MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics

8.01

Problem Solving Session 7 Fixed Axis Rotational Dynamics and Static Equilibrium

W9D3-1 Group Problem Atwood Machine Solutions

A pulley of mass

mp , radius R , and moment of inertia about the center of mass Icm , is suspended from a ceiling. An inextensible string of negligible mass is wrapped around the pulley and attached on one end to an object of mass m1 and on the other end to an object of mass with m1 > m2 . At time t = 0 , the objects are released from rest. How long

does it take the objects to move a distance d? If Icm =12mpR

2 , what does your answer

reduce to?

Solution: We will use Newton’s Second Law, and the Rotational Law to find the acceleration of all the objects in the system. We can then use kinematics to calculate the time it takes for each object to move a certain distance. As a strategy we will answer the following questions.

a) What are the free body force diagrams on the pulley and the two objects. Is the tension in the string constant?

b) Write down Newton’s Second Law for the pulley and the two objects.

c) Write down the rotational equation of motion for the pulley.

d) What is the constraint condition between the magnitude of the angular

acceleration of the pulley and the magnitude of the acceleration of either object.

e) Find the direction and magnitude of the acceleration of the two objects.

f) How long does it take the objects to move a distance d?

g) If Icm =12mpR

2 , what does your answer reduce to?

a) Free body diagrams:

The two tensions in the string each produce a torque on the pulley in opposite directions. The pulley has rotational inertial: the moment of inertia is not zero. Hence if the pulley undergoes angular acceleration there must be a net non-zero torque. (Note the pulley is not massless. Earlier in the semester we always assumed that the pulley is massless to avoid the complication we are considering here.) The two tensions in the string each produce a torque on the pulley in opposite directions. The net torque is proportional to the difference in tensions in the string on the two sides of the pulley. Therefore the tensions cannot be equal when the pulley undergoes angular acceleration. b) Equation of linear motion for pulley (Pulley is at rest):

T3 ! T1 ! T2 ! mpg = 0 . (1) Equation of linear motion for m1 :

m1g ! T1 = m1a . (2) Equation of linear motion for m2 :

T2 ! m2g = m2a . (3)

c) Equation of rotational motion for pulley:

R T1 ! T2( ) = Icm" . (4) d) For pure rotation,

a = R! . (5) e) From (2) and (3):

T1 ! T2 = m1 ! m2( )g ! (m1 + m2 )a . (6) From (4) and Eq (5) along with the moment of inertia:

T1 ! T2 =Icm"R

=IcmaR2

. (7)

Equating the expressions for T T2 1! in (6) and (7) we obtain:

IcmaR2

= m1 ! m2( )g ! (m1 + m2 )a , (8)

which we can solve for the acceleration.

a =m1 ! m2( )g

IcmR2

+ (m1 + m2 )"#$

%&'

. (9)

f) d at=12

2 and the results from part e) =>

t = 2da

=2d Icm

R2+ (m1 + m2 )

!"#

$%&

m1 ' m2( )g . (10)

If Icm =12mpR

2 , then

t = 2da

=d mp + 2(m1 + m2 )( )

m1 ! m2( )g . (11)

IC-W09D3-2 Group Problem Person on a Ladder Solution A person of mass pm is standing on a rung, one third of the way up a ladder of length d . The mass of the ladder is lm , uniformly distributed. The ladder is initially inclined at an angle ! with respect to the horizontal. Assume that there is no friction between the ladder and the wall but that there is friction between the base of the ladder and the floor with a coefficient of static friction sµ . In this problem you will try to find the minimum coefficient of friction between the ladder and the floor so that the person and ladder do not slip.

a) Draw a free-body force diagrams for the person and the ladder separately, explicitly showing where the forces act remembering that the gravitational force acts at the center of mass of the ladder. Include unit vectors on the diagram. Identify any possible third law interaction pairs. Write down the Newton’s Second Law equations for both person and ladder.

b) Determine which point to calculate torque about. Remember that if you choose a

point S where a force acts then that force has zero torque about that point. When calculating torque about a chosen point, you can always formally calculate the cross product

!!S = !rS ,F "

!F . You may also argue geometrically if the given

information of the problem makes it easier to compute the moment arm of the force about your chosen point or the perpendicular component of the force with respect to a line drawn from your chosen point to the point where the force acts. You still need to determine the direction of the torque.

