +F= Ma

2Nt-2Mg=0 ft is the friction force at the top both on the plank and on the rollers just in

opposite directions

F=600 N M=6kg

As each roller moves forward by the distance s the plank moves twice as far This

means that the acceleration of each roller is frac12 the acceleration of the plank

fb

fb

ft

ft

mg

mg

N

t

N

t

FW physics 10192011 843 PM page 18 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

18

Moment of inertia tensor

For your personal enjoyment only it will not be required in a test

For a more general derivation of the moment of inertia let us proceed without any

assumptions as to the kind of rotation We assume only that we are dealing with a rigid

body

(147)

i iv v

i i i

i

i i i i i i i i

i i

x y z

L r m r

L r m r m r r r r

r x y z

We need to calculate the three components for the vector L

(148)

2

Omitting the running index i in the x and components

x i x

i

L m x

2 2

xy z x x 2 2

2 2

y z i x y z

i

y i y

i

y z m y z x y z

L m x y

2

x yz y x y 2 2

2 2 2

z i y x z

i

z i z

i

z m x z y x z

L m x y z

x y zz x y z 2 2

i z x y

i

m x y z x y

This 3 by 3 ldquomatrixrdquo is called the moment of inertia tensor It reveals among other things

that the angular momentum L is not necessarily parallel to the angular velocity vector ω

(149)

2 2

2 2

2 2

2 2 2 2

1 etc

etc

x x

y i y

i

z z

xx i i i

i body

xy i i i

i body

L y z xy xz

L m yx x z yz

L zx zy x y

I I m y z y z dm

I m x y xydm

If the symmetry axes of a uniform symmetric body coincide with the coordinate axes the

mixed terms xy zx and so on vanish and the tensor has a simple diagonal form These

axes are then called principal axes

Remarkably it is always possible for a body of any shape and mass distribution to find a

set of three orthogonal axes such that the off-diagonal elements (the so-called products of

inertia) vanish

FW physics 10192011 843 PM page 19 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

19

(150)

0 0

0 0

0 0

x xx x

y yy y

z zz z

L I

L I

L I

For a uniform solid sphere or a shell any three orthogonal axes are principle axes and

we have

(151)

2

2

1

2 2 2

2 2 2 2

4 5 5 3 3

2 1 2 1

4

3 2 4

2 4 2 44

3 3 5 3

xx yy zz

zz

sphere

xx yy zz

spherer

R

R

I I I

I x y dm dm dV r dr

I I I I x y z dm dm r dr

I r dr R R M R R

For the inner radius equal 0 we get 22

5I MR which is the moment of inertia of a solid

sphere rotating about any diameter

For a thin spherical shell of mass m we get (no integration necessary)

(152) 2

2 2 2 2

2

2 2 2

3 2 4 4

2 24

3 3

xx yy zz

sphereR

mI I I I x y z dm m R

R

I R R mR

To calculate the moment of inertia of a thin hoop for a rotation around its center

The hoop has a line density ρ every point of the hoop has the same distance R from the

center

2

2 2 2 3 2

0

2I r dm r Rd R Rd R MR

since M=2πRρ

Calculating various moments of inertia

If we want to find the moment of inertia I for a rotating solid cylinder we have to choose

a cylindrical volume element which has a variable distance r (from 0 to R) from the axis

of rotation

(21) with 0 r R 0 2 and 0 z hdm rdrd dz

We can integrate over θ and z to give a volume element of

(22) 2dm rhdr

to arrive at the simple integral

FW physics 10192011 843 PM page 20 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

20

Ri ri

D A C

mi

(23)

43

2

0

2

M2 2 and with =

4 V

1

2

RR M

I h r dr hR h

I MR

The same can be achieved more readily with one simple integral if one chooses as

volume element a thin cylindrical disk of height h and thickness dr

(24)

42 2 3 2

0

2

12 2 2

4 2

with M= R

RR

I r dm r rhdr h r dr h MR

h

If we integrate r from an inner radius A to an outer radius B we get the moment of

inertia for a cylindrical shell with thickness B-A

(25)

4 43

2 2

2 2

( ) M2 2 and with =

4 V

1( )

2

B

A

B A MI h r dr h

B A h

I M B A

To calculate the moment of inertia of a rectangular stick of width A height H and

length L we put the axis of rotation at one end of the stick at the edge of the width A

The volume element

(26) dm= Hmiddotdxmiddotdy 0 A and 0y x L

(27) 3 3

2 2

0 0

( )3 3

L A

x y

L AI H x y dxdy H A L

The volume of the stick is H∙L∙A and its density is MH∙L∙A and therefore

(28) 2 2( )

3

M L AI

which is the moment of inertia for a rotation around one end (corner) To get

the moment of inertia around the center of mass we use the parallel axis

theorem

(29) 2 21( )

12cmI M L A

Proof of the parallel axis theorem the axes of rotation

are at A and C(center of mass) perpendicular to the

plane of writing

FW physics 10192011 843 PM page 21 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

21

(210)

2

2 2 2

2 2

2

2

2

0 because 0 why

A i i i i i i i

i i i i i

cm i i

I m R m D r m D r r D

D m m r D m r

MD I m r

Thus

(211) 2 where D is the distance between the two axesA cmI MD I

For a circular cone rotating around the z

axis we choose the volume element in

cylindrical coordinates

(212)

2 2 2

dV=rmiddotdrmiddotd middotdz 0 r 0

hr h z= and dz= and x +y =r

R R

R z h

R hdr

r z

(213)2

2 2

0 0 0

R h

cm

r z

I r dV r rdrdz d

As z depends on r we must first change the

boundaries of z which varies from hrR to h

and integrate it before we integrate over r

The integration over θ is simply 2π

(214)

542 3

0

0 0

4 4 4 42

2

2 2 24 5

32 2

14 5 20 10 10

3

R h R

R

cm

z hr R

hr hr hrI dzr rdr h r dr

R R

hR hR R R Mh h MR

R h

r

h

R

x

z

y

FW physics 10192011 843 PM page 22 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

22

Cycloids optional The speed of the center of mass is equal to the speed of a point on the rim (rolling

without slipping at constant velocity) with respect to the center For the x and y

coordinates of a point on the rim we get therefore

(215) r R

(216) 0

sin( ) and cos( )x t x R t R t y t R R t

If we start our motion with the point on the rim at x=0 and y=0 we get

(217) sin and cosx t R t R t y t R R t

For a point inside of the circle (RrsquoltR) the coordinates are given by

(218) sin and cosx t R t R t y t R R t

We see that the horizontal distance traveled remains equal to 2πR (not 2πRrsquo)

Assume for example that Rrsquo=05R

x0

In the time t the point on the rim of the circle (starting at the contact point

of the circle with the x-axis moves through the angle ωt to the left while

at the same time the center of the circle moves to the right by ωtR A point

inside the circle travels a shorter distance

x

FW physics 10192011 843 PM page 23 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

23

(219)

2 22 2 2 2

2 2

2

2 2

0

2 2

2

0 0

cos sin

(1 cos 2cos ) sin

2 1 cos 2 1 cos

1 12 1 cos sin 1 cos 2 sin 1 cos

2 2 2

2 2sin 2 sin 2 2 sin2 2

dx R dt R tdt dy R tdt

dx R dt t t dy R dt t

ds dx dy R t dt Rd

xS R d x x x

S R d R d R zdz

2

04 cos 82

R R

For a point inside the circle we get

(220)

222 2 2 2 2 2 2

2 2 2 2

22 2

cos sin

( cos 2 cos ) sin

2 cos

1 5for example R= we get ds= cos cos

2 4 4

dx R dt R tdt dy R tdt

dx dt R R t RR t dy R dt t

ds dx dy R R RR tdt

RR R R tdt R tdt

End of cycloids

FW physics 10192011 843 PM page 11 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

11

Spinning top A spinning top provides a nice example in which the vector nature of angular momentum

and torque become indispensable for the understanding of the physical situation

The cone shaped top with mass m rotates around its own axis with the angular speed

ω The exterior force mg creates a torque around the point at which the top rests on the

ground

This torque is equal to the vector change dL

dt

The change in the direction of the angular momentum creates a precession rotation with

the angular precession frequency Ωp

2 1L L L is a vector in the tangential direction perpendicular to L It describes a

circular arc of the magnitude of L L where is the angle between the vectors

1 2L and L pt

is the angular

velocity of precession of the vector L

(134)

) sin sin sin

sin)

sin

p

p

A

da mgR L L

dt

mgR mgRb

L I

The horizontal circle described by the

angular momentum L has a radius of

Lsinβ Therefore ΔL=LsinβΔθ

The precession frequency is then given by

(134) b)

Another approach is to look at the fact that we are dealing with a vector L I which

rotates in a circle For any such vector an equation similar to

dr

rdt

is valid

In our case this means that

(135)

p

dLL

dt

This is of course equal to the torque created by the force of gravity on the center of mass

of the rotating top which is given by

A

CM

FW physics 10192011 843 PM page 12 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

12

(136)

R mg

p

dLL R mg

dt

Both cross-products have the same direction which is to say that the angles in both

formulas are the same This is why we can write the equality for the magnitudes and

cancel sinβ

(137)

P P

Rmg RmgL Rmg

L I

FW physics 10192011 843 PM page 13 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

13

The rest is optional

Rolling of a spool

Consider the rolling spool above It consists of a set of two disks of radius R=1m with

mass M=30kg The interior disk serves as an axle and has a mass m=05kg and radius

r1=02 m It has a cord tightly wound around it A person pulls this cord to the right with

a constant force F=3N causing the disk to roll without friction to the right The person

pulls the string for x=15 m Find the velocity of the center of the disk at the end point of

15m

The interesting thing is that the distance x by which the cord is pulled to the right is not

equal to the distance covered by the center of the disk This can be understood easily

when one considers that the point C moves at a speed of 1vC R r with respect to

the ground In the time t the cord unfolds to a distance 1 1( ) ( )x R r t R r which

means that the angle in question is 1

x

R r

To find the speed of the rolling cylinder we need to use the work energy theorem

(138)

2 2 2 2 2

1

2

2 2 2 2

2

1 1

2

1 1 1 1 05( )v 15 004 15

2 2 2 2 2

v 1 1 v 1545 ( ) v 1125 v 123v

2 2 12

45v v 0605

123

W Fx M m I I MR mr kgm

Fx M m IR r R r

m

s

Spools of wire A spool of wire of mass M and radius R is unwound under a constant force F applied

horizontally to the right at the top of the spool Assuming that the spool is a solid cylinder

that does not slip show that (a) the acceleration of the center of mass is 4F3M and b)

D

C

A

B r1

R

FW physics 10192011 843 PM page 14 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

14

that the force of friction is to the right and equal in magnitude to F3 c) If the cylinder

starts from rest and rolls without slipping what is the speed of its center of mass after it

has rolled through a distance d 4

v3

Fd

M

We take the torques around point A

(139) 2 21 3 42 2

2 2 3

FRF MR MR F Ma a

M

Take the torques which provide the rolling around the center

(140) 21 1 1 1 4

2 2 2 2 3 3s s s

F FRF f R MR F f Ma f F Ma F M

M

c) work energy theorem

(141) 2 2 23 3 42 v v

2 2 3

FdW F d K MR M

M

The force must be applied for the distance 2d while the center moves to the

right by distance d

D

A

B

R

FW physics 10192011 843 PM page 15 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

15

The falling spool A string is wound around a uniform disk of radius R and mass M The disk is suspended

from a vertical string at its rim The string is fixed at a ceiling

a) Show that the tension in the string is 13 of the weight b) that the acceleration of the

center of the disk is 23 g and c) that the speed of the center of the disk after falling a

distance d is equal to 4

v=3

gd Verify this by

using energy considerations

a)

(142)

2

torques around the center

1 1 1RT=I

2 2 2

1

2

2

3

1 2 1

2 3 3

cm

mg T ma

mR maR T ma

mg ma ma

a g

T m g mg

To obtain the speed of the center of the disk we use the kinematic formulas

(143) 2 2 4v 2 2 v

3 3ad d g gd

To obtain the same result using energy conservation we need to realize that we have

conservation of energy The tension does not do any work

(144)

2 2 2 21 1 3 3v

2 2 2 4

4v

3

Amgd I mR m

gd

The dragged spool A spool of thread consists of a cylinder of radius R1with end-caps of radius R2 The mass

of the spool including the thread is m and its moment of inertia around the axis through

its center is I The spool is placed on a rough horizontal surface so it rolls without

slipping when a force T acting to the left is applied to the free end of the thread around

R1 Show that the magnitude of the friction force exerted by the surface on the spool is

given by 1 2

2

2

I mR Rf T

I mR

Determine the direction of the force of friction Calculate the

acceleration of the spool

The spool will roll to the right

f Tf T ma a

m m

R2

R1

FW physics 10192011 843 PM page 16 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

16

We take the torques around the center to get a clockwise rotation of the spool with an

acceleration of the center to the right

(145)

2

1 2 1 2 2

2

IaTR fR TR R fR Ia

R

Eliminate a from the two equations

(146)

2 2

1 2 2 1 2 2

1 2

2

2

1 2 1 2

2 2

2 2

1

mTR R mfR If IT mTR R IT If mfR

I mR Rf T

I mR

I mR R I mR RT T Ta

I mR m m m I mR

One can see that there is an acceleration as long as R1ltR2

Why do rolling objects slow down If rolling objects had perfect circular shapes they would never slow down while rolling

on a flat horizontal surface However a sphere resting or rolling on a surface is slightly

flattened at the contact area This results in a net torque due to the normal force of the

surface on the sphere opposing the direction of rolling

direction of rolling

creates a torque around the center point

opposing the rolling and bringing it to a

stop

FW physics 10192011 843 PM page 17 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

