Physics 111: Lecture 19 Today’s Agenda

Physics 111: Lecture 19, Pg 1

Physics 111: Lecture 19

Today’s Agenda

Review Many body dynamics Weight and massive pulley Rolling and sliding examples Rotation around a moving axis: Puck on ice Rolling down an incline Bowling ball: sliding to rolling Atwood’s Machine with a massive pulley


Review: Direction & The Right Hand Rule

To figure out in which direction the rotation vector points, curl the fingers of your right hand the same way the object turns, and your thumb will point in the direction of the rotation vector!

We normally pick the z-axis to be the rotation axis as shown.= z

= z

= z

For simplicity we omit the subscripts unless explicitly needed.

x

y

z

x

y

z


Review: Torque and Angular Acceleration

NET = I

This is the rotational analogue of FNET = ma

Torque is the rotational analogue of force:The amount of “twist” provided by a force.

Moment of inertia I is the rotational analogue of massIf I is big, more torque is required to achieve a given

angular acceleration.


Lecture 19, Act 1Rotations

Two wheels can rotate freely about fixed axles through their centers. The wheels have the same mass, but one has twice the radius of the other.

Forces F1 and F2 are applied as shown. What is F2 / F1 if the angular acceleration of the wheels is the same?

(a) 1

(b) 2

(c) 4

F1

F2


Lecture 19, Act 1Solution

We know

I mR2but and FR

I

so

mRF

mRFR 2

1

2

1

2

1

2

RR

mRmR

FF

F1

F2

Since R2 = 2 R12

FF

1

2


Review: Work & Energy

The work done by a torque acting through a displacement is given by:

The power provided by a constant torque is therefore given by:

W

PdW

dt

d

dt


Falling weight & pulley

A mass m is hung by a string that is wrapped around a pulley of radius R attached to a heavy flywheel. The moment of inertia of the pulley + flywheel is I. The string does not slip on the pulley.

Starting at rest, how long does it take for the mass to fall a distance L.

I

m

R

T

mg

a

L


Falling weight & pulley...

For the hanging mass use F = mamg - T = ma

For the pulley + flywheel use = I = TR = I

Realize that a = R

Now solve for a using the above equations.

I

m

R

T

mg

a

L

amR

mRg

2

2 I

TRa

RI


Falling weight & pulley...

Using 1-D kinematics (Lecture 1) we can solve for the time required for the weight to fall a distance L: I

m

R

T

mg

a

L

amR

mRg

2

2 I

L at1

22

tL

a

2

where

Flywheel

w/ weight


Rotation around a moving axis.

A string is wound around a puck (disk) of mass M and radius R. The puck is initially lying at rest on a frictionless horizontal surface. The string is pulled with a force F and does not slip as it unwinds.

What length of string L has unwound after the puck has moved a distance D?

F

RM

Top view


Rotation around a moving axis...

The CM moves according to F = MA

F

M A

AF

M

D AtF

Mt

1

2 22 2 The distance moved by the CM is thus

RI

1

22MR

MRF2

MR21

RFI 2

===

The disk will rotate about its CM according to = I

1

22 2t

F

MRt So the angular displacement is


Rotation around a moving axis...

So we know both the distance moved by the CM and the angle of rotation about the CM as a function of time:

D

F

DF

Mt

22

F

MRt 2

F

Divide (b) by (a):

(a) (b)

D R

2

R D 2

L

The length of stringpulled out is L = R:

L D2


Comments on CM acceleration:

We just used = I for rotation about an axis through the CM even though the CM was accelerating! The CM is not an inertial reference frame! Is this OK??

(After all, we can only use F = ma in an inertial reference frame).

YES! We can always write = I for an axis through the CM.This is true even if the CM is accelerating.We will prove this when we discuss angular momentum!

F

R

M A


Rolling

An object with mass M, radius R, and moment of inertia I rolls without slipping down a plane inclined at an angle with respect to horizontal. What is its acceleration?

Consider CM motion and rotation about the CM separately when solving this problem (like we did with the lastproblem)...

R

I

M


Rolling...

Static friction f causes rolling. It is an unknown, so we must solve for it.