c) Determine the minimum coefficient of friction between the ladder and the floor so

that the person and ladder do not slip. Solution: We shall apply the two conditions for static equilibrium on the ladder,

(1) The sum of the forces acting on the rigid body is zero, total 1 2= + + ! ! ! =F F F 0

!! ! !. (12)

(2) The vector sum of the torques about any point S on a rigid body is zero,

total

,1 ,2S S S! ! != + + " " " = 0!! ! ! . (13)

a) Draw a free-body force diagrams for the person and the ladder separately,

explicitly showing where the forces act remembering that the gravitational force acts at the center of mass of the ladder. Include unit vectors on the diagram. Identify any possible third law interaction pairs.

Consider the forces acting on the person. The gravitational force acts at the center of mass of the person and the force on the person due to the contact between the person and the ladder is an upward normal force, denoted lpN

! in the diagrams in these solutions. The

force diagram on the person is shown below.

The equation for static equilibrium of forces on the person is

ˆ : 0lp pN m g! =j (14) The normal force plN

! that the person exerts on the ladder is part of a third law interaction

pair. From Newton’s Third Law, lp pl= !N N

! !. (15)

Denote the magnitude of plN

! by plN . Equation (14) then becomes

pl pN m g= . (16)

There are four forces acting on the ladder. The person exerts a downward contact force on the ladder at the point of contact on the rung of the ladder, plN

!, a distance / 3d from

the contact point with the floor. The gravitational force between the earth and the ladder,

lm g!

, acts at the center of mass of the ladder, which is a distance / 2d from the contact point with the floor since the mass of the ladder is assumed to be uniformly distributed. At the point where the ladder is in contact with the wall, the contact force of the wall with ladder wlN

! is only perpendicular because we have assumed that the contact surface is

frictionless. At the point where the ladder is contact with the floor, the contact force has both a vertical component, the normal force glN

! and a horizontal component pointing

toward the wall, the static friction, sf!

. The force diagram is shown in the figure below

Key point: The magnitude of the static friction force depends on the other forces and where they act. As the person walks up the ladder, the normal force due to the person,

plN!

, changes position and hence the friction force will change in magnitude possibly causing the ladder to slip. The equations for static equilibrium of forces on the ladder (using Equation (16) for the magnitude of the normal force of the person on the ladder) becomes

ˆ : 0gl p lN m g m g! ! =j (17)

ˆ : 0s wlf N! =i . (18) Solve Equation. (17) for the upward normal force of the ground on the ladder, which has magnitude

Ngl = mpg + ml g . (19)

We can use Equation (18) to find a relationship between the friction force of the ground on the ladder and the normal force of the wall on the ladder,

s wlf N= . (20)

b) Determine which point to calculate torque about. Remember that if you choose a point S where a force acts then that force has zero torque about that point. When calculating torque about a chosen point, you can always formally calculate the cross product

!!S = !rS ,F "

!F . You may also argue geometrically if the given

information of the problem makes it easier to compute the moment arm of the force about your chosen point or the perpendicular component of the force with respect to a line drawn from your chosen point to the point where the force acts. You still need to determine the direction of the torque.

Because the sum of the forces are zero, the torque about any point will be zero so we shall calculate the torque about several points: the point of contact between the ladder and the ground, the point of contact between the wall and the ladder, and the center of mass of the ladder.

a) Torque about the contact point between the ladder and the ground The torque equation about the contact point S1 between the ladder and the floor is given by

!!S1

= !rS1 , p "!N pl +

!rS1 ,cm " ml!g + !rS1 ,w "

!Nwl =

!0 . (21)

We show the relevant vectors in the figures below.