17

Moving plank on rollers To illustrate the point about the difficulty of discerning the correct direction of fs in the

context of rolling motion let us look at the following problem

A plank with a mass of M=600 kg rides on top of two identical solid cylindrical rollers

that have radii of R=500 cm and masses m=200 kg The plank is pulled by a constant

horizontal force of magnitude F=600 N (see picture) The cylinders roll without slipping

on a flat surface and no slipping occurs between the rollers and the plank Find the

acceleration of the plank and the rollers find the frictional forces

The reason why the plank moves twice the distance as the center of mass of the rollers is

that with respect to the ground the top of each roller moves with twice the speed

v 2top R and 2 2top tops R a R

The forces on the rollers ft-fb=mar and (ft+ fb )R=Iα=05 mR2 α=05Rmar

add the last two equations to get

2 ft=15mar=3ar ft=0752ar=15ar

fb=-025 mar=

from the plank equation we get -3ar+6=6a=12 ar ar=0400 a=0800

fb=-025 204=-0200N

ft=0600N

The interesting point here is that we have chosen fb in the wrong direction The friction

force fb points to the right

forces on the plank -2ft +F= Ma

2Nt-2Mg=0 ft is the friction force at the top both on the plank and on the rollers just in

opposite directions

F=600 N M=6kg

As each roller moves forward by the distance s the plank moves twice as far This

means that the acceleration of each roller is frac12 the acceleration of the plank

fb

fb

ft

ft

mg

mg

N

t

N

t

FW physics 10192011 843 PM page 18 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

18

Moment of inertia tensor

For your personal enjoyment only it will not be required in a test

For a more general derivation of the moment of inertia let us proceed without any

assumptions as to the kind of rotation We assume only that we are dealing with a rigid

body

(147)

i iv v

i i i

i

i i i i i i i i

i i

x y z

L r m r

L r m r m r r r r

r x y z

We need to calculate the three components for the vector L

(148)

2

Omitting the running index i in the x and components

x i x

i

L m x

2 2

xy z x x 2 2

2 2

y z i x y z

i

y i y

i

y z m y z x y z

L m x y

2

x yz y x y 2 2

2 2 2

z i y x z

i

z i z

i

z m x z y x z

L m x y z

x y zz x y z 2 2

i z x y

i

m x y z x y

This 3 by 3 ldquomatrixrdquo is called the moment of inertia tensor It reveals among other things

that the angular momentum L is not necessarily parallel to the angular velocity vector ω

(149)

2 2

2 2

2 2

2 2 2 2

1 etc

etc

x x

y i y

i

z z

xx i i i

i body

xy i i i

i body

L y z xy xz

L m yx x z yz

L zx zy x y

I I m y z y z dm

I m x y xydm

If the symmetry axes of a uniform symmetric body coincide with the coordinate axes the

mixed terms xy zx and so on vanish and the tensor has a simple diagonal form These

axes are then called principal axes

Remarkably it is always possible for a body of any shape and mass distribution to find a

set of three orthogonal axes such that the off-diagonal elements (the so-called products of

inertia) vanish

FW physics 10192011 843 PM page 19 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

19

(150)

0 0

0 0

0 0

x xx x

y yy y

z zz z

L I

L I

L I

For a uniform solid sphere or a shell any three orthogonal axes are principle axes and

we have

(151)

2

2

1

2 2 2

2 2 2 2

4 5 5 3 3

2 1 2 1

4

3 2 4

2 4 2 44

3 3 5 3

xx yy zz

zz

sphere

xx yy zz

spherer

R

R

I I I

I x y dm dm dV r dr

I I I I x y z dm dm r dr

I r dr R R M R R

For the inner radius equal 0 we get 22

5I MR which is the moment of inertia of a solid

sphere rotating about any diameter

For a thin spherical shell of mass m we get (no integration necessary)

(152) 2

2 2 2 2

2

2 2 2

3 2 4 4

2 24

3 3

xx yy zz

sphereR

mI I I I x y z dm m R

R

I R R mR

To calculate the moment of inertia of a thin hoop for a rotation around its center

The hoop has a line density ρ every point of the hoop has the same distance R from the

center

2

2 2 2 3 2

0

2I r dm r Rd R Rd R MR

since M=2πRρ

Calculating various moments of inertia

If we want to find the moment of inertia I for a rotating solid cylinder we have to choose

a cylindrical volume element which has a variable distance r (from 0 to R) from the axis

of rotation

(21) with 0 r R 0 2 and 0 z hdm rdrd dz

We can integrate over θ and z to give a volume element of

(22) 2dm rhdr

to arrive at the simple integral

FW physics 10192011 843 PM page 20 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

20

Ri ri

D A C

mi

(23)

43

2

0

2

M2 2 and with =

4 V

1

2

RR M

I h r dr hR h

I MR

The same can be achieved more readily with one simple integral if one chooses as

volume element a thin cylindrical disk of height h and thickness dr

(24)

42 2 3 2

0

2

12 2 2

4 2

with M= R

RR

I r dm r rhdr h r dr h MR

h

If we integrate r from an inner radius A to an outer radius B we get the moment of

inertia for a cylindrical shell with thickness B-A

(25)

4 43

2 2

2 2

( ) M2 2 and with =

4 V

1( )

2

B

A

B A MI h r dr h

B A h

I M B A

To calculate the moment of inertia of a rectangular stick of width A height H and

length L we put the axis of rotation at one end of the stick at the edge of the width A

The volume element

(26) dm= Hmiddotdxmiddotdy 0 A and 0y x L

(27) 3 3

2 2

0 0

( )3 3

L A

x y

L AI H x y dxdy H A L

The volume of the stick is H∙L∙A and its density is MH∙L∙A and therefore

(28) 2 2( )

3

M L AI

which is the moment of inertia for a rotation around one end (corner) To get

the moment of inertia around the center of mass we use the parallel axis

theorem

(29) 2 21( )

12cmI M L A

Proof of the parallel axis theorem the axes of rotation

are at A and C(center of mass) perpendicular to the

plane of writing

FW physics 10192011 843 PM page 21 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

21

(210)

2

2 2 2

2 2

2

2

2

0 because 0 why

A i i i i i i i

i i i i i

cm i i

I m R m D r m D r r D

D m m r D m r

MD I m r

Thus

(211) 2 where D is the distance between the two axesA cmI MD I

For a circular cone rotating around the z

axis we choose the volume element in

cylindrical coordinates

(212)

2 2 2

dV=rmiddotdrmiddotd middotdz 0 r 0

hr h z= and dz= and x +y =r

R R

R z h

R hdr

r z

(213)2

2 2

0 0 0

R h

cm

r z

I r dV r rdrdz d

As z depends on r we must first change the

boundaries of z which varies from hrR to h

and integrate it before we integrate over r

The integration over θ is simply 2π

(214)

542 3

0

0 0

4 4 4 42

2

2 2 24 5

32 2

14 5 20 10 10

3

R h R

R

cm

z hr R

hr hr hrI dzr rdr h r dr

R R

hR hR R R Mh h MR

R h

r

h

R

x

z

y

FW physics 10192011 843 PM page 22 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

22

Cycloids optional The speed of the center of mass is equal to the speed of a point on the rim (rolling

without slipping at constant velocity) with respect to the center For the x and y

coordinates of a point on the rim we get therefore

(215) r R

(216) 0

sin( ) and cos( )x t x R t R t y t R R t

If we start our motion with the point on the rim at x=0 and y=0 we get

(217) sin and cosx t R t R t y t R R t

For a point inside of the circle (RrsquoltR) the coordinates are given by

(218) sin and cosx t R t R t y t R R t

We see that the horizontal distance traveled remains equal to 2πR (not 2πRrsquo)

Assume for example that Rrsquo=05R

x0

In the time t the point on the rim of the circle (starting at the contact point

of the circle with the x-axis moves through the angle ωt to the left while

at the same time the center of the circle moves to the right by ωtR A point

inside the circle travels a shorter distance

x

FW physics 10192011 843 PM page 23 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

23

(219)

2 22 2 2 2

2 2

2

2 2

0

2 2

2

0 0

cos sin

(1 cos 2cos ) sin

2 1 cos 2 1 cos

1 12 1 cos sin 1 cos 2 sin 1 cos

2 2 2

2 2sin 2 sin 2 2 sin2 2

dx R dt R tdt dy R tdt

dx R dt t t dy R dt t

ds dx dy R t dt Rd

xS R d x x x

S R d R d R zdz

2

04 cos 82

R R

For a point inside the circle we get

(220)

222 2 2 2 2 2 2

2 2 2 2

22 2

cos sin

( cos 2 cos ) sin

2 cos

1 5for example R= we get ds= cos cos

2 4 4

dx R dt R tdt dy R tdt

dx dt R R t RR t dy R dt t

ds dx dy R R RR tdt

RR R R tdt R tdt

End of cycloids

FW physics 10192011 843 PM page 12 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

12

(136)

R mg

p

dLL R mg

dt

Both cross-products have the same direction which is to say that the angles in both

formulas are the same This is why we can write the equality for the magnitudes and

cancel sinβ

(137)

P P

Rmg RmgL Rmg

L I

FW physics 10192011 843 PM page 13 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

13

The rest is optional

Rolling of a spool

Consider the rolling spool above It consists of a set of two disks of radius R=1m with

mass M=30kg The interior disk serves as an axle and has a mass m=05kg and radius

r1=02 m It has a cord tightly wound around it A person pulls this cord to the right with

a constant force F=3N causing the disk to roll without friction to the right The person

pulls the string for x=15 m Find the velocity of the center of the disk at the end point of

15m

The interesting thing is that the distance x by which the cord is pulled to the right is not

equal to the distance covered by the center of the disk This can be understood easily

when one considers that the point C moves at a speed of 1vC R r with respect to

the ground In the time t the cord unfolds to a distance 1 1( ) ( )x R r t R r which

means that the angle in question is 1

x

R r

To find the speed of the rolling cylinder we need to use the work energy theorem

(138)

2 2 2 2 2

1

2

2 2 2 2

2

1 1

2

1 1 1 1 05( )v 15 004 15

2 2 2 2 2

v 1 1 v 1545 ( ) v 1125 v 123v

2 2 12

45v v 0605

123

W Fx M m I I MR mr kgm

Fx M m IR r R r

m

s

Spools of wire A spool of wire of mass M and radius R is unwound under a constant force F applied

horizontally to the right at the top of the spool Assuming that the spool is a solid cylinder

that does not slip show that (a) the acceleration of the center of mass is 4F3M and b)

D

C

A

B r1

R

FW physics 10192011 843 PM page 14 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

14

that the force of friction is to the right and equal in magnitude to F3 c) If the cylinder

starts from rest and rolls without slipping what is the speed of its center of mass after it

has rolled through a distance d 4

v3

Fd

M

We take the torques around point A

(139) 2 21 3 42 2

2 2 3

FRF MR MR F Ma a

M

Take the torques which provide the rolling around the center

(140) 21 1 1 1 4

2 2 2 2 3 3s s s

F FRF f R MR F f Ma f F Ma F M

M

c) work energy theorem

(141) 2 2 23 3 42 v v

2 2 3

FdW F d K MR M

M

The force must be applied for the distance 2d while the center moves to the

right by distance d

D

A

B

R

FW physics 10192011 843 PM page 15 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

15

The falling spool A string is wound around a uniform disk of radius R and mass M The disk is suspended

from a vertical string at its rim The string is fixed at a ceiling

a) Show that the tension in the string is 13 of the weight b) that the acceleration of the

center of the disk is 23 g and c) that the speed of the center of the disk after falling a

distance d is equal to 4

v=3

gd Verify this by

using energy considerations

a)

(142)

2

torques around the center

1 1 1RT=I

2 2 2

1

2

2

3

1 2 1

2 3 3

cm

mg T ma

mR maR T ma

mg ma ma

a g

T m g mg

To obtain the speed of the center of the disk we use the kinematic formulas

(143) 2 2 4v 2 2 v

3 3ad d g gd

To obtain the same result using energy conservation we need to realize that we have

conservation of energy The tension does not do any work

(144)

2 2 2 21 1 3 3v

2 2 2 4

4v

3

Amgd I mR m

gd

The dragged spool A spool of thread consists of a cylinder of radius R1with end-caps of radius R2 The mass

of the spool including the thread is m and its moment of inertia around the axis through

its center is I The spool is placed on a rough horizontal surface so it rolls without

slipping when a force T acting to the left is applied to the free end of the thread around

R1 Show that the magnitude of the friction force exerted by the surface on the spool is

given by 1 2

2

2

I mR Rf T

I mR

Determine the direction of the force of friction Calculate the

acceleration of the spool

The spool will roll to the right

f Tf T ma a

m m

R2

R1

FW physics 10192011 843 PM page 16 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

16

We take the torques around the center to get a clockwise rotation of the spool with an

acceleration of the center to the right

(145)

2

1 2 1 2 2

2

IaTR fR TR R fR Ia

R

Eliminate a from the two equations

(146)

2 2

1 2 2 1 2 2

1 2

2

2

1 2 1 2

2 2

2 2

1

mTR R mfR If IT mTR R IT If mfR

I mR Rf T

I mR

I mR R I mR RT T Ta

I mR m m m I mR

One can see that there is an acceleration as long as R1ltR2

Why do rolling objects slow down If rolling objects had perfect circular shapes they would never slow down while rolling

on a flat horizontal surface However a sphere resting or rolling on a surface is slightly

flattened at the contact area This results in a net torque due to the normal force of the

surface on the sphere opposing the direction of rolling

direction of rolling

creates a torque around the center point

opposing the rolling and bringing it to a

stop

FW physics 10192011 843 PM page 17 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

17

Moving plank on rollers To illustrate the point about the difficulty of discerning the correct direction of fs in the

context of rolling motion let us look at the following problem

A plank with a mass of M=600 kg rides on top of two identical solid cylindrical rollers

that have radii of R=500 cm and masses m=200 kg The plank is pulled by a constant

horizontal force of magnitude F=600 N (see picture) The cylinders roll without slipping

on a flat surface and no slipping occurs between the rollers and the plank Find the

acceleration of the plank and the rollers find the frictional forces

The reason why the plank moves twice the distance as the center of mass of the rollers is

that with respect to the ground the top of each roller moves with twice the speed

v 2top R and 2 2top tops R a R

The forces on the rollers ft-fb=mar and (ft+ fb )R=Iα=05 mR2 α=05Rmar

add the last two equations to get

2 ft=15mar=3ar ft=0752ar=15ar

fb=-025 mar=

from the plank equation we get -3ar+6=6a=12 ar ar=0400 a=0800

fb=-025 204=-0200N

ft=0600N

The interesting point here is that we have chosen fb in the wrong direction The friction

force fb points to the right

forces on the plank -2ft +F= Ma

2Nt-2Mg=0 ft is the friction force at the top both on the plank and on the rollers just in

opposite directions

F=600 N M=6kg

As each roller moves forward by the distance s the plank moves twice as far This

means that the acceleration of each roller is frac12 the acceleration of the plank

fb

fb

ft

ft

mg

mg

N

t

N

t

FW physics 10192011 843 PM page 18 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

