Massive Pulleys Review

2 kg

Massive pulleys have rotational inertia! It takes a net torque to change their angular velocity ω .

Torque on a pulley is caused by the rope tension T. If there is no slipping, its magnitude is T times the radius the rope goes around the pulley.

τ net = Iα

In solving massive pulley problems you can relate linear motion of rope to angular motion of pulley.

2

Clicker question -2 Set frequency to BA A mass m hangs from string wrapped around a pulley of radius R. The pulley has a moment of inertia I and its pivot is frictionless. Because of gravity the mass falls and the pulley rotates. The magnitude of the torque on the pulley is…

A.  greater than mgR B.  less than mgR C.  equal to mgR

R

m

(Hint: Is the tension in the string = mg?)

Free body diagram of mass gives T=mg-ma when a is positive so T<mg so RmgrT <=τ

T

mgFG =

3

Clicker question -1 Set frequency to BA For the 2 kg weight attached to 6 kg pulley with radius of 0.10 m, what is the kinetic energy of the weight after dropping for 1 second. Remember a=4 m/s2 and K=½mv2, K=½Iω2,and Idisk=½MR2.

A.  2 J B.  4 J C.  8 J D.  16 J E.  32 J

2 kg

After 1 second v = v0 + at = 0+ 4 m/s2 ⋅1 s = 4 m/s

K = 12 mv

2 = 12 2 kg ⋅ 4 m/s( )2

=16 JKinetic energy for the weight is

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Clicker question 0 Set frequency to BA For 2 kg weight attached to 6 kg pulley with R=0.10 m, what is the kinetic energy of the pulley after the weight drops for 1 second. Remember v=4 m/s for the weight after 1 second and K=½mv2 and K=½Iω2 and Idisk=½MR2.

A.  4 J B.  8 J C.  16 J D.  24 J E.  48 J

2 kg

K = 12 Iω

2 = 12

12 MR

2( ) vR!

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2

= 14 Mv

2 = 14 6 kg ⋅ 4 m/s( )2

= 24 J

The weight velocity and the outside pulley velocity are the same since they are connected by the rope.

Can get angular velocity from rim velocity ω =

vR

and then find kinetic energy

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Massive pulleys

2 kg

A 2 kg weight is attached to the end of a rope coiled around a pulley with mass of 6 kg and radius of 0.1 m.

The acceleration of the weight is found to be 4 m/s2.

After 1 second the total kinetic energy was found to be K total = Kweight +Kpulley =16 J+ 24 J = 40 J

After 1 second the weight will descend Δy = v0 yt +

12 ayt

2 = 0+ 12 ⋅ 4 m/s2 ⋅ (1 s)2 = 2 m

losing gravitational potential energy Ugrav =mgh = 2 kg ⋅10 m/s2 ⋅2 m = 40 J Conservation of

energy still holds!

From Newton’s 2nd law:

Clicker question 1 Set frequency to BA Two unequal weights are attached to a rope that is looped around a pulley. The rope does not slide with respect to the pulley. The pulley is massive and has a frictionless pivot. After the blocks are released, what is the relationship between T1 and T2? A.  T1 > T2 B.  T1 = T2 C.  T1 < T2 D.  Impossible to tell

2 kg

τ net = RT2 − RT1 = R T2 −T1( )

There is a related CAPA question due next week.

1kg

T1T2

Consider just the pulley:

T2R

T1

What are the torques acting on the pulley and what is the net torque?

τ net = IαSince the pulley will start rotating counter-clockwise, there must be positive angular acceleration so T2 > T1.

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Possible pitfall Sometimes more than one R in the problem.

2 kg

R is found in several places (choose correctly)

1. Torque calculation τ = rF sinφ

2. Moment of inertia calculation Idisk =12MR

2

3. Conversion between rope linear velocity and pulley angular velocity or rope linear acceleration and pulley angular acceleration

v = Rω

a = Rα

8

Very important notes on gravity The potential energy due to gravity for an extended body is measured from the center of mass.

