Physics 201: Lecture 19, Pg 1 Lecture 19 Goals: Specify rolling motion (center of mass velocity to angular velocity Compare kinetic and rotational energies

Physics 201: Lecture 19, Pg 1

Lecture 19Goals:Goals:

• Specify rolling motion (center of mass velocity to angular velocity

• Compare kinetic and rotational energies with rolling

• Work combined force and torque problems

• Revisit vector cross product

• Introduce angular momentum


Connection with Center-of-mass motion

If an object of mass M is moving linearly at velocity VCM without rotating then its kinetic energy is

CM21

RK I

2CM2

12CM2

1 VK MI If the object is both moving linearly and rotating then

If an object with moment of inertia ICM is rotating in place about its center of mass at angular velocity then its kinetic energy is

2CM2

1T VK M


But what of Rolling Motion Consider a cylinder rolling at a constant speed. Contact point has zero velocity in the laboratory frame. Center of wheel has a velocity of VCM

Top of wheel has a velocity of 2VCM

VCMCM

2VCM


Rolling Motion

Now consider a cylinder rolling at a constant speed.

VCM CM

The cylinder is rotating about CM and its CM is moving at constant speed (VCM). Thus its total kinetic energy is given by :

2CM2

12CM2

1TOT VK MI


Motion Again consider a cylinder rolling at a constant speed.

VCM

CM

Sliding only

CM

Rotation only

vt = R

CM

2VCM

VCM

Both with

|vt| = |vCM |


Concept question

For a hoop of mass M and radius R that is rolling without slipping, which is greater, its translational or its rotational kinetic energy?

A. Translational energy is greater

B. Rotational energy is greater

C. They are equal

D. The answer depends on the radius

E. The answer depends on the mass


Example : Rolling Motion A solid cylinder is about to roll down an inclined plane.

What is its speed at the bottom of the plane ? Use Work-Energy theorem

M

h

Mv ?

Disk has radius R

M M

M

M

M

Mgh = ½ Mv2 + ½ ICM 2 and v =R

Mgh = ½ Mv2 + ½ (½ M R2 )(v/R)2 = ¾ Mv2

v = 2(gh/3)½


Question: The Loop-the-Loop … with rolling

To complete the loop-the-loop, how does the height h compare with rolling as to that with frictionless sliding?

A.hrolling > hsliding

B.hrolling = hsliding

C.hrolling < hsliding

h ?

R

ball has mass m & r <<R

U=mg2R


Example: Loop-the-Loop with rolling

To complete the loop the loop, how high do we have to release a ball with radius r (r <<R) ?

Condition for completing the loop the loop: Circular motion at the top of the loop (ac = v2 / R)

Use fact that E = U + K = constant (or work energy)

h ?

R

ball has mass m & r <<R

Recall that “g” is the source of the centripetal acceleration and N just goes to zero is the limiting case.

Also recall the minimum speed at the top is

gRTangentialv

Ub=mgh

U=mg2R


Example: Loop-the-Loop with rolling

Work energy (K = W) Kf = KCM + KRot = mg (h-2R) (at top) Kf = ½ mv2 + ½ 2/5 mr2 2 = mg (h-2R) & v=r ½ mgR + 1/5 mgR = mg(h-2R) h = 5/2R+1/5R

h ?

R

ball has mass m & r <<R gR

Tangentialv

Ub=mgh

U=mg2R

Just a little bit more….


Torque : Rolling Motion (center of mass point) A solid cylinder, with mass m and radius R, is rolling

without slipping down an inclined plane. What is its angular acceleration ?

Fx = m acm = mg sin - fs m Racm = mgR sin - Rfs

Fy = 0 = N – mg cos (not used) = -R fs = Icm 3. = - acm / R

-mR2 = mgR sin + I-mR2 – I= mgR sin

= -mgR sin (mR2 + I) = -2g sin / 3R

mg

N

fs


Torque : Rolling Motion (contact point) A solid cylinder, with mass m and radius R, is rolling

without slipping down an inclined plane. What is its angular acceleration ?

= -R mg sin = I 2. I = mR2 + Icm

= -mgR sin (mR2 + Icm) = 2g sin / 3R

mg

N

fs


Torque : Come back spool (contact point) A solid cylinder, with mass m, inner radius r and outer radius

R, being pulled by a horizontal force F at radius r? If the cylinder does not slip then what is the angular acceleration?

mg

N

fs

F

= -rF = I 2. I = mR2 + Icm = 3/2 mR2 = -rF(mR2 + Icm) = -2Fr/ 3R

mg

N

fs

F

= 0 = I

= 0

mg

N

fs

F

= rF = I 2. I = mR2 + Icm

= 2Fr/ 3R


Torque : Limit of Rolling (center of mass) A solid cylinder, with mass m and radius R, being

pulled by a horizontal force F on its axis of rotation. If s is the coefficient of static friction at the contact point, what is the maximum force that can be applied before slipping occurs?

mg

N

fs

F1Fx = m acm = F - fs 2. Fy = 0 = N – mg 3. = -R fs = Icm

- acm- fs mR2 / Icm

4. fs ≤ s N = s mg (maximum)5. acm = - R

m acm = fs mR2 / Icm = F - fs

fs (1+ mR2 / Icm ) = s mg (1+ mR2

/ Icm ) = FF = s mg (1+ mR2

/ ( ½ mR2 ) ) = 3 s mg


Modified Atwood’s Machine (More torques) Two blocks, as shown, are attached by a massless string which

passes over a pulley with radius R and rotational inertia I = ½ MR2. The string moves past the pulley without slipping. The surface of the table is frictionless. What are the tensions in the strings ?

m1g

NT T

T1

T1 m2g

mass 1

Fy = m1 ay= -m1g +T1

mass 2

Fx = m2 ax = –Tpulley

= I = R T1 – R T

ay = ax= -R = a

m1 a = -m1g +T1

m2 a = –T

-I a / R = R T1 – R T


Modified Atwood’s Machine (More torques) Two blocks, as shown, are attached by a massless string which

passes over a pulley with radius R and rotational inertia I. The string moves past the pulley without slipping. The surface of the table is frictionless. What are the tensions in the strings?

m1g

NT T

T1

T1 m2g

1. m1 a = – m1g +T1

2. m2 a = –T

3. -I a / R = R T1 – R T

Solve for a

-I a / R2 = m1 a +m1g + m2 a

a = – m1g / (m1 + ½ M + m2)

T= - m2 a T1= m1 a + m1g


Angular Momentum

0p

Ext dt

d

F

We have shown that for a system of particles,

momentum

is conserved if

Are the rotational equivalents?

angular momentum

is conserved if

vm

p

Ipr L

0Ext dtLdFr

IFr


For Thursday

Read all of chapter 10 (gyroscopes will not be tested) HW9

Documents

Physics 201: Lecture 19, Pg 1 Lecture 19 Goals: Specify rolling motion (center of mass velocity to angular velocity Compare kinetic and rotational energies