18

Moment of inertia tensor

For your personal enjoyment only it will not be required in a test

For a more general derivation of the moment of inertia let us proceed without any

assumptions as to the kind of rotation We assume only that we are dealing with a rigid

body

(147)

i iv v

i i i

i

i i i i i i i i

i i

x y z

L r m r

L r m r m r r r r

r x y z

We need to calculate the three components for the vector L

(148)

2

Omitting the running index i in the x and components

x i x

i

L m x

2 2

xy z x x 2 2

2 2

y z i x y z

i

y i y

i

y z m y z x y z

L m x y

2

x yz y x y 2 2

2 2 2

z i y x z

i

z i z

i

z m x z y x z

L m x y z

x y zz x y z 2 2

i z x y

i

m x y z x y

This 3 by 3 ldquomatrixrdquo is called the moment of inertia tensor It reveals among other things

that the angular momentum L is not necessarily parallel to the angular velocity vector ω

(149)

2 2

2 2

2 2

2 2 2 2

1 etc

etc

x x

y i y

i

z z

xx i i i

i body

xy i i i

i body

L y z xy xz

L m yx x z yz

L zx zy x y

I I m y z y z dm

I m x y xydm

If the symmetry axes of a uniform symmetric body coincide with the coordinate axes the

mixed terms xy zx and so on vanish and the tensor has a simple diagonal form These

axes are then called principal axes

Remarkably it is always possible for a body of any shape and mass distribution to find a

set of three orthogonal axes such that the off-diagonal elements (the so-called products of

inertia) vanish

FW physics 10192011 843 PM page 19 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

19

(150)

0 0

0 0

0 0

x xx x

y yy y

z zz z

L I

L I

L I

For a uniform solid sphere or a shell any three orthogonal axes are principle axes and

we have

(151)

2

2

1

2 2 2

2 2 2 2

4 5 5 3 3

2 1 2 1

4

3 2 4

2 4 2 44

3 3 5 3

xx yy zz

zz

sphere

xx yy zz

spherer

R

R

I I I

I x y dm dm dV r dr

I I I I x y z dm dm r dr

I r dr R R M R R

For the inner radius equal 0 we get 22

5I MR which is the moment of inertia of a solid

sphere rotating about any diameter

For a thin spherical shell of mass m we get (no integration necessary)

(152) 2

2 2 2 2

2

2 2 2

3 2 4 4

2 24

3 3

xx yy zz

sphereR

mI I I I x y z dm m R

R

I R R mR

To calculate the moment of inertia of a thin hoop for a rotation around its center

The hoop has a line density ρ every point of the hoop has the same distance R from the

center

2

2 2 2 3 2

0

2I r dm r Rd R Rd R MR

since M=2πRρ

Calculating various moments of inertia

If we want to find the moment of inertia I for a rotating solid cylinder we have to choose

a cylindrical volume element which has a variable distance r (from 0 to R) from the axis

of rotation

(21) with 0 r R 0 2 and 0 z hdm rdrd dz

We can integrate over θ and z to give a volume element of

(22) 2dm rhdr

to arrive at the simple integral

FW physics 10192011 843 PM page 20 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

20

Ri ri

D A C

mi

(23)

43

2

0

2

M2 2 and with =

4 V

1

2

RR M

I h r dr hR h

I MR

The same can be achieved more readily with one simple integral if one chooses as

volume element a thin cylindrical disk of height h and thickness dr

(24)

42 2 3 2

0

2

12 2 2

4 2

with M= R

RR

I r dm r rhdr h r dr h MR

h

If we integrate r from an inner radius A to an outer radius B we get the moment of

inertia for a cylindrical shell with thickness B-A

(25)

4 43

2 2

2 2

( ) M2 2 and with =

4 V

1( )

2

B

A

B A MI h r dr h

B A h

I M B A

To calculate the moment of inertia of a rectangular stick of width A height H and

length L we put the axis of rotation at one end of the stick at the edge of the width A

The volume element

(26) dm= Hmiddotdxmiddotdy 0 A and 0y x L

(27) 3 3

2 2

0 0

( )3 3

L A

x y

L AI H x y dxdy H A L

The volume of the stick is H∙L∙A and its density is MH∙L∙A and therefore

(28) 2 2( )

3

M L AI

which is the moment of inertia for a rotation around one end (corner) To get

the moment of inertia around the center of mass we use the parallel axis

theorem

(29) 2 21( )

12cmI M L A

Proof of the parallel axis theorem the axes of rotation

are at A and C(center of mass) perpendicular to the

plane of writing

FW physics 10192011 843 PM page 21 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

21

(210)

2

2 2 2

2 2

2

2

2

0 because 0 why

A i i i i i i i

i i i i i

cm i i

I m R m D r m D r r D

D m m r D m r

MD I m r

Thus

(211) 2 where D is the distance between the two axesA cmI MD I

For a circular cone rotating around the z

axis we choose the volume element in

cylindrical coordinates

(212)

2 2 2

dV=rmiddotdrmiddotd middotdz 0 r 0

hr h z= and dz= and x +y =r

R R

R z h

R hdr

r z

(213)2

2 2

0 0 0

R h

cm

r z

I r dV r rdrdz d

As z depends on r we must first change the

boundaries of z which varies from hrR to h

and integrate it before we integrate over r

The integration over θ is simply 2π

(214)

542 3

0

0 0

4 4 4 42

2

2 2 24 5

32 2

14 5 20 10 10

3

R h R

R

cm

z hr R

hr hr hrI dzr rdr h r dr

R R

hR hR R R Mh h MR

R h

r

h

R

x

z

y

FW physics 10192011 843 PM page 22 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

22

Cycloids optional The speed of the center of mass is equal to the speed of a point on the rim (rolling

without slipping at constant velocity) with respect to the center For the x and y

coordinates of a point on the rim we get therefore

(215) r R

(216) 0

sin( ) and cos( )x t x R t R t y t R R t

If we start our motion with the point on the rim at x=0 and y=0 we get

(217) sin and cosx t R t R t y t R R t

For a point inside of the circle (RrsquoltR) the coordinates are given by

(218) sin and cosx t R t R t y t R R t

We see that the horizontal distance traveled remains equal to 2πR (not 2πRrsquo)

Assume for example that Rrsquo=05R

x0

In the time t the point on the rim of the circle (starting at the contact point

of the circle with the x-axis moves through the angle ωt to the left while

at the same time the center of the circle moves to the right by ωtR A point

inside the circle travels a shorter distance

x

FW physics 10192011 843 PM page 23 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

23

(219)

2 22 2 2 2

2 2

2

2 2

0

2 2

2

0 0

cos sin

(1 cos 2cos ) sin

2 1 cos 2 1 cos

1 12 1 cos sin 1 cos 2 sin 1 cos

2 2 2

2 2sin 2 sin 2 2 sin2 2

dx R dt R tdt dy R tdt

dx R dt t t dy R dt t

ds dx dy R t dt Rd

xS R d x x x

S R d R d R zdz

2

04 cos 82

R R

For a point inside the circle we get

(220)

222 2 2 2 2 2 2

2 2 2 2

22 2

cos sin

( cos 2 cos ) sin

2 cos

1 5for example R= we get ds= cos cos

2 4 4

dx R dt R tdt dy R tdt

dx dt R R t RR t dy R dt t

ds dx dy R R RR tdt

RR R R tdt R tdt

End of cycloids

FW physics 10192011 843 PM page 13 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

13

The rest is optional

Rolling of a spool

Consider the rolling spool above It consists of a set of two disks of radius R=1m with

mass M=30kg The interior disk serves as an axle and has a mass m=05kg and radius

r1=02 m It has a cord tightly wound around it A person pulls this cord to the right with

a constant force F=3N causing the disk to roll without friction to the right The person

pulls the string for x=15 m Find the velocity of the center of the disk at the end point of

15m

The interesting thing is that the distance x by which the cord is pulled to the right is not

equal to the distance covered by the center of the disk This can be understood easily

when one considers that the point C moves at a speed of 1vC R r with respect to

the ground In the time t the cord unfolds to a distance 1 1( ) ( )x R r t R r which

means that the angle in question is 1

x

R r

To find the speed of the rolling cylinder we need to use the work energy theorem

(138)

2 2 2 2 2

1

2

2 2 2 2

2

1 1

2

1 1 1 1 05( )v 15 004 15

2 2 2 2 2

v 1 1 v 1545 ( ) v 1125 v 123v

2 2 12

45v v 0605

123

W Fx M m I I MR mr kgm

Fx M m IR r R r

m

s

Spools of wire A spool of wire of mass M and radius R is unwound under a constant force F applied

horizontally to the right at the top of the spool Assuming that the spool is a solid cylinder

that does not slip show that (a) the acceleration of the center of mass is 4F3M and b)

D

C

A

B r1

R

FW physics 10192011 843 PM page 14 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

14

that the force of friction is to the right and equal in magnitude to F3 c) If the cylinder

starts from rest and rolls without slipping what is the speed of its center of mass after it

has rolled through a distance d 4

v3

Fd

M

We take the torques around point A

(139) 2 21 3 42 2

2 2 3

FRF MR MR F Ma a

M

Take the torques which provide the rolling around the center

(140) 21 1 1 1 4

2 2 2 2 3 3s s s

F FRF f R MR F f Ma f F Ma F M

M

c) work energy theorem

(141) 2 2 23 3 42 v v

2 2 3

FdW F d K MR M

M

The force must be applied for the distance 2d while the center moves to the

right by distance d

D

A

B

R

FW physics 10192011 843 PM page 15 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

15

The falling spool A string is wound around a uniform disk of radius R and mass M The disk is suspended

from a vertical string at its rim The string is fixed at a ceiling

a) Show that the tension in the string is 13 of the weight b) that the acceleration of the

center of the disk is 23 g and c) that the speed of the center of the disk after falling a

distance d is equal to 4

v=3

gd Verify this by

using energy considerations

a)

(142)

2

torques around the center

1 1 1RT=I

2 2 2

1

2

2

3

1 2 1

2 3 3

cm

mg T ma

mR maR T ma

mg ma ma

a g

T m g mg

To obtain the speed of the center of the disk we use the kinematic formulas

(143) 2 2 4v 2 2 v

3 3ad d g gd

To obtain the same result using energy conservation we need to realize that we have

conservation of energy The tension does not do any work

(144)

2 2 2 21 1 3 3v

2 2 2 4

4v

3

Amgd I mR m

gd

The dragged spool A spool of thread consists of a cylinder of radius R1with end-caps of radius R2 The mass

of the spool including the thread is m and its moment of inertia around the axis through

its center is I The spool is placed on a rough horizontal surface so it rolls without

slipping when a force T acting to the left is applied to the free end of the thread around

R1 Show that the magnitude of the friction force exerted by the surface on the spool is

given by 1 2

2

2

I mR Rf T

I mR

Determine the direction of the force of friction Calculate the

acceleration of the spool

The spool will roll to the right

f Tf T ma a

m m

R2

R1

FW physics 10192011 843 PM page 16 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

16

We take the torques around the center to get a clockwise rotation of the spool with an

acceleration of the center to the right

(145)

2

1 2 1 2 2

2

IaTR fR TR R fR Ia

R

Eliminate a from the two equations

(146)

2 2

1 2 2 1 2 2

1 2

2

2

1 2 1 2

2 2

2 2

1

mTR R mfR If IT mTR R IT If mfR

I mR Rf T

I mR

I mR R I mR RT T Ta

I mR m m m I mR

One can see that there is an acceleration as long as R1ltR2

Why do rolling objects slow down If rolling objects had perfect circular shapes they would never slow down while rolling

on a flat horizontal surface However a sphere resting or rolling on a surface is slightly

flattened at the contact area This results in a net torque due to the normal force of the

surface on the sphere opposing the direction of rolling

direction of rolling

creates a torque around the center point

opposing the rolling and bringing it to a

stop

FW physics 10192011 843 PM page 17 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

17

Moving plank on rollers To illustrate the point about the difficulty of discerning the correct direction of fs in the

context of rolling motion let us look at the following problem

A plank with a mass of M=600 kg rides on top of two identical solid cylindrical rollers

that have radii of R=500 cm and masses m=200 kg The plank is pulled by a constant

horizontal force of magnitude F=600 N (see picture) The cylinders roll without slipping

on a flat surface and no slipping occurs between the rollers and the plank Find the

acceleration of the plank and the rollers find the frictional forces

The reason why the plank moves twice the distance as the center of mass of the rollers is

that with respect to the ground the top of each roller moves with twice the speed

v 2top R and 2 2top tops R a R

The forces on the rollers ft-fb=mar and (ft+ fb )R=Iα=05 mR2 α=05Rmar

add the last two equations to get

2 ft=15mar=3ar ft=0752ar=15ar

fb=-025 mar=

from the plank equation we get -3ar+6=6a=12 ar ar=0400 a=0800

fb=-025 204=-0200N

ft=0600N

The interesting point here is that we have chosen fb in the wrong direction The friction

force fb points to the right

forces on the plank -2ft +F= Ma

2Nt-2Mg=0 ft is the friction force at the top both on the plank and on the rollers just in

opposite directions

F=600 N M=6kg

As each roller moves forward by the distance s the plank moves twice as far This

means that the acceleration of each roller is frac12 the acceleration of the plank

fb

fb

ft

ft

mg

mg

N

t

N

t

FW physics 10192011 843 PM page 18 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

18

Moment of inertia tensor

For your personal enjoyment only it will not be required in a test

For a more general derivation of the moment of inertia let us proceed without any

assumptions as to the kind of rotation We assume only that we are dealing with a rigid

body

(147)

i iv v

i i i

i

i i i i i i i i

i i

x y z

L r m r

L r m r m r r r r

r x y z

We need to calculate the three components for the vector L

(148)