We now explicitly write out the vectors from our choice of point to where the forces are acting:

!rS1 , p =

d3

cos! i + d3

sin! j , (22)

!rS1 ,cm =

d2

cos! i + d2

sin! j , (23)

!rS1 ,w = d cos! i + d sin! j . (24)

The forces are

!N pl = !mpgj , (25)

!Nwl = !Nwl i , (26)

ml

!g = !ml gj . (27) Then the torque about 1S is

!0 =!!S1

=d3

cos" i + d3

sin" j#$%

&'() *mpgj+ d

2cos" i + d

2sin" j

#$%

&'() *ml gj

+ d cos" i + d sin" j( ) ) *Nwl i. (28)

Calculating the cross products we get

!!S1

= "d3

cos#mpg k "d2

cos#ml g k + d sin#Nwlk = 0 . (29)

We can solve this last equation for the magnitude of the normal force of the wall on the ladder

Nwl = gcotan!

mp

3+

ml

2

"

#$

%

&' . (30)

We can now substitute Equation (30) into Equation (20) and solve for the magnitude of the friction force

( )2 1s 2 1(1/ 3) cos (1/ 2) cos

cot / 3 / 2sin

m g m gf g m m! ! !!

+= = + . (31)

c) Determine the minimum coefficient of friction between the ladder and the floor so that the person and ladder do not slip.

The minimum coefficient of friction between the ladder and the floor so that the person and ladder do not slip is given by the condition that

fs = µs Ngl . (32)

Substituting Equation (19) into Equation (32) gives

fs = µs(mpg + ml g) . (33)

We can now equate the expressions in Equations (33) and (31) and solve for the minimum coefficient of static friction such that the ladder just starts to slip,

µs =

mp / 3+ ml / 2

mp + ml

cotan! . (34)

Alternative Choices for Calculating the Torque: Torque about the center of mass: Let’s compute the torque about the center of mass cm . Then

!!cm = !rcm,g " (

!Ngl +

!fgl ) +

!rcm, p "!N pl +

!rcm,w "!Nwl =

!0 , (35)

where

!rcm,g = !

d2

cos" i ! d2

sin" j , (36)

!rcm, p = !

d6

cos" i ! d6

sin" j , (37)

!rcm,w =

d2

cos! i + d2

sin! j . (38)

Noting that

!Ngl = Ngl j , (39)

!fgl = fs i . (40)

The torque about the center of mass is then

!0 =!!cm = "

d2

cos# i " d2

sin# j$%&

'()* (Ngl j+ fs i) + "

d6

cos# i " d6

sin# j$%&

'()* "mpgj

+d2

cos# i + d2

sin# j$%&

'()* "Nwl i

.(41)

We now calculate the cross products and get

!!cm = "

d2

cos#Ngl +d2

sin# fs

$%&

'()

k +d6

cos#mpg$%&

'()

k +d2

sin#Nwl

$%&

'()

k =!0 .(42)

Recall from Eqs. (19) and (20)that gl p lN m g m g= + and s wlf N= , Eq. becomes

!!cm = "

d2

cos# mpg + ml g( ) + d2

sin#Nwl

$%&

'()

k +d6

cos#mpg$%&

'()

k +d2

sin#Nwl

$%&

'()

k =!0

(43) We can solve this equation for wlN

Nwl = cotan!g

mp

3+

ml

2

"

#$

%

&' (44)

identical to our previous result (Eq. (30)). Torque about the contact point 2S between the wall and the ladder

The torque equation about 2S is given by

!!S2

= !rS2 , p "!N pl +

!rS2 ,cm " ml!g + !rS2 ,g " (

!Ngl +

!fgl ) =

!0 . (45)

We now explicitly write out the vectors from our choice of point to where the forces are acting

!rS2 , p = !

2d3

cos" i ! 2d3

sin" j , (46)

!rS2 ,cm = !

d2

cos" i + !d2

sin" j , (47)

!rS2 ,g = !d cos" i ! d sin" j . (48)

Then the torque about 2S is

!0 =!!S2

= "2d3

cos# i " 2d3

sin# j$%&

'()* "mpgj+ "

d2

cos# i " d2

sin# j$%&

'()* "ml gj

+ ("d cos# i " d sin# j) * (Ngl j+ fs i).(49)

Calculating the cross products we get

!0 =

2d3

cos!mpgk +d2

cos!ml gk " d cos!Nglk + d sin! fsk . (50)

We need to use Eqs. (19) and (20): gl p lN m g m g= + and s wlf N= , so that Eq. (50) becomes

!0 =

2d3

cos!mpgk +d2

cos!ml gk " d cos!(mpg + ml g)k + d sin!Nwlk . (51)

We can solve this last equation for the magnitude of the normal force of the wall on the ladder, finding our same result that

Nwl = gcotan!

mp

3+

ml

2

"

#$

%

&' . (52)