Physics 2211: Lecture 38

Lecture 38, Page 1 Physics 2211 Spring 2005© 2005 Dr. Bill Holm

Physics 2211: Lecture 38Physics 2211: Lecture 38

Rolling Motion Mass rolling down incline plane

Energy solution Newton’s 2nd Law solution

Angular Momentum


For a solid object which rotates about its center or mass and whose CM is moving:

VCM

Translational & rotational motion Translational & rotational motion combinedcombined

2 21 12 2NET CM CMK MV I


Rolling MotionRolling Motion

Rolling without slipping

R

s

R

svcm

s Rds d

R Rdt dt

But CM

dsv

dt

CMv R



Objects of different I rolling down an inclined plane: find their speed at the bottom of the plane and their acceleration on the plane.

h

R

v = R

M

2 21 12 2E Mv Mgy I

2 21 12 2finalE Mv I

v = 0

= 0

K = 0initalE Mgh


12

1

v gh

c

The rolling speed is always lower than in the case of simple sliding

since the kinetic energy is shared between CM motion and rotation.

2 21 12 2finalE Mv I

2

2 21 12 2

final

vE MR Mv

Rc 21

2 1 Mvc

final initialE E

212 1 Mvc Mgh

Use v = R and I = cMR2

1 hoop

1/2 disk

2/5 spherec =



2 2v a x

12

1

v gh

c

v

h

R

M

2sin

ha

sin

1

g

ac

1 hoop

1/2 disk

2/5 spherec =


x


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 MotionRolling Motion(alternate approach)(alternate approach)


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

RM

y

x

Mg

f

sinMg

sin MMg af

Now consider rotation about the CMand use = I realizing that

= Rf and a = R


Ia

RfR

I2

af

R


We have two equations:

aR

I

M

I2

af

R

I

2

2

sinMRa g

MR

For a sphere, c = :

5

725

sinsin

1

ga g

25

2sin

1

I cc

ga for MR

We can combine these to eliminate f:


sin - Mg f ma


ExampleExampleRotationsRotations

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?

(1) bigger one

(2) smaller one

(3) same


ExampleExample SolutionSolution

h

12

1

v gh

c

1 hoop

1/2 disk

2/5 spherec =

Answer:

(3) same

speed does not depend on size,

as long as the shape is the same!!

43v gh


Direction of Rotational MotionDirection of Rotational Motion In general, the rotation variables (etc) are vectors

(have direction) If the plane of rotation is in the x-y plane, then the convention is

x

y

z

x

y

z

CCW rotation is in the + z direction

CW rotation is in the - z direction


Direction of Rotation:Direction of Rotation:The Right Hand RuleThe 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


Angular momentum of a rigid bodyAngular momentum of a rigid bodyabout a fixed axis:about a fixed axis:

Consider a rigid distribution of point particles rotating in the x-y plane around the z axis, as shown below. The total angular momentum around the origin is the sum of the angular momenta of each particle:

i

j m2

m1

m3

1r

3r2r

3v

1v

2v

We see that is in the z direction.L

Using vi = ri , we get

2 ˆi i

i

L m r k

Analog of !IL

p mv

ˆi i i i i i i i

i i i

L r p m r v m r v k (since and are

perpendicular) iv

ir


Angular momentum of a rigid bodyAngular momentum of a rigid bodyabout a fixed axis:about a fixed axis:

In general, for an object rotating about a fixed (z) axis we can write LZ = I

The direction of LZ is given by theright hand rule (same as ).

We will omit the Z subscript for simplicity,and write L = I

z

ZL I


Conservation of Angular MomentumConservation of Angular Momentum

Total angular momentum is conservedTotal angular momentum is conserved

where EXT

dL

dt

L I

In the absence of external torquesIn the absence of external torques 0EXT

dL

dt


Example: Two DisksExample: Two Disks

A disk of mass M and radius R rotates around the z axis with angular velocity i. A second identical disk, initially not rotating, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity f.

i

z

f

z



First realize that there are no external torques acting on the two-disk system.Angular momentum will be conserved!

Initially, the total angular momentum is due only to the disk on the bottom:

i

z

2

1I 211 2i i iL MR



First realize that there are no external torques acting on the two-disk system.Angular momentum will be conserved!

Finally, the total angular momentum is dueto both disks spinning:

f

z

21

I I 2 2 21 11 2 2 2f f f f fL MR MR MR

or

I 2 212 2f f f f fL M R MR



Since Li = Lf

z

f

z

Li Lf

if 2

1

An inelastic collision, since K is not conserved (friction)!

I 2 2 2 2 21 1 1 12 2 2 82f f f f iK M R MR

I 2 2 2 2 21 1 1 12 2 2 4i i i I iK MR MR

2 212 i fMR MR


Example: Rotating TableExample: Rotating Table

A student sits on a rotating stool with his arms extended and a weight in each hand. The total moment of inertia is Ii, and he is rotating with angular speed i. He then pulls his hands in toward his body so that the moment of inertia reduces to If. What is his final angular speed f?

i

Ii

f

If


Example: Rotating TableExample: Rotating