2

Omitting the running index i in the x and components

x i x

i

L m x

2 2

xy z x x 2 2

2 2

y z i x y z

i

y i y

i

y z m y z x y z

L m x y

2

x yz y x y 2 2

2 2 2

z i y x z

i

z i z

i

z m x z y x z

L m x y z

x y zz x y z 2 2

i z x y

i

m x y z x y

This 3 by 3 ldquomatrixrdquo is called the moment of inertia tensor It reveals among other things

that the angular momentum L is not necessarily parallel to the angular velocity vector ω

(149)

2 2

2 2

2 2

2 2 2 2

1 etc

etc

x x

y i y

i

z z

xx i i i

i body

xy i i i

i body

L y z xy xz

L m yx x z yz

L zx zy x y

I I m y z y z dm

I m x y xydm

If the symmetry axes of a uniform symmetric body coincide with the coordinate axes the

mixed terms xy zx and so on vanish and the tensor has a simple diagonal form These

axes are then called principal axes

Remarkably it is always possible for a body of any shape and mass distribution to find a

set of three orthogonal axes such that the off-diagonal elements (the so-called products of

inertia) vanish

FW physics 10192011 843 PM page 19 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

19

(150)

0 0

0 0

0 0

x xx x

y yy y

z zz z

L I

L I

L I

For a uniform solid sphere or a shell any three orthogonal axes are principle axes and

we have

(151)

2

2

1

2 2 2

2 2 2 2

4 5 5 3 3

2 1 2 1

4

3 2 4

2 4 2 44

3 3 5 3

xx yy zz

zz

sphere

xx yy zz

spherer

R

R

I I I

I x y dm dm dV r dr

I I I I x y z dm dm r dr

I r dr R R M R R

For the inner radius equal 0 we get 22

5I MR which is the moment of inertia of a solid

sphere rotating about any diameter

For a thin spherical shell of mass m we get (no integration necessary)

(152) 2

2 2 2 2

2

2 2 2

3 2 4 4

2 24

3 3

xx yy zz

sphereR

mI I I I x y z dm m R

R

I R R mR

To calculate the moment of inertia of a thin hoop for a rotation around its center

The hoop has a line density ρ every point of the hoop has the same distance R from the

center

2

2 2 2 3 2

0

2I r dm r Rd R Rd R MR

since M=2πRρ

Calculating various moments of inertia

If we want to find the moment of inertia I for a rotating solid cylinder we have to choose

a cylindrical volume element which has a variable distance r (from 0 to R) from the axis

of rotation

(21) with 0 r R 0 2 and 0 z hdm rdrd dz

We can integrate over θ and z to give a volume element of

(22) 2dm rhdr

to arrive at the simple integral

FW physics 10192011 843 PM page 20 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

20

Ri ri

D A C

mi

(23)

43

2

0

2

M2 2 and with =

4 V

1

2

RR M

I h r dr hR h

I MR

The same can be achieved more readily with one simple integral if one chooses as

volume element a thin cylindrical disk of height h and thickness dr

(24)

42 2 3 2

0

2

12 2 2

4 2

with M= R

RR

I r dm r rhdr h r dr h MR

h

If we integrate r from an inner radius A to an outer radius B we get the moment of

inertia for a cylindrical shell with thickness B-A

(25)

4 43

2 2

2 2

( ) M2 2 and with =

4 V

1( )

2

B

A

B A MI h r dr h

B A h

I M B A

To calculate the moment of inertia of a rectangular stick of width A height H and

length L we put the axis of rotation at one end of the stick at the edge of the width A

The volume element

(26) dm= Hmiddotdxmiddotdy 0 A and 0y x L

(27) 3 3

2 2

0 0

( )3 3

L A

x y

L AI H x y dxdy H A L

The volume of the stick is H∙L∙A and its density is MH∙L∙A and therefore

(28) 2 2( )

3

M L AI

which is the moment of inertia for a rotation around one end (corner) To get

the moment of inertia around the center of mass we use the parallel axis

theorem

(29) 2 21( )

12cmI M L A

Proof of the parallel axis theorem the axes of rotation

are at A and C(center of mass) perpendicular to the

plane of writing

FW physics 10192011 843 PM page 21 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

21

(210)

2

2 2 2

2 2

2

2

2

0 because 0 why

A i i i i i i i

i i i i i

cm i i

I m R m D r m D r r D

D m m r D m r

MD I m r

Thus

(211) 2 where D is the distance between the two axesA cmI MD I

For a circular cone rotating around the z

axis we choose the volume element in

cylindrical coordinates

(212)

2 2 2

dV=rmiddotdrmiddotd middotdz 0 r 0

hr h z= and dz= and x +y =r

R R

R z h

R hdr

r z

(213)2

2 2

0 0 0

R h

cm

r z

I r dV r rdrdz d

As z depends on r we must first change the

boundaries of z which varies from hrR to h

and integrate it before we integrate over r

The integration over θ is simply 2π

(214)

542 3

0

0 0

4 4 4 42

2

2 2 24 5

32 2

14 5 20 10 10

3

R h R

R

cm

z hr R

hr hr hrI dzr rdr h r dr

R R

hR hR R R Mh h MR

R h

r

h

R

x

z

y

FW physics 10192011 843 PM page 22 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

22

Cycloids optional The speed of the center of mass is equal to the speed of a point on the rim (rolling

without slipping at constant velocity) with respect to the center For the x and y

coordinates of a point on the rim we get therefore

(215) r R

(216) 0

sin( ) and cos( )x t x R t R t y t R R t

If we start our motion with the point on the rim at x=0 and y=0 we get

(217) sin and cosx t R t R t y t R R t

For a point inside of the circle (RrsquoltR) the coordinates are given by

(218) sin and cosx t R t R t y t R R t

We see that the horizontal distance traveled remains equal to 2πR (not 2πRrsquo)

Assume for example that Rrsquo=05R

x0

In the time t the point on the rim of the circle (starting at the contact point

of the circle with the x-axis moves through the angle ωt to the left while

at the same time the center of the circle moves to the right by ωtR A point

inside the circle travels a shorter distance

x

FW physics 10192011 843 PM page 23 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

23

(219)

2 22 2 2 2

2 2

2

2 2

0

2 2

2

0 0

cos sin

(1 cos 2cos ) sin

2 1 cos 2 1 cos

1 12 1 cos sin 1 cos 2 sin 1 cos

2 2 2

2 2sin 2 sin 2 2 sin2 2

dx R dt R tdt dy R tdt

dx R dt t t dy R dt t

ds dx dy R t dt Rd

xS R d x x x

S R d R d R zdz

2

04 cos 82

R R

For a point inside the circle we get

(220)

222 2 2 2 2 2 2

2 2 2 2

22 2

cos sin

( cos 2 cos ) sin

2 cos

1 5for example R= we get ds= cos cos

2 4 4

dx R dt R tdt dy R tdt

dx dt R R t RR t dy R dt t

ds dx dy R R RR tdt

RR R R tdt R tdt

End of cycloids

FW physics 10192011 843 PM page 14 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

14

that the force of friction is to the right and equal in magnitude to F3 c) If the cylinder

starts from rest and rolls without slipping what is the speed of its center of mass after it

has rolled through a distance d 4

v3

Fd

M

We take the torques around point A

(139) 2 21 3 42 2

2 2 3

FRF MR MR F Ma a

M

Take the torques which provide the rolling around the center

(140) 21 1 1 1 4

2 2 2 2 3 3s s s

F FRF f R MR F f Ma f F Ma F M

M

c) work energy theorem

(141) 2 2 23 3 42 v v

2 2 3

FdW F d K MR M

M

The force must be applied for the distance 2d while the center moves to the

right by distance d

D

A

B

R

FW physics 10192011 843 PM page 15 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

15

The falling spool A string is wound around a uniform disk of radius R and mass M The disk is suspended

from a vertical string at its rim The string is fixed at a ceiling

a) Show that the tension in the string is 13 of the weight b) that the acceleration of the

center of the disk is 23 g and c) that the speed of the center of the disk after falling a

distance d is equal to 4

v=3

gd Verify this by

using energy considerations

a)

(142)

2

torques around the center

1 1 1RT=I

2 2 2

1

2

2

3

1 2 1

2 3 3

cm

mg T ma

mR maR T ma

mg ma ma

a g

T m g mg

To obtain the speed of the center of the disk we use the kinematic formulas

(143) 2 2 4v 2 2 v

3 3ad d g gd

To obtain the same result using energy conservation we need to realize that we have

conservation of energy The tension does not do any work

(144)

2 2 2 21 1 3 3v

2 2 2 4

4v

3

Amgd I mR m

gd

The dragged spool A spool of thread consists of a cylinder of radius R1with end-caps of radius R2 The mass

of the spool including the thread is m and its moment of inertia around the axis through

its center is I The spool is placed on a rough horizontal surface so it rolls without

slipping when a force T acting to the left is applied to the free end of the thread around

R1 Show that the magnitude of the friction force exerted by the surface on the spool is

given by 1 2

2

2

I mR Rf T

I mR

Determine the direction of the force of friction Calculate the

acceleration of the spool

The spool will roll to the right

f Tf T ma a

m m

R2

R1

FW physics 10192011 843 PM page 16 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

16

We take the torques around the center to get a clockwise rotation of the spool with an

acceleration of the center to the right

(145)

2

1 2 1 2 2

2

IaTR fR TR R fR Ia

R

Eliminate a from the two equations

(146)

2 2

1 2 2 1 2 2

1 2

2

2

1 2 1 2

2 2

2 2

1

mTR R mfR If IT mTR R IT If mfR

I mR Rf T

I mR

I mR R I mR RT T Ta

I mR m m m I mR

One can see that there is an acceleration as long as R1ltR2

Why do rolling objects slow down If rolling objects had perfect circular shapes they would never slow down while rolling

on a flat horizontal surface However a sphere resting or rolling on a surface is slightly

flattened at the contact area This results in a net torque due to the normal force of the

surface on the sphere opposing the direction of rolling

direction of rolling

creates a torque around the center point

opposing the rolling and bringing it to a

stop

FW physics 10192011 843 PM page 17 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

17

Moving plank on rollers To illustrate the point about the difficulty of discerning the correct direction of fs in the

context of rolling motion let us look at the following problem

A plank with a mass of M=600 kg rides on top of two identical solid cylindrical rollers

that have radii of R=500 cm and masses m=200 kg The plank is pulled by a constant

horizontal force of magnitude F=600 N (see picture) The cylinders roll without slipping

on a flat surface and no slipping occurs between the rollers and the plank Find the

acceleration of the plank and the rollers find the frictional forces

The reason why the plank moves twice the distance as the center of mass of the rollers is

that with respect to the ground the top of each roller moves with twice the speed

v 2top R and 2 2top tops R a R

The forces on the rollers ft-fb=mar and (ft+ fb )R=Iα=05 mR2 α=05Rmar

add the last two equations to get

2 ft=15mar=3ar ft=0752ar=15ar

fb=-025 mar=

from the plank equation we get -3ar+6=6a=12 ar ar=0400 a=0800

fb=-025 204=-0200N

ft=0600N

The interesting point here is that we have chosen fb in the wrong direction The friction

force fb points to the right

forces on the plank -2ft +F= Ma

2Nt-2Mg=0 ft is the friction force at the top both on the plank and on the rollers just in

opposite directions

F=600 N M=6kg

As each roller moves forward by the distance s the plank moves twice as far This

means that the acceleration of each roller is frac12 the acceleration of the plank

fb

fb

ft

ft

mg

mg

N

t

N

t

FW physics 10192011 843 PM page 18 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

18

Moment of inertia tensor

For your personal enjoyment only it will not be required in a test

For a more general derivation of the moment of inertia let us proceed without any

assumptions as to the kind of rotation We assume only that we are dealing with a rigid

body

(147)

i iv v

i i i

i

i i i i i i i i

i i

x y z

L r m r

L r m r m r r r r

r x y z

We need to calculate the three components for the vector L

(148)

2

Omitting the running index i in the x and components

x i x

i

L m x

2 2

xy z x x 2 2

2 2

y z i x y z

i

y i y

i

y z m y z x y z

L m x y

2

x yz y x y 2 2

2 2 2

z i y x z

i

z i z

i

z m x z y x z

L m x y z

x y zz x y z 2 2

i z x y

i

m x y z x y

This 3 by 3 ldquomatrixrdquo is called the moment of inertia tensor It reveals among other things

that the angular momentum L is not necessarily parallel to the angular velocity vector ω

(149)

2 2

2 2

2 2

2 2 2 2

1 etc

etc

x x

y i y

i

z z

xx i i i

i body

xy i i i

i body

L y z xy xz

L m yx x z yz

L zx zy x y

I I m y z y z dm

I m x y xydm

If the symmetry axes of a uniform symmetric body coincide with the coordinate axes the

mixed terms xy zx and so on vanish and the tensor has a simple diagonal form These

axes are then called principal axes

Remarkably it is always possible for a body of any shape and mass distribution to find a

set of three orthogonal axes such that the off-diagonal elements (the so-called products of

inertia) vanish

FW physics 10192011 843 PM page 19 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

19

(150)

0 0

0 0

0 0

x xx x

y yy y

z zz z

L I

L I

L I

For a uniform solid sphere or a shell any three orthogonal axes are principle axes and

we have

(151)

2

2

1

2 2 2

2 2 2 2

4 5 5 3 3

2 1 2 1

4

3 2 4

2 4 2 44

3 3 5 3

xx yy zz

zz

sphere

xx yy zz

spherer

R

R

I I I

I x y dm dm dV r dr

I I I I x y z dm dm r dr

I r dr R R M R R

For the inner radius equal 0 we get 22

5I MR which is the moment of inertia of a solid

sphere rotating about any diameter

For a thin spherical shell of mass m we get (no integration necessary)

(152) 2

2 2 2 2

2

2 2 2

3 2 4 4

2 24

3 3

xx yy zz

sphereR

mI I I I x y z dm m R

R

I R R mR

To calculate the moment of inertia of a thin hoop for a rotation around its center

The hoop has a line density ρ every point of the hoop has the same distance R from the

center

2

2 2 2 3 2

0

2I r dm r Rd R Rd R MR

since M=2πRρ

Calculating various moments of inertia

If we want to find the moment of inertia I for a rotating solid cylinder we have to choose

a cylindrical volume element which has a variable distance r (from 0 to R) from the axis

of rotation

(21) with 0 r R 0 2 and 0 z hdm rdrd dz

We can integrate over θ and z to give a volume element of

(22) 2dm rhdr

to arrive at the simple integral

FW physics 10192011 843 PM page 20 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