First consider the free body diagram of the object and use FNET = MACM :

In the x direction Mg sin - f = MA

Now consider rotation about the CMand use = I realizing that = Rf and A = R

R

M

f

Mg

y

x

RfA

RI f

A

RI 2


Rolling...

We have two equations:

We can combine these to eliminate f:

f IA

R2

+=

IMR

sinMRgA 2

2

A R

I

M

sing75

MR52

MR

sinMRgA

22

2

=+

=

For a sphere:

mafsinMg =-



Two uniform cylinders are machined out of solid aluminum. One has twice the radius of the other.If both are placed at the top of the same ramp and released,

which is moving faster at the bottom?

(a) bigger one

(b) smaller one

(c) same



Consider one of them. Say it has radius R, mass M and falls a height H.

H

Energy conservation: - DU = DK MgH MV 12

12

2 2I

I 12

2MR VR

but and

MgH MRV

RMV

12

12

12

22

22

MgH MV MV MV 14

12

34

2 2 2


Lecture 19, Act 2 Solution

H

MgH MV34

2So: gH V34

2

V gH 43

So, (c) does not depend on size,

as long as the shape is the same!!


Sliding to Rolling

A bowling ball of mass M and radius R is thrown with initial velocity v0. It is initially not rotating. After sliding with kinetic friction along the lane for a distance D it finally rolls without slipping and has a new velocity vf. The coefficient of kinetic friction between the ball and the lane is . What is the final velocity, vf, of the ball?

vf= R

f = Mgv0

D

Roll bowling ball


Sliding to Rolling...

While sliding, the force of friction will accelerate the ball in the -x direction: F = -Mg = Ma so a = -g

The speed of the ball is therefore v = v0 - gt (a) Friction also provides a torque about the CM of the ball.

Using = I and remembering that I = 2/5MR2 for a solid sphere about an axis through its CM:

D

x

2MR52

MgR ==R2g5

=

f = Mg

tR2g5

t0

=+= (b)

v f= R

v0


Sliding to Rolling...

We have two equations:

Using (b) we can solve for t as a function of

Plugging this into (a) and using vf = R (the condition for rolling without slipping):

D

x

tR2g5

=v v gt 0 (a) (b)

tR

g

2

5

v vf 5

7 0

f = Mg

Doesn’t depend

on , M, g!!

vf= R

v0



A bowling ball (uniform solid sphere) rolls along the floor without slipping.What is the ratio of its rotational kinetic energy to its

translational kinetic energy?

I 25

2MRRecall that for a solid sphere about

an axis through its CM:

(a) (b) (c)

255

1 12



The total kinetic energy is partly due to rotation and partly due to translation (CM motion).

rotational

K

translational

K

12

2I 12

2MVK =


Lecture 19, Act 3 Solution

VR

Since it rolls without slipping:

rotational

K

Translational

K

12

2I 12

2MVK =

2

2

TRANS

ROT

MV21

I21

KK

52

MVRV

MR52

2

2

22


Atwoods Machine with Massive Pulley:

A pair of masses are hung over a massive disk-shaped pulley as shown.Find the acceleration of the blocks.

m2m1

R

M

y

x

m2g

aT1

m1g

a

T2

For the hanging masses use F = ma -m1g + T1 = -m1a -m2g + T2 = m2a

Ia

RMRa

1

2

Ia

R

I 1

22MR(Since for a disk)

For the pulley use = I

T1R - T2R


Atwoods Machine with Massive Pulley...

We have three equations and three unknowns (T1, T2, a). Solve for a.

-m1g + T1 = -m1a (1)

-m2g + T2 = m2a (2)

T1 - T2 (3)

am m

m m Mg

1 2

1 2 2

1

2Ma

Large and small pulleys

m2m1

R

M

y

x

m2m1

m2g

aT1

m1g

a

T2


Recap of today’s lecture

Review (Text: 9-1 to 9-6) Many body dynamics Weight and massive pulley (Text: 9-4) Rolling and sliding examples (Text: 9-6) Rotation around a moving axis: Puck on ice (Text: 9-4) Rolling down an incline (Text: 9-6) Bowling ball: sliding to rolling Atwood’s Machine with a massive pulley (Text: 9-4)

Look at textbook problems Chapter 9: # 53, 89, 92, 113, 125

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Physics 111: Lecture 19 Today’s Agenda