When calculating torque due to gravity, the location of the force of gravity can be assumed to be the center of mass.

°= 60sin2 MgLτ°== 30sin2grav

LMgMghU

MgFG =°= 30sin2

LhM

°30

LNote the torque will change as the rod falls due to the angle changing

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A few more tips

If torques and forces aren’t helping, try conservation of energy (and vice versa)

Newton’s 2nd law is always true (for inertial reference frames)

Everything you have already learned continues to work

Conservation of energy still applies.

Be careful when relating signs of angular acceleration and linear (tangential) acceleration or angular velocity and linear velocity

Time for a Torque Wave

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11

Rotation & Translation Problems so far had a stationary axis.

Other problems have a moving axis like a boomerang, bowling ball, or yo-yo.

This is a combination of translational motion and rotational motion

Split motion into two parts

1. Translational motion of the center of mass

2. Rotational motion around the center of mass

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Separate kinetic energy terms Separate kinetic energy into translational and rotational parts 2

CM212

CM21

rotCM ωIMvKKK +=+=

Sometimes center of mass velocity & angular velocity are related.

Rolling without slipping: ωRv =CM

CMv

translational

ω+rotational

CMv=total

Note this requires static friction to prevent slipping

13 So hoop has the most kinetic energy:

Clicker question 2 Set frequency to BA A sphere, hoop, and cylinder, all with the same mass M and radius R, are rolling along with the same speed v. Which one has the most kinetic energy? A.  Sphere B.  Hoop C.  Disk D.  All have the same kinetic energy

v v v

All have the same translational kinetic energy: KCM =12MvCM

2

The hoop has the largest moment of inertia so it will have the most rotational kinetic energy: Krot =

12 ICMω

2

All have the same angular velocity: ω = vCM / R

K = KCM +Krot

Hoop vs Disk

14

A hoop, and disk, with the same mass M and radius R, start rolling down at ramp, starting from rest. Which one reaches the bottom first?

h Ug =Mgh

K = 12MvCM

2 + 12 ICMω

2

vCM = Rω

K =12MvCM

2 +12I( ) vCM

R!

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=12vCM2 (M +

IR2)=Mgh

vCM =2Mgh

M +IR2

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Yo-yo A yo-yo of mass M and radius R starts from rest and is dropped, attached by an unrolling string. What is the center of mass velocity after falling a height h?

We know that falling a height loses gravitational potential energy which in this case is Ug =Mgh

This energy goes into kinetic energy: K = 12MvCM

2 + 12 ICMω

2

Since this is rolling without slipping vCM = Rω

K =12MvCM

2 +1212MR2

!

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!

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=34MvCM

2 Setting Ug = K

vCM =43gh

Asking for velocity suggests conservation of energy.

16

Pulling the yo-yo Yo-yo with inner radius R1 and outer radius R2 is on a table. String is pulled with force F as shown. There is enough static friction to prevent the yo-yo from sliding. Which way does the yo-yo move?

FfF

Fnet = F −Ff =MaCM τ net = R2Ff − R1F = ICMα

Rolling without slipping: and vCM = Rω aCM = Rα

so R2Ff − R1F =12MR2

2 aCMR2

=12MR2aCM

F

Substituting Ff = F −MaCM gives R2 F −MaCM( )− R1F = 12MR2aCM

Rearranging: R2F − R1F = 32MR2aCM aCM =

2 R2 − R1( )F3MR2

but you need to make sure you get the signs right!

Given forces and asking direction suggests Newton’s 2nd law.

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Pulling the yo-yo

FfFFnet = F −Ff =MaCM

τ net = R2Ff − R1F = ICMα

Applying rolling without slipping: and vCM = Rω aCM = Rα

Using the extended free body diagram and Newton’s 2nd law we found the equations:

aCM =2 R2 − R1( )F3MR2

After some math: If R2 > R1 then aCM > 0(which is to the left)

If R1 = 0 then F exerts no torque and aCM =2F3M

2R1R

Next Week – Chapter 11

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Angular Momentum!!!