20

Ri ri

D A C

mi

(23)

43

2

0

2

M2 2 and with =

4 V

1

2

RR M

I h r dr hR h

I MR

The same can be achieved more readily with one simple integral if one chooses as

volume element a thin cylindrical disk of height h and thickness dr

(24)

42 2 3 2

0

2

12 2 2

4 2

with M= R

RR

I r dm r rhdr h r dr h MR

h

If we integrate r from an inner radius A to an outer radius B we get the moment of

inertia for a cylindrical shell with thickness B-A

(25)

4 43

2 2

2 2

( ) M2 2 and with =

4 V

1( )

2

B

A

B A MI h r dr h

B A h

I M B A

To calculate the moment of inertia of a rectangular stick of width A height H and

length L we put the axis of rotation at one end of the stick at the edge of the width A

The volume element

(26) dm= Hmiddotdxmiddotdy 0 A and 0y x L

(27) 3 3

2 2

0 0

( )3 3

L A

x y

L AI H x y dxdy H A L

The volume of the stick is H∙L∙A and its density is MH∙L∙A and therefore

(28) 2 2( )

3

M L AI

which is the moment of inertia for a rotation around one end (corner) To get

the moment of inertia around the center of mass we use the parallel axis

theorem

(29) 2 21( )

12cmI M L A

Proof of the parallel axis theorem the axes of rotation

are at A and C(center of mass) perpendicular to the

plane of writing

FW physics 10192011 843 PM page 21 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

21

(210)

2

2 2 2

2 2

2

2

2

0 because 0 why

A i i i i i i i

i i i i i

cm i i

I m R m D r m D r r D

D m m r D m r

MD I m r

Thus

(211) 2 where D is the distance between the two axesA cmI MD I

For a circular cone rotating around the z

axis we choose the volume element in

cylindrical coordinates

(212)

2 2 2

dV=rmiddotdrmiddotd middotdz 0 r 0

hr h z= and dz= and x +y =r

R R

R z h

R hdr

r z

(213)2

2 2

0 0 0

R h

cm

r z

I r dV r rdrdz d

As z depends on r we must first change the

boundaries of z which varies from hrR to h

and integrate it before we integrate over r

The integration over θ is simply 2π

(214)

542 3

0

0 0

4 4 4 42

2

2 2 24 5

32 2

14 5 20 10 10

3

R h R

R

cm

z hr R

hr hr hrI dzr rdr h r dr

R R

hR hR R R Mh h MR

R h

r

h

R

x

z

y

FW physics 10192011 843 PM page 22 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

22

Cycloids optional The speed of the center of mass is equal to the speed of a point on the rim (rolling

without slipping at constant velocity) with respect to the center For the x and y

coordinates of a point on the rim we get therefore

(215) r R

(216) 0

sin( ) and cos( )x t x R t R t y t R R t

If we start our motion with the point on the rim at x=0 and y=0 we get

(217) sin and cosx t R t R t y t R R t

For a point inside of the circle (RrsquoltR) the coordinates are given by

(218) sin and cosx t R t R t y t R R t

We see that the horizontal distance traveled remains equal to 2πR (not 2πRrsquo)

Assume for example that Rrsquo=05R

x0

In the time t the point on the rim of the circle (starting at the contact point

of the circle with the x-axis moves through the angle ωt to the left while

at the same time the center of the circle moves to the right by ωtR A point

inside the circle travels a shorter distance

x

FW physics 10192011 843 PM page 23 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

23

(219)

2 22 2 2 2

2 2

2

2 2

0

2 2

2

0 0

cos sin

(1 cos 2cos ) sin

2 1 cos 2 1 cos

1 12 1 cos sin 1 cos 2 sin 1 cos

2 2 2

2 2sin 2 sin 2 2 sin2 2

dx R dt R tdt dy R tdt

dx R dt t t dy R dt t

ds dx dy R t dt Rd

xS R d x x x

S R d R d R zdz

2

04 cos 82

R R

For a point inside the circle we get

(220)

222 2 2 2 2 2 2

2 2 2 2

22 2

cos sin

( cos 2 cos ) sin

2 cos

1 5for example R= we get ds= cos cos

2 4 4

dx R dt R tdt dy R tdt

dx dt R R t RR t dy R dt t

ds dx dy R R RR tdt

RR R R tdt R tdt

End of cycloids

FW physics 10192011 843 PM page 15 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

15

The falling spool A string is wound around a uniform disk of radius R and mass M The disk is suspended

from a vertical string at its rim The string is fixed at a ceiling

a) Show that the tension in the string is 13 of the weight b) that the acceleration of the

center of the disk is 23 g and c) that the speed of the center of the disk after falling a

distance d is equal to 4

v=3

gd Verify this by

using energy considerations

a)

(142)

2

torques around the center

1 1 1RT=I

2 2 2

1

2

2

3

1 2 1

2 3 3

cm

mg T ma

mR maR T ma

mg ma ma

a g

T m g mg

To obtain the speed of the center of the disk we use the kinematic formulas

(143) 2 2 4v 2 2 v

3 3ad d g gd

To obtain the same result using energy conservation we need to realize that we have

conservation of energy The tension does not do any work

(144)

2 2 2 21 1 3 3v

2 2 2 4

4v

3

Amgd I mR m

gd

The dragged spool A spool of thread consists of a cylinder of radius R1with end-caps of radius R2 The mass

of the spool including the thread is m and its moment of inertia around the axis through

its center is I The spool is placed on a rough horizontal surface so it rolls without

slipping when a force T acting to the left is applied to the free end of the thread around

R1 Show that the magnitude of the friction force exerted by the surface on the spool is

given by 1 2

2

2

I mR Rf T

I mR

Determine the direction of the force of friction Calculate the

acceleration of the spool

The spool will roll to the right

f Tf T ma a

m m

R2

R1

FW physics 10192011 843 PM page 16 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

16

We take the torques around the center to get a clockwise rotation of the spool with an

acceleration of the center to the right

(145)

2

1 2 1 2 2

2

IaTR fR TR R fR Ia

R

Eliminate a from the two equations

(146)

2 2

1 2 2 1 2 2

1 2

2

2

1 2 1 2

2 2

2 2

1

mTR R mfR If IT mTR R IT If mfR

I mR Rf T

I mR

I mR R I mR RT T Ta

I mR m m m I mR

One can see that there is an acceleration as long as R1ltR2

Why do rolling objects slow down If rolling objects had perfect circular shapes they would never slow down while rolling

on a flat horizontal surface However a sphere resting or rolling on a surface is slightly

flattened at the contact area This results in a net torque due to the normal force of the

surface on the sphere opposing the direction of rolling

direction of rolling

creates a torque around the center point

opposing the rolling and bringing it to a

stop

FW physics 10192011 843 PM page 17 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

17

Moving plank on rollers To illustrate the point about the difficulty of discerning the correct direction of fs in the

context of rolling motion let us look at the following problem

A plank with a mass of M=600 kg rides on top of two identical solid cylindrical rollers

that have radii of R=500 cm and masses m=200 kg The plank is pulled by a constant

horizontal force of magnitude F=600 N (see picture) The cylinders roll without slipping

on a flat surface and no slipping occurs between the rollers and the plank Find the

acceleration of the plank and the rollers find the frictional forces

The reason why the plank moves twice the distance as the center of mass of the rollers is

that with respect to the ground the top of each roller moves with twice the speed

v 2top R and 2 2top tops R a R

The forces on the rollers ft-fb=mar and (ft+ fb )R=Iα=05 mR2 α=05Rmar

add the last two equations to get

2 ft=15mar=3ar ft=0752ar=15ar

fb=-025 mar=

from the plank equation we get -3ar+6=6a=12 ar ar=0400 a=0800

fb=-025 204=-0200N

ft=0600N

The interesting point here is that we have chosen fb in the wrong direction The friction

force fb points to the right

forces on the plank -2ft +F= Ma

2Nt-2Mg=0 ft is the friction force at the top both on the plank and on the rollers just in

opposite directions

F=600 N M=6kg

As each roller moves forward by the distance s the plank moves twice as far This

means that the acceleration of each roller is frac12 the acceleration of the plank

fb

fb

ft

ft

mg

mg

N

t

N

t

FW physics 10192011 843 PM page 18 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

18

Moment of inertia tensor

For your personal enjoyment only it will not be required in a test

For a more general derivation of the moment of inertia let us proceed without any

assumptions as to the kind of rotation We assume only that we are dealing with a rigid

body

(147)

i iv v

i i i

i

i i i i i i i i

i i

x y z

L r m r

L r m r m r r r r

r x y z

We need to calculate the three components for the vector L

(148)

2

Omitting the running index i in the x and components

x i x

i

L m x

2 2

xy z x x 2 2

2 2

y z i x y z

i

y i y

i

y z m y z x y z

L m x y

2

x yz y x y 2 2

2 2 2

z i y x z

i

z i z

i

z m x z y x z

L m x y z

x y zz x y z 2 2

i z x y

i

m x y z x y

This 3 by 3 ldquomatrixrdquo is called the moment of inertia tensor It reveals among other things

that the angular momentum L is not necessarily parallel to the angular velocity vector ω

(149)

2 2

2 2

2 2

2 2 2 2

1 etc

etc

x x

y i y

i

z z

xx i i i

i body

xy i i i

i body

L y z xy xz

L m yx x z yz

L zx zy x y

I I m y z y z dm

I m x y xydm

If the symmetry axes of a uniform symmetric body coincide with the coordinate axes the

mixed terms xy zx and so on vanish and the tensor has a simple diagonal form These

axes are then called principal axes

Remarkably it is always possible for a body of any shape and mass distribution to find a

set of three orthogonal axes such that the off-diagonal elements (the so-called products of

inertia) vanish

FW physics 10192011 843 PM page 19 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

19

(150)

0 0

0 0

0 0

x xx x

y yy y

z zz z

L I

L I

L I

For a uniform solid sphere or a shell any three orthogonal axes are principle axes and

we have

(151)

2

2

1

2 2 2

2 2 2 2

4 5 5 3 3

2 1 2 1

4

3 2 4

2 4 2 44

3 3 5 3

xx yy zz

zz

sphere

xx yy zz

spherer

R

R

I I I

I x y dm dm dV r dr

I I I I x y z dm dm r dr

I r dr R R M R R

For the inner radius equal 0 we get 22

5I MR which is the moment of inertia of a solid

sphere rotating about any diameter

For a thin spherical shell of mass m we get (no integration necessary)

(152) 2

2 2 2 2

2

2 2 2

3 2 4 4

2 24

3 3

xx yy zz

sphereR

mI I I I x y z dm m R

R

I R R mR

To calculate the moment of inertia of a thin hoop for a rotation around its center

The hoop has a line density ρ every point of the hoop has the same distance R from the

center

2

2 2 2 3 2

0

2I r dm r Rd R Rd R MR

since M=2πRρ

Calculating various moments of inertia

If we want to find the moment of inertia I for a rotating solid cylinder we have to choose

a cylindrical volume element which has a variable distance r (from 0 to R) from the axis

of rotation

(21) with 0 r R 0 2 and 0 z hdm rdrd dz

We can integrate over θ and z to give a volume element of

(22) 2dm rhdr

to arrive at the simple integral

FW physics 10192011 843 PM page 20 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

20

Ri ri

D A C

mi

(23)

43

2

0

2

M2 2 and with =

4 V

1

2

RR M

I h r dr hR h

I MR

The same can be achieved more readily with one simple integral if one chooses as

volume element a thin cylindrical disk of height h and thickness dr

(24)

42 2 3 2

0

2

12 2 2

4 2

with M= R

RR

I r dm r rhdr h r dr h MR

h

If we integrate r from an inner radius A to an outer radius B we get the moment of

inertia for a cylindrical shell with thickness B-A

(25)

4 43

2 2

2 2

( ) M2 2 and with =

4 V

1( )

2

B

A

B A MI h r dr h

B A h

I M B A

To calculate the moment of inertia of a rectangular stick of width A height H and

length L we put the axis of rotation at one end of the stick at the edge of the width A

The volume element

(26) dm= Hmiddotdxmiddotdy 0 A and 0y x L

(27) 3 3

2 2

0 0

( )3 3

L A

x y

L AI H x y dxdy H A L

The volume of the stick is H∙L∙A and its density is MH∙L∙A and therefore

(28) 2 2( )

3

M L AI

which is the moment of inertia for a rotation around one end (corner) To get

the moment of inertia around the center of mass we use the parallel axis

theorem

(29) 2 21( )

12cmI M L A

Proof of the parallel axis theorem the axes of rotation

are at A and C(center of mass) perpendicular to the

plane of writing

FW physics 10192011 843 PM page 21 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

21

(210)

2

2 2 2

2 2

2

2

2

0 because 0 why

A i i i i i i i

i i i i i

cm i i

I m R m D r m D r r D

D m m r D m r

MD I m r

Thus

(211) 2 where D is the distance between the two axesA cmI MD I

For a circular cone rotating around the z

axis we choose the volume element in

cylindrical coordinates

(212)

2 2 2

dV=rmiddotdrmiddotd middotdz 0 r 0

hr h z= and dz= and x +y =r

R R

R z h

R hdr

r z

(213)2

2 2

0 0 0

R h

cm

r z

I r dV r rdrdz d

As z depends on r we must first change the

boundaries of z which varies from hrR to h

and integrate it before we integrate over r

The integration over θ is simply 2π

(214)

542 3

0

0 0

4 4 4 42

2

2 2 24 5

32 2

14 5 20 10 10

3

R h R

R

cm

z hr R

hr hr hrI dzr rdr h r dr

R R

hR hR R R Mh h MR

R h

r

h

R

x

z

y

FW physics 10192011 843 PM page 22 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

22

Cycloids optional The speed of the center of mass is equal to the speed of a point on the rim (rolling

without slipping at constant velocity) with respect to the center For the x and y

coordinates of a point on the rim we get therefore

(215) r R

(216) 0

sin( ) and cos( )x t x R t R t y t R R t

If we start our motion with the point on the rim at x=0 and y=0 we get

(217) sin and cosx t R t R t y t R R t

For a point inside of the circle (RrsquoltR) the coordinates are given by

(218) sin and cosx t R t R t y t R R t

We see that the horizontal distance traveled remains equal to 2πR (not 2πRrsquo)

