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Smooth Rolling

Smooth rolling, if a wheel does not slip at the point of contact

cm

cm

s R

v R

a R

θ

ωα

=

==

θso, because arc length s is the same as linear distance moved forward,

s

s

rolling without slipping: translation + rotation

cmR vω =

vcm

pure translation

+

pure rotation

0

2 cmv

rolling

As seen by an observer at rest:

= vcm

Rolling and Kinetic Energy

ω

2cm2

22

22cm

rotTtot

)(2

1

)(2

12

1

2

1

vR

Im

ImR

Imv

KKK

+=

+⇒

+=

+=

ω

ω

If the object rolls without slipping then vcm = Rω.

4

Example: Two objects (a solid disk and a solid sphere) are rolling down a ramp. Both objects start from rest and from the same height. Which object reaches the bottom of the ramp first?

h

θ

The object with the largest linear velocity (v) at the bottom of the ramp will win the race.

Rolling Example

5

+=

+=

+=++=+

+=+

=

2

22

2222

2

2

1

2

1

2

1

2

1

2

100

RI

m

mghv

vR

Immgh

R

vImvImvmgh

KUKU

EE

ffii

fi

ω

Apply conservation of mechanical energy:

Example continued:

Solving for v:

6

2sphere

2disk

5

22

1

mRI

mRI

=

=

For the disk:

For the sphere:

ghv3

4disk =

ghv7

10sphere =

Since Vsphere> Vdisk the sphere wins the race.

ghv 2box =Compare these to a box sliding down the frictionless ramp.

Example continued:

The moments of inertia are:

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7

How do objects in the previous example roll?

FBD:

Both the normal force and the weight act through the center of mass so Στ = 0. This means that the object cannot rotate when only these two forces are applied.

N

w

y

x

Acceleration

8

FBD:

Add friction:

0cos

sin cm

=−=

=−=

==

∑∑∑

θ

θ

ατ

wNF

maFwF

IrF

y

sx

s

y

x

N

w

Fs

θAlso need acm = αR and

The above system of equations can be solved for v at the bottom of the ramp. The result is the same as when using energy methods. (See text example 8.13.)

xavv ∆+= 220

2

It is static friction that makes an object roll.

angular momentum

REMINDERS

Linear Motion Rotations

coordinate xvelocity v

acceleration a

mass mforce F

1st Newton’s Law F=0: v=const2nd Newton’s Law F = ma

Kinetic Energy K = ½mv2

Momentum p = mv

θθθθ angleωωωω angular velocity αααα angular acceleration

ΙΙΙΙ rotational inertiaττττ torqueττττ=0: ωωωω=constττττ = Iαααα

K = ½Iωωωω2 Kinetic Energy??? Angular Momentum

10

Angular Momentum and AMC

vp

pF

mtt

=∆∆=

→∆ 0net lim

ωL

L

Itt

=∆∆=

→∆ 0net limτ

Units of p are kg m/s Units of L are kg m2/s

When no net external forces act, the momentum of a system remains constant (pi = pf)

When no net external torques act, the angular momentum of a system remains constant (Li = Lf).

x-axis

r

P

m

p θθθθ

y-axis p⊥⊥⊥⊥ Angular momentum of a moving particle:

prrprpL ⊥⊥ === θsin

11

Example (text problem 8.69): A turntable of mass 5.00 kg has a radius of 0.100 m and spins with a frequency of 0.500 rev/sec. What is the angular momentum? Assume a uniform disk.

rad/sec 14.3rev 1

rad 2

sec

rev500.0 =

= πω

/sm kg 079.02

1 22 =

== ωω MRIL

Example

12

Example (text problem 8.75): A skater is initially spinning at a rate of 10.0 rad/sec with I=2.50 kg m2 when her arms are extended. What is her angular velocity after she pulls her arms in and reduces I to 1.60 kg m2?

The skater is on ice, so we can ignore external torques.

( ) rad/sec 6.15rad/sec 0.10m kg 60.1

m kg 50.22

2

=

=

=

=

=

if

if

ffii

fi

I

I

II

LL

ωω

ωω

AMC Example

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University of Florida PHY 2053 Page 13

Exam 2 Fall 2011: Problem 17

R FT

Mg

y-axis

• A cloth tape is wound around the outside of a non-uniform solid cylinder (mass M, radius R) and fastened to the ceiling as shown in the figure. The cylinder is held with the tape vertical and then release from rest. If the acceleration of the center-of-mass of the cylinder is 3g/5, what is its moment of inertia about its symmetry axis?

Answer: % Right: 48%

232 MR

R

yT MaFMg =−

R

IaIRF y

T === ατ)( yT agMF −=

2322

53

53

22

MRMRg

ggMR

a

ag

a

FRI

y

y

y

T =

−=

−==

University of Florida PHY 2053 Page 14

Exam 2 Fall 2011: Problem 31

• A uniform solid disk with mass M and radius R is mounted on a vertical shaft with negligible rotational inertia and is initially rotating with angular speed ωωωω. A non-rotating uniform solid disk with mass M and radius R/2 is suddenly dropped onto the same shaft as shown in the figure. The two disks stick together and rotate at the same angluar speed. What is the new angular speed of the two disk system? Answer: 4ωωωω/5% Right: 47%

R M

M

ffffiii MRMRILMRIL ωωωω )( 222

1212

1212

1 +=====

ωωωω 54

22

2

22

21

21

)2/(=

+=

+=

RR

R

RR

Rf

University of Florida PHY 2053 Page 15

Exam 2 Fall 2010: Problem 16

Initial

P

R

M

m

• A mouse of mass M/6 lies on the rim of a solid uniform disk of mass M that can rotate freely about its center like a merry-go-round. Initially the mouse and disk rotate together with an angular velocity of ωωωω. If the mouse walks to a new position that is at the center of the disk what is the new angular velocity of the mouse-disk system?Answer: 4ωωωω/3% Right: 58%

Final

P

R

M

m ωω )( 2mRIIL diskiii +==fdiskfff IIL ωω ==

ωω )( 2mRII disknewdisk +=fi LL =

ωωωωω 34

221

22 2111 =

+=

+=

+=

M

m

MR

mR

I

mR

disknew

2291

22

22ωMR

I

L

I

L

EEE

i

i

f

f

if

=−=

−=∆Note that energy is not conserved in

this problem!