Assume for example that Rrsquo=05R

x0

In the time t the point on the rim of the circle (starting at the contact point

of the circle with the x-axis moves through the angle ωt to the left while

at the same time the center of the circle moves to the right by ωtR A point

inside the circle travels a shorter distance

x

FW physics 10192011 843 PM page 23 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

23

(219)

2 22 2 2 2

2 2

2

2 2

0

2 2

2

0 0

cos sin

(1 cos 2cos ) sin

2 1 cos 2 1 cos

1 12 1 cos sin 1 cos 2 sin 1 cos

2 2 2

2 2sin 2 sin 2 2 sin2 2

dx R dt R tdt dy R tdt

dx R dt t t dy R dt t

ds dx dy R t dt Rd

xS R d x x x

S R d R d R zdz

2

04 cos 82

R R

For a point inside the circle we get

(220)

222 2 2 2 2 2 2

2 2 2 2

22 2

cos sin

( cos 2 cos ) sin

2 cos

1 5for example R= we get ds= cos cos

2 4 4

dx R dt R tdt dy R tdt

dx dt R R t RR t dy R dt t

ds dx dy R R RR tdt

RR R R tdt R tdt

End of cycloids

FW physics 10192011 843 PM page 16 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

16

We take the torques around the center to get a clockwise rotation of the spool with an

acceleration of the center to the right

(145)

2

1 2 1 2 2

2

IaTR fR TR R fR Ia

R

Eliminate a from the two equations

(146)

2 2

1 2 2 1 2 2

1 2

2

2

1 2 1 2

2 2

2 2

1

mTR R mfR If IT mTR R IT If mfR

I mR Rf T

I mR

I mR R I mR RT T Ta

I mR m m m I mR

One can see that there is an acceleration as long as R1ltR2

Why do rolling objects slow down If rolling objects had perfect circular shapes they would never slow down while rolling

on a flat horizontal surface However a sphere resting or rolling on a surface is slightly

flattened at the contact area This results in a net torque due to the normal force of the

surface on the sphere opposing the direction of rolling

direction of rolling

creates a torque around the center point

opposing the rolling and bringing it to a

stop

FW physics 10192011 843 PM page 17 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

17

Moving plank on rollers To illustrate the point about the difficulty of discerning the correct direction of fs in the

context of rolling motion let us look at the following problem

A plank with a mass of M=600 kg rides on top of two identical solid cylindrical rollers

that have radii of R=500 cm and masses m=200 kg The plank is pulled by a constant

horizontal force of magnitude F=600 N (see picture) The cylinders roll without slipping

on a flat surface and no slipping occurs between the rollers and the plank Find the

acceleration of the plank and the rollers find the frictional forces

The reason why the plank moves twice the distance as the center of mass of the rollers is

that with respect to the ground the top of each roller moves with twice the speed

v 2top R and 2 2top tops R a R

The forces on the rollers ft-fb=mar and (ft+ fb )R=Iα=05 mR2 α=05Rmar

add the last two equations to get

2 ft=15mar=3ar ft=0752ar=15ar

fb=-025 mar=

from the plank equation we get -3ar+6=6a=12 ar ar=0400 a=0800

fb=-025 204=-0200N

ft=0600N

The interesting point here is that we have chosen fb in the wrong direction The friction

force fb points to the right

forces on the plank -2ft +F= Ma

2Nt-2Mg=0 ft is the friction force at the top both on the plank and on the rollers just in

opposite directions

F=600 N M=6kg

As each roller moves forward by the distance s the plank moves twice as far This

means that the acceleration of each roller is frac12 the acceleration of the plank

fb

fb

ft

ft

mg

mg

N

t

N

t

FW physics 10192011 843 PM page 18 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

18

Moment of inertia tensor

For your personal enjoyment only it will not be required in a test

For a more general derivation of the moment of inertia let us proceed without any

assumptions as to the kind of rotation We assume only that we are dealing with a rigid

body

(147)

i iv v

i i i

i

i i i i i i i i

i i

x y z

L r m r

L r m r m r r r r

r x y z

We need to calculate the three components for the vector L

(148)

2

Omitting the running index i in the x and components

x i x

i

L m x

2 2

xy z x x 2 2

2 2

y z i x y z

i

y i y

i

y z m y z x y z

L m x y

2

x yz y x y 2 2

2 2 2

z i y x z

i

z i z

i

z m x z y x z

L m x y z

x y zz x y z 2 2

i z x y

i

m x y z x y

This 3 by 3 ldquomatrixrdquo is called the moment of inertia tensor It reveals among other things

that the angular momentum L is not necessarily parallel to the angular velocity vector ω

(149)

2 2

2 2

2 2

2 2 2 2

1 etc

etc

x x

y i y

i

z z

xx i i i

i body

xy i i i

i body

L y z xy xz

L m yx x z yz

L zx zy x y

I I m y z y z dm

I m x y xydm

If the symmetry axes of a uniform symmetric body coincide with the coordinate axes the

mixed terms xy zx and so on vanish and the tensor has a simple diagonal form These

axes are then called principal axes

Remarkably it is always possible for a body of any shape and mass distribution to find a

set of three orthogonal axes such that the off-diagonal elements (the so-called products of

inertia) vanish

FW physics 10192011 843 PM page 19 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

19

(150)

0 0

0 0

0 0

x xx x

y yy y

z zz z

L I

L I

L I

For a uniform solid sphere or a shell any three orthogonal axes are principle axes and

we have

(151)

2

2

1

2 2 2

2 2 2 2

4 5 5 3 3

2 1 2 1

4

3 2 4

2 4 2 44

3 3 5 3

xx yy zz

zz

sphere

xx yy zz

spherer

R

R

I I I

I x y dm dm dV r dr

I I I I x y z dm dm r dr

I r dr R R M R R

For the inner radius equal 0 we get 22

5I MR which is the moment of inertia of a solid

sphere rotating about any diameter

For a thin spherical shell of mass m we get (no integration necessary)

(152) 2

2 2 2 2

2

2 2 2

3 2 4 4

2 24

3 3

xx yy zz

sphereR

mI I I I x y z dm m R

R

I R R mR

To calculate the moment of inertia of a thin hoop for a rotation around its center

The hoop has a line density ρ every point of the hoop has the same distance R from the

center

2

2 2 2 3 2

0

2I r dm r Rd R Rd R MR

since M=2πRρ

Calculating various moments of inertia

If we want to find the moment of inertia I for a rotating solid cylinder we have to choose

a cylindrical volume element which has a variable distance r (from 0 to R) from the axis

of rotation

(21) with 0 r R 0 2 and 0 z hdm rdrd dz

We can integrate over θ and z to give a volume element of

(22) 2dm rhdr

to arrive at the simple integral

FW physics 10192011 843 PM page 20 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

20

Ri ri

D A C

mi

(23)

43

2

0

2

M2 2 and with =

4 V

1

2

RR M

I h r dr hR h

I MR

The same can be achieved more readily with one simple integral if one chooses as

volume element a thin cylindrical disk of height h and thickness dr

(24)

42 2 3 2

0

2

12 2 2

4 2

with M= R

RR

I r dm r rhdr h r dr h MR

h

If we integrate r from an inner radius A to an outer radius B we get the moment of

inertia for a cylindrical shell with thickness B-A

(25)

4 43

2 2

2 2

( ) M2 2 and with =

4 V

1( )

2

B

A

B A MI h r dr h

B A h

I M B A

To calculate the moment of inertia of a rectangular stick of width A height H and

length L we put the axis of rotation at one end of the stick at the edge of the width A

The volume element

(26) dm= Hmiddotdxmiddotdy 0 A and 0y x L

(27) 3 3

2 2

0 0

( )3 3

L A

x y

L AI H x y dxdy H A L

The volume of the stick is H∙L∙A and its density is MH∙L∙A and therefore

(28) 2 2( )

3

M L AI

which is the moment of inertia for a rotation around one end (corner) To get

the moment of inertia around the center of mass we use the parallel axis

theorem

(29) 2 21( )

12cmI M L A

Proof of the parallel axis theorem the axes of rotation

are at A and C(center of mass) perpendicular to the

plane of writing

FW physics 10192011 843 PM page 21 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

21

(210)

2

2 2 2

2 2

2

2

2

0 because 0 why

A i i i i i i i

i i i i i

cm i i

I m R m D r m D r r D

D m m r D m r

MD I m r

Thus

(211) 2 where D is the distance between the two axesA cmI MD I

For a circular cone rotating around the z

axis we choose the volume element in

cylindrical coordinates

(212)

2 2 2

dV=rmiddotdrmiddotd middotdz 0 r 0

hr h z= and dz= and x +y =r

R R

R z h

R hdr

r z

(213)2

2 2

0 0 0

R h

cm

r z

I r dV r rdrdz d

As z depends on r we must first change the

boundaries of z which varies from hrR to h

and integrate it before we integrate over r

The integration over θ is simply 2π

(214)

542 3

0

0 0

4 4 4 42

2

2 2 24 5

32 2

14 5 20 10 10

3

R h R

R

cm

z hr R

hr hr hrI dzr rdr h r dr

R R

hR hR R R Mh h MR

R h

r

h

R

x

z

y

FW physics 10192011 843 PM page 22 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

22

Cycloids optional The speed of the center of mass is equal to the speed of a point on the rim (rolling

without slipping at constant velocity) with respect to the center For the x and y

coordinates of a point on the rim we get therefore

(215) r R

(216) 0

sin( ) and cos( )x t x R t R t y t R R t

If we start our motion with the point on the rim at x=0 and y=0 we get

(217) sin and cosx t R t R t y t R R t

For a point inside of the circle (RrsquoltR) the coordinates are given by

(218) sin and cosx t R t R t y t R R t

We see that the horizontal distance traveled remains equal to 2πR (not 2πRrsquo)

Assume for example that Rrsquo=05R

x0

In the time t the point on the rim of the circle (starting at the contact point

of the circle with the x-axis moves through the angle ωt to the left while

at the same time the center of the circle moves to the right by ωtR A point

inside the circle travels a shorter distance

x

FW physics 10192011 843 PM page 23 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

23

(219)

2 22 2 2 2

2 2

2

2 2

0

2 2

2

0 0

cos sin

(1 cos 2cos ) sin

2 1 cos 2 1 cos

1 12 1 cos sin 1 cos 2 sin 1 cos

2 2 2

2 2sin 2 sin 2 2 sin2 2

dx R dt R tdt dy R tdt

dx R dt t t dy R dt t

ds dx dy R t dt Rd

xS R d x x x

S R d R d R zdz

2

04 cos 82

R R

For a point inside the circle we get

(220)

222 2 2 2 2 2 2

2 2 2 2

22 2

cos sin

( cos 2 cos ) sin

2 cos

1 5for example R= we get ds= cos cos

2 4 4

dx R dt R tdt dy R tdt

dx dt R R t RR t dy R dt t

ds dx dy R R RR tdt

RR R R tdt R tdt

End of cycloids

FW physics 10192011 843 PM page 17 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

17

Moving plank on rollers To illustrate the point about the difficulty of discerning the correct direction of fs in the

context of rolling motion let us look at the following problem

A plank with a mass of M=600 kg rides on top of two identical solid cylindrical rollers

that have radii of R=500 cm and masses m=200 kg The plank is pulled by a constant

horizontal force of magnitude F=600 N (see picture) The cylinders roll without slipping

on a flat surface and no slipping occurs between the rollers and the plank Find the

acceleration of the plank and the rollers find the frictional forces

The reason why the plank moves twice the distance as the center of mass of the rollers is

that with respect to the ground the top of each roller moves with twice the speed

v 2top R and 2 2top tops R a R

The forces on the rollers ft-fb=mar and (ft+ fb )R=Iα=05 mR2 α=05Rmar

add the last two equations to get

2 ft=15mar=3ar ft=0752ar=15ar

fb=-025 mar=

from the plank equation we get -3ar+6=6a=12 ar ar=0400 a=0800

fb=-025 204=-0200N

ft=0600N

The interesting point here is that we have chosen fb in the wrong direction The friction

force fb points to the right

forces on the plank -2ft +F= Ma

2Nt-2Mg=0 ft is the friction force at the top both on the plank and on the rollers just in

opposite directions

F=600 N M=6kg

As each roller moves forward by the distance s the plank moves twice as far This

means that the acceleration of each roller is frac12 the acceleration of the plank

fb

fb

ft

ft

mg

mg

N

t

N

t

FW physics 10192011 843 PM page 18 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

18

Moment of inertia tensor

For your personal enjoyment only it will not be required in a test

For a more general derivation of the moment of inertia let us proceed without any

assumptions as to the kind of rotation We assume only that we are dealing with a rigid

body

(147)

i iv v

i i i

i

i i i i i i i i

i i

x y z

L r m r

L r m r m r r r r

r x y z

We need to calculate the three components for the vector L

(148)

2

Omitting the running index i in the x and components

x i x

i

L m x

2 2

xy z x x 2 2

2 2

y z i x y z

i

y i y

i

y z m y z x y z

L m x y

2

x yz y x y 2 2

2 2 2

z i y x z

i

z i z

i

z m x z y x z

L m x y z

x y zz x y z 2 2

i z x y

i

m x y z x y

This 3 by 3 ldquomatrixrdquo is called the moment of inertia tensor It reveals among other things

that the angular momentum L is not necessarily parallel to the angular velocity vector ω

(149)

2 2

2 2

2 2

2 2 2 2

1 etc

etc

x x

y i y

i

z z

xx i i i

i body

xy i i i

i body

L y z xy xz

L m yx x z yz

L zx zy x y

I I m y z y z dm

I m x y xydm

If the symmetry axes of a uniform symmetric body coincide with the coordinate axes the

mixed terms xy zx and so on vanish and the tensor has a simple diagonal form These

axes are then called principal axes

Remarkably it is always possible for a body of any shape and mass distribution to find a

set of three orthogonal axes such that the off-diagonal elements (the so-called products of

inertia) vanish

FW physics 10192011 843 PM page 19 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

19

(150)

0 0

0 0

0 0

x xx x

y yy y

z zz z

L I

L I

L I

For a uniform solid sphere or a shell any three orthogonal axes are principle axes and

we have

(151)

2

2

1

2 2 2

2 2 2 2

4 5 5 3 3

2 1 2 1

4

3 2 4

2 4 2 44

3 3 5 3

xx yy zz

zz

sphere

xx yy zz

spherer

R

R

I I I

I x y dm dm dV r dr

I I I I x y z dm dm r dr

I r dr R R M R R

For the inner radius equal 0 we get 22

5I MR which is the moment of inertia of a solid

sphere rotating about any diameter

For a thin spherical shell of mass m we get (no integration necessary)

(152) 2

2 2 2 2

2

2 2 2

3 2 4 4

2 24

3 3

xx yy zz

sphereR

mI I I I x y z dm m R

R

I R R mR

To calculate the moment of inertia of a thin hoop for a rotation around its center

The hoop has a line density ρ every point of the hoop has the same distance R from the

center

2

2 2 2 3 2

0

2I r dm r Rd R Rd R MR

since M=2πRρ

Calculating various moments of inertia

If we want to find the moment of inertia I for a rotating solid cylinder we have to choose

a cylindrical volume element which has a variable distance r (from 0 to R) from the axis

of rotation

(21) with 0 r R 0 2 and 0 z hdm rdrd dz

We can integrate over θ and z to give a volume element of

(22) 2dm rhdr

to arrive at the simple integral

FW physics 10192011 843 PM page 20 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

20

Ri ri

D A C

mi

(23)

43

2

0

2

M2 2 and with =

4 V

1

2

RR M

I h r dr hR h

I MR

The same can be achieved more readily with one simple integral if one chooses as

volume element a thin cylindrical disk of height h and thickness dr

(24)

42 2 3 2

0

2

12 2 2

4 2

with M= R

RR

I r dm r rhdr h r dr h MR

h

If we integrate r from an inner radius A to an outer radius B we get the moment of

inertia for a cylindrical shell with thickness B-A

(25)

4 43

2 2

2 2

( ) M2 2 and with =

4 V

1( )

2

B

A

B A MI h r dr h

B A h

I M B A

To calculate the moment of inertia of a rectangular stick of width A height H and

length L we put the axis of rotation at one end of the stick at the edge of the width A

The volume element

(26) dm= Hmiddotdxmiddotdy 0 A and 0y x L

(27) 3 3

2 2

0 0

( )3 3

L A

x y

L AI H x y dxdy H A L

The volume of the stick is H∙L∙A and its density is MH∙L∙A and therefore

(28) 2 2( )

3

M L AI

which is the moment of inertia for a rotation around one end (corner) To get

the moment of inertia around the center of mass we use the parallel axis

theorem

(29) 2 21( )

12cmI M L A

Proof of the parallel axis theorem the axes of rotation

are at A and C(center of mass) perpendicular to the

plane of writing

FW physics 10192011 843 PM page 21 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

21

(210)

2

2 2 2

2 2

2

2

2

0 because 0 why

A i i i i i i i

i i i i i

cm i i

I m R m D r m D r r D

D m m r D m r

MD I m r

Thus

(211) 2 where D is the distance between the two axesA cmI MD I

For a circular cone rotating around the z

axis we choose the volume element in

cylindrical coordinates

(212)

2 2 2

dV=rmiddotdrmiddotd middotdz 0 r 0

hr h z= and dz= and x +y =r

R R

R z h

R hdr

r z

(213)2

2 2

0 0 0

R h

cm

r z

I r dV r rdrdz d

As z depends on r we must first change the

boundaries of z which varies from hrR to h

and integrate it before we integrate over r

The integration over θ is simply 2π

(214)

542 3

0

0 0

4 4 4 42

2

2 2 24 5

32 2

14 5 20 10 10

3

R h R

R

cm

z hr R

hr hr hrI dzr rdr h r dr

R R

hR hR R R Mh h MR

R h

r

h

R

x

z

y

FW physics 10192011 843 PM page 22 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

22

Cycloids optional The speed of the center of mass is equal to the speed of a point on the rim (rolling

without slipping at constant velocity) with respect to the center For the x and y

coordinates of a point on the rim we get therefore

(215) r R

(216) 0

sin( ) and cos( )x t x R t R t y t R R t

If we start our motion with the point on the rim at x=0 and y=0 we get

(217) sin and cosx t R t R t y t R R t

For a point inside of the circle (RrsquoltR) the coordinates are given by

(218) sin and cosx t R t R t y t R R t

We see that the horizontal distance traveled remains equal to 2πR (not 2πRrsquo)

Assume for example that Rrsquo=05R

x0

In the time t the point on the rim of the circle (starting at the contact point

of the circle with the x-axis moves through the angle ωt to the left while

at the same time the center of the circle moves to the right by ωtR A point

inside the circle travels a shorter distance

x

FW physics 10192011 843 PM page 23 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

23

(219)

2 22 2 2 2

2 2

2

2 2

0

2 2

2

0 0

cos sin

(1 cos 2cos ) sin

2 1 cos 2 1 cos

1 12 1 cos sin 1 cos 2 sin 1 cos

2 2 2

2 2sin 2 sin 2 2 sin2 2

dx R dt R tdt dy R tdt

dx R dt t t dy R dt t

ds dx dy R t dt Rd

xS R d x x x

S R d R d R zdz

2

04 cos 82

R R

For a point inside the circle we get

(220)

222 2 2 2 2 2 2

2 2 2 2

22 2

cos sin

( cos 2 cos ) sin

2 cos

1 5for example R= we get ds= cos cos

2 4 4

dx R dt R tdt dy R tdt

dx dt R R t RR t dy R dt t

ds dx dy R R RR tdt

RR R R tdt R tdt

End of cycloids

FW physics 10192011 843 PM page 18 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

18

Moment of inertia tensor

For your personal enjoyment only it will not be required in a test

For a more general derivation of the moment of inertia let us proceed without any

assumptions as to the kind of rotation We assume only that we are dealing with a rigid

body

(147)

i iv v

i i i

i

i i i i i i i i

i i

x y z

L r m r

L r m r m r r r r

r x y z

We need to calculate the three components for the vector L

(148)

2

Omitting the running index i in the x and components

x i x

i

L m x

2 2

xy z x x 2 2

2 2

y z i x y z

i

y i y

i

y z m y z x y z

L m x y

2

x yz y x y 2 2

2 2 2

z i y x z

i

z i z

i

z m x z y x z

L m x y z

x y zz x y z 2 2

i z x y

i

m x y z x y

This 3 by 3 ldquomatrixrdquo is called the moment of inertia tensor It reveals among other things

that the angular momentum L is not necessarily parallel to the angular velocity vector ω

(149)

2 2

2 2

2 2

2 2 2 2

1 etc

etc

x x

y i y

i

z z

xx i i i

i body

xy i i i

i body

L y z xy xz

L m yx x z yz

L zx zy x y

I I m y z y z dm

I m x y xydm

If the symmetry axes of a uniform symmetric body coincide with the coordinate axes the

mixed terms xy zx and so on vanish and the tensor has a simple diagonal form These

axes are then called principal axes

Remarkably it is always possible for a body of any shape and mass distribution to find a

set of three orthogonal axes such that the off-diagonal elements (the so-called products of

inertia) vanish

FW physics 10192011 843 PM page 19 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

19

(150)

0 0

0 0

0 0

x xx x

y yy y

z zz z

L I

L I

L I

For a uniform solid sphere or a shell any three orthogonal axes are principle axes and

we have

(151)

2

2

1

2 2 2

2 2 2 2

4 5 5 3 3

2 1 2 1

4

3 2 4

2 4 2 44

3 3 5 3

xx yy zz

zz

sphere

xx yy zz

spherer

R

R

I I I

I x y dm dm dV r dr

I I I I x y z dm dm r dr

I r dr R R M R R

For the inner radius equal 0 we get 22

5I MR which is the moment of inertia of a solid

sphere rotating about any diameter

For a thin spherical shell of mass m we get (no integration necessary)

(152) 2

2 2 2 2

2

2 2 2

3 2 4 4

2 24

3 3

xx yy zz

sphereR

mI I I I x y z dm m R

R

I R R mR

To calculate the moment of inertia of a thin hoop for a rotation around its center

The hoop has a line density ρ every point of the hoop has the same distance R from the

center

2

2 2 2 3 2

0

2I r dm r Rd R Rd R MR

since M=2πRρ

Calculating various moments of inertia

If we want to find the moment of inertia I for a rotating solid cylinder we have to choose

a cylindrical volume element which has a variable distance r (from 0 to R) from the axis

of rotation

(21) with 0 r R 0 2 and 0 z hdm rdrd dz

We can integrate over θ and z to give a volume element of

(22) 2dm rhdr

to arrive at the simple integral

FW physics 10192011 843 PM page 20 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

20

Ri ri

D A C

mi

(23)

43

2

0

2

M2 2 and with =

4 V

1

2

RR M

I h r dr hR h

I MR

The same can be achieved more readily with one simple integral if one chooses as

volume element a thin cylindrical disk of height h and thickness dr

(24)

42 2 3 2

0

2

12 2 2

4 2

with M= R

RR

I r dm r rhdr h r dr h MR

h

If we integrate r from an inner radius A to an outer radius B we get the moment of

inertia for a cylindrical shell with thickness B-A

(25)

4 43

2 2

2 2

( ) M2 2 and with =

4 V

1( )

2

B

A

B A MI h r dr h

B A h

I M B A

To calculate the moment of inertia of a rectangular stick of width A height H and

length L we put the axis of rotation at one end of the stick at the edge of the width A

The volume element

(26) dm= Hmiddotdxmiddotdy 0 A and 0y x L

(27) 3 3

2 2

0 0

( )3 3

L A

x y

L AI H x y dxdy H A L

The volume of the stick is H∙L∙A and its density is MH∙L∙A and therefore

(28) 2 2( )

3

M L AI

which is the moment of inertia for a rotation around one end (corner) To get

the moment of inertia around the center of mass we use the parallel axis

theorem

(29) 2 21( )

12cmI M L A

Proof of the parallel axis theorem the axes of rotation

are at A and C(center of mass) perpendicular to the

plane of writing

FW physics 10192011 843 PM page 21 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

21

(210)

2

2 2 2

2 2

2

2

2

0 because 0 why

A i i i i i i i

i i i i i

cm i i

I m R m D r m D r r D

D m m r D m r

MD I m r

Thus

(211) 2 where D is the distance between the two axesA cmI MD I

For a circular cone rotating around the z

axis we choose the volume element in

cylindrical coordinates

(212)

2 2 2

dV=rmiddotdrmiddotd middotdz 0 r 0

hr h z= and dz= and x +y =r

R R

R z h

R hdr

r z

(213)2

2 2

0 0 0

R h

cm

r z

I r dV r rdrdz d

As z depends on r we must first change the

boundaries of z which varies from hrR to h

and integrate it before we integrate over r

The integration over θ is simply 2π

(214)

542 3

0

0 0

4 4 4 42

2

2 2 24 5

32 2

14 5 20 10 10

3

R h R

R

cm

z hr R

hr hr hrI dzr rdr h r dr

R R

hR hR R R Mh h MR

R h

r

h

R

x

z

y

FW physics 10192011 843 PM page 22 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

22

Cycloids optional The speed of the center of mass is equal to the speed of a point on the rim (rolling

without slipping at constant velocity) with respect to the center For the x and y

coordinates of a point on the rim we get therefore

(215) r R

(216) 0

sin( ) and cos( )x t x R t R t y t R R t

If we start our motion with the point on the rim at x=0 and y=0 we get

(217) sin and cosx t R t R t y t R R t

For a point inside of the circle (RrsquoltR) the coordinates are given by

(218) sin and cosx t R t R t y t R R t

We see that the horizontal distance traveled remains equal to 2πR (not 2πRrsquo)

Assume for example that Rrsquo=05R

x0

In the time t the point on the rim of the circle (starting at the contact point

of the circle with the x-axis moves through the angle ωt to the left while

at the same time the center of the circle moves to the right by ωtR A point

inside the circle travels a shorter distance

x

FW physics 10192011 843 PM page 23 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

23

(219)

2 22 2 2 2

2 2

2

2 2

0

2 2

2

0 0

cos sin

(1 cos 2cos ) sin

2 1 cos 2 1 cos

1 12 1 cos sin 1 cos 2 sin 1 cos

2 2 2

2 2sin 2 sin 2 2 sin2 2

dx R dt R tdt dy R tdt

dx R dt t t dy R dt t

ds dx dy R t dt Rd

xS R d x x x

S R d R d R zdz

2

04 cos 82

R R

For a point inside the circle we get

(220)

222 2 2 2 2 2 2

2 2 2 2

22 2

cos sin

( cos 2 cos ) sin

2 cos

1 5for example R= we get ds= cos cos

2 4 4

dx R dt R tdt dy R tdt

dx dt R R t RR t dy R dt t

ds dx dy R R RR tdt

RR R R tdt R tdt

End of cycloids

FW physics 10192011 843 PM page 19 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

19

(150)

0 0

0 0

0 0

x xx x

y yy y

z zz z

L I

L I

L I

For a uniform solid sphere or a shell any three orthogonal axes are principle axes and

we have

(151)

2

2

1

2 2 2

2 2 2 2

4 5 5 3 3

2 1 2 1

4

3 2 4

2 4 2 44

3 3 5 3

xx yy zz

zz

sphere

xx yy zz

spherer

R

R

I I I

I x y dm dm dV r dr

I I I I x y z dm dm r dr

I r dr R R M R R

For the inner radius equal 0 we get 22

5I MR which is the moment of inertia of a solid

sphere rotating about any diameter

For a thin spherical shell of mass m we get (no integration necessary)

(152) 2

2 2 2 2

2

2 2 2

3 2 4 4

2 24

3 3

xx yy zz

sphereR

mI I I I x y z dm m R

R

I R R mR

To calculate the moment of inertia of a thin hoop for a rotation around its center

The hoop has a line density ρ every point of the hoop has the same distance R from the

center

2

2 2 2 3 2

0

2I r dm r Rd R Rd R MR

since M=2πRρ

Calculating various moments of inertia

If we want to find the moment of inertia I for a rotating solid cylinder we have to choose

a cylindrical volume element which has a variable distance r (from 0 to R) from the axis

of rotation

(21) with 0 r R 0 2 and 0 z hdm rdrd dz

We can integrate over θ and z to give a volume element of

(22) 2dm rhdr

to arrive at the simple integral

FW physics 10192011 843 PM page 20 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

20

Ri ri

D A C

mi

(23)

43

2

0

2

M2 2 and with =

4 V

1

2

RR M

I h r dr hR h

I MR

The same can be achieved more readily with one simple integral if one chooses as

volume element a thin cylindrical disk of height h and thickness dr

(24)

42 2 3 2

0

2

12 2 2

4 2

with M= R

RR

I r dm r rhdr h r dr h MR

h

If we integrate r from an inner radius A to an outer radius B we get the moment of

inertia for a cylindrical shell with thickness B-A

(25)

4 43

2 2

2 2

( ) M2 2 and with =

4 V

1( )

2

B

A

B A MI h r dr h

B A h

I M B A

To calculate the moment of inertia of a rectangular stick of width A height H and

length L we put the axis of rotation at one end of the stick at the edge of the width A

The volume element

(26) dm= Hmiddotdxmiddotdy 0 A and 0y x L

(27) 3 3

2 2

0 0

( )3 3

L A

x y

L AI H x y dxdy H A L

The volume of the stick is H∙L∙A and its density is MH∙L∙A and therefore

(28) 2 2( )

3

M L AI

which is the moment of inertia for a rotation around one end (corner) To get

the moment of inertia around the center of mass we use the parallel axis

theorem

(29) 2 21( )

12cmI M L A

Proof of the parallel axis theorem the axes of rotation

are at A and C(center of mass) perpendicular to the

plane of writing

FW physics 10192011 843 PM page 21 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

21

(210)

2

2 2 2

2 2

2

2

2

0 because 0 why

A i i i i i i i

i i i i i

cm i i

I m R m D r m D r r D

D m m r D m r

MD I m r

Thus

(211) 2 where D is the distance between the two axesA cmI MD I

For a circular cone rotating around the z

axis we choose the volume element in

cylindrical coordinates

(212)

2 2 2

dV=rmiddotdrmiddotd middotdz 0 r 0

hr h z= and dz= and x +y =r

R R

R z h

R hdr

r z

(213)2

2 2

0 0 0

R h

cm

r z

I r dV r rdrdz d

As z depends on r we must first change the

boundaries of z which varies from hrR to h

and integrate it before we integrate over r

The integration over θ is simply 2π

(214)

542 3

0

0 0

4 4 4 42

2

2 2 24 5

32 2

14 5 20 10 10

3

R h R

R

cm

z hr R

hr hr hrI dzr rdr h r dr

R R

hR hR R R Mh h MR

R h

r

h

R

x

z

y

FW physics 10192011 843 PM page 22 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

22

Cycloids optional The speed of the center of mass is equal to the speed of a point on the rim (rolling

without slipping at constant velocity) with respect to the center For the x and y

coordinates of a point on the rim we get therefore

(215) r R

(216) 0

sin( ) and cos( )x t x R t R t y t R R t

If we start our motion with the point on the rim at x=0 and y=0 we get

(217) sin and cosx t R t R t y t R R t

For a point inside of the circle (RrsquoltR) the coordinates are given by

(218) sin and cosx t R t R t y t R R t

We see that the horizontal distance traveled remains equal to 2πR (not 2πRrsquo)

Assume for example that Rrsquo=05R

x0

In the time t the point on the rim of the circle (starting at the contact point

of the circle with the x-axis moves through the angle ωt to the left while

at the same time the center of the circle moves to the right by ωtR A point

inside the circle travels a shorter distance

x

FW physics 10192011 843 PM page 23 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

23

(219)

2 22 2 2 2

2 2

2

2 2

0

2 2

2

0 0

cos sin

(1 cos 2cos ) sin

2 1 cos 2 1 cos

1 12 1 cos sin 1 cos 2 sin 1 cos

2 2 2

2 2sin 2 sin 2 2 sin2 2

dx R dt R tdt dy R tdt

dx R dt t t dy R dt t

ds dx dy R t dt Rd

xS R d x x x

S R d R d R zdz

2

04 cos 82

R R

For a point inside the circle we get

(220)

222 2 2 2 2 2 2

2 2 2 2

22 2

cos sin

( cos 2 cos ) sin

2 cos

1 5for example R= we get ds= cos cos

2 4 4

dx R dt R tdt dy R tdt

dx dt R R t RR t dy R dt t

ds dx dy R R RR tdt

RR R R tdt R tdt

End of cycloids

FW physics 10192011 843 PM page 20 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

20

Ri ri

D A C

mi

(23)

43

2

0

2

M2 2 and with =

4 V

1

2

RR M

I h r dr hR h

I MR

The same can be achieved more readily with one simple integral if one chooses as

volume element a thin cylindrical disk of height h and thickness dr

(24)

42 2 3 2

0

2

12 2 2

4 2

with M= R

RR

I r dm r rhdr h r dr h MR

h

If we integrate r from an inner radius A to an outer radius B we get the moment of

inertia for a cylindrical shell with thickness B-A

(25)

4 43

2 2

2 2

( ) M2 2 and with =

4 V

1( )

2

B

A

B A MI h r dr h

B A h

I M B A

To calculate the moment of inertia of a rectangular stick of width A height H and

length L we put the axis of rotation at one end of the stick at the edge of the width A

The volume element

(26) dm= Hmiddotdxmiddotdy 0 A and 0y x L

(27) 3 3

2 2

0 0

( )3 3

L A

x y

L AI H x y dxdy H A L

The volume of the stick is H∙L∙A and its density is MH∙L∙A and therefore

(28) 2 2( )

3

M L AI

which is the moment of inertia for a rotation around one end (corner) To get

the moment of inertia around the center of mass we use the parallel axis

theorem

(29) 2 21( )

12cmI M L A

Proof of the parallel axis theorem the axes of rotation

are at A and C(center of mass) perpendicular to the

plane of writing

FW physics 10192011 843 PM page 21 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

21

(210)

2

2 2 2

2 2

2

2

2

0 because 0 why

A i i i i i i i

i i i i i

cm i i

I m R m D r m D r r D

D m m r D m r

MD I m r

Thus

(211) 2 where D is the distance between the two axesA cmI MD I

For a circular cone rotating around the z

axis we choose the volume element in

cylindrical coordinates

(212)

2 2 2

dV=rmiddotdrmiddotd middotdz 0 r 0

hr h z= and dz= and x +y =r

R R

R z h

R hdr

r z

(213)2

2 2

0 0 0

R h

cm

r z

I r dV r rdrdz d

As z depends on r we must first change the

boundaries of z which varies from hrR to h

and integrate it before we integrate over r

The integration over θ is simply 2π

(214)

542 3

0

0 0

4 4 4 42

2

2 2 24 5

32 2

14 5 20 10 10

3

R h R

R

cm

z hr R

hr hr hrI dzr rdr h r dr

R R

hR hR R R Mh h MR

R h

r

h

R

x

z

y

FW physics 10192011 843 PM page 22 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

22

Cycloids optional The speed of the center of mass is equal to the speed of a point on the rim (rolling

without slipping at constant velocity) with respect to the center For the x and y

coordinates of a point on the rim we get therefore

(215) r R

(216) 0

sin( ) and cos( )x t x R t R t y t R R t

If we start our motion with the point on the rim at x=0 and y=0 we get

(217) sin and cosx t R t R t y t R R t

For a point inside of the circle (RrsquoltR) the coordinates are given by

(218) sin and cosx t R t R t y t R R t

We see that the horizontal distance traveled remains equal to 2πR (not 2πRrsquo)

Assume for example that Rrsquo=05R

x0

In the time t the point on the rim of the circle (starting at the contact point

of the circle with the x-axis moves through the angle ωt to the left while

at the same time the center of the circle moves to the right by ωtR A point

inside the circle travels a shorter distance

x

FW physics 10192011 843 PM page 23 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

23

(219)

2 22 2 2 2

2 2

2

2 2

0

2 2

2

0 0

cos sin

(1 cos 2cos ) sin

2 1 cos 2 1 cos

1 12 1 cos sin 1 cos 2 sin 1 cos

2 2 2

2 2sin 2 sin 2 2 sin2 2

dx R dt R tdt dy R tdt

dx R dt t t dy R dt t

ds dx dy R t dt Rd

xS R d x x x

S R d R d R zdz

2

04 cos 82

R R

For a point inside the circle we get

(220)

222 2 2 2 2 2 2

2 2 2 2

22 2

cos sin

( cos 2 cos ) sin

2 cos

1 5for example R= we get ds= cos cos

2 4 4

dx R dt R tdt dy R tdt

dx dt R R t RR t dy R dt t

ds dx dy R R RR tdt

RR R R tdt R tdt

End of cycloids

FW physics 10192011 843 PM page 21 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

21

(210)

2

2 2 2

2 2

2

2

2

0 because 0 why

A i i i i i i i

i i i i i

cm i i

I m R m D r m D r r D

D m m r D m r

MD I m r

Thus

(211) 2 where D is the distance between the two axesA cmI MD I

For a circular cone rotating around the z

axis we choose the volume element in

cylindrical coordinates

(212)

2 2 2

dV=rmiddotdrmiddotd middotdz 0 r 0

hr h z= and dz= and x +y =r

R R

R z h

R hdr

r z

(213)2

2 2

0 0 0

R h

cm

r z

I r dV r rdrdz d

As z depends on r we must first change the

boundaries of z which varies from hrR to h

and integrate it before we integrate over r

The integration over θ is simply 2π

(214)

542 3

0

0 0

4 4 4 42

2

2 2 24 5

32 2

14 5 20 10 10

3

R h R

R

cm

z hr R

hr hr hrI dzr rdr h r dr

R R

hR hR R R Mh h MR

R h

r

h

R

x

z

y

FW physics 10192011 843 PM page 22 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

22

Cycloids optional The speed of the center of mass is equal to the speed of a point on the rim (rolling

without slipping at constant velocity) with respect to the center For the x and y

coordinates of a point on the rim we get therefore

(215) r R

(216) 0

sin( ) and cos( )x t x R t R t y t R R t

If we start our motion with the point on the rim at x=0 and y=0 we get

(217) sin and cosx t R t R t y t R R t

For a point inside of the circle (RrsquoltR) the coordinates are given by

(218) sin and cosx t R t R t y t R R t

We see that the horizontal distance traveled remains equal to 2πR (not 2πRrsquo)

Assume for example that Rrsquo=05R

x0

In the time t the point on the rim of the circle (starting at the contact point

of the circle with the x-axis moves through the angle ωt to the left while

at the same time the center of the circle moves to the right by ωtR A point

inside the circle travels a shorter distance

x

FW physics 10192011 843 PM page 23 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

23

(219)

2 22 2 2 2

2 2

2

2 2

0

2 2

2

0 0

cos sin

(1 cos 2cos ) sin

2 1 cos 2 1 cos

1 12 1 cos sin 1 cos 2 sin 1 cos

2 2 2

2 2sin 2 sin 2 2 sin2 2

dx R dt R tdt dy R tdt

dx R dt t t dy R dt t

ds dx dy R t dt Rd

xS R d x x x

S R d R d R zdz

2

04 cos 82

R R

For a point inside the circle we get

(220)

222 2 2 2 2 2 2

2 2 2 2

22 2

cos sin

( cos 2 cos ) sin

2 cos

1 5for example R= we get ds= cos cos

2 4 4

dx R dt R tdt dy R tdt

dx dt R R t RR t dy R dt t

ds dx dy R R RR tdt

RR R R tdt R tdt

End of cycloids

FW physics 10192011 843 PM page 22 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

22

Cycloids optional The speed of the center of mass is equal to the speed of a point on the rim (rolling

without slipping at constant velocity) with respect to the center For the x and y

coordinates of a point on the rim we get therefore

(215) r R

(216) 0

sin( ) and cos( )x t x R t R t y t R R t

If we start our motion with the point on the rim at x=0 and y=0 we get

(217) sin and cosx t R t R t y t R R t

For a point inside of the circle (RrsquoltR) the coordinates are given by

(218) sin and cosx t R t R t y t R R t

We see that the horizontal distance traveled remains equal to 2πR (not 2πRrsquo)

Assume for example that Rrsquo=05R

x0

In the time t the point on the rim of the circle (starting at the contact point

of the circle with the x-axis moves through the angle ωt to the left while

at the same time the center of the circle moves to the right by ωtR A point

inside the circle travels a shorter distance

x

FW physics 10192011 843 PM page 23 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

23

(219)

2 22 2 2 2

2 2

2

2 2

0

2 2

2

0 0

cos sin

(1 cos 2cos ) sin

2 1 cos 2 1 cos

1 12 1 cos sin 1 cos 2 sin 1 cos

2 2 2

2 2sin 2 sin 2 2 sin2 2

dx R dt R tdt dy R tdt

dx R dt t t dy R dt t

ds dx dy R t dt Rd

xS R d x x x

S R d R d R zdz

2

04 cos 82

R R

For a point inside the circle we get

(220)

222 2 2 2 2 2 2

2 2 2 2

22 2

cos sin

( cos 2 cos ) sin

2 cos

1 5for example R= we get ds= cos cos

2 4 4

dx R dt R tdt dy R tdt

dx dt R R t RR t dy R dt t

ds dx dy R R RR tdt

RR R R tdt R tdt

End of cycloids

FW physics 10192011 843 PM page 23 of 23

Cphysics130 lecture-Giancolich 10 11 torques and work rolling L top tens Idocx

23

(219)

2 22 2 2 2

2 2

2

2 2

0

2 2

2

0 0

cos sin

(1 cos 2cos ) sin

2 1 cos 2 1 cos

1 12 1 cos sin 1 cos 2 sin 1 cos

2 2 2

2 2sin 2 sin 2 2 sin2 2

dx R dt R tdt dy R tdt

dx R dt t t dy R dt t

ds dx dy R t dt Rd

xS R d x x x

S R d R d R zdz

2

04 cos 82

R R

For a point inside the circle we get

(220)

222 2 2 2 2 2 2

2 2 2 2

22 2

cos sin

( cos 2 cos ) sin

2 cos

1 5for example R= we get ds= cos cos

2 4 4

dx R dt R tdt dy R tdt

dx dt R R t RR t dy R dt t

ds dx dy R R RR tdt

RR R R tdt R tdt

End of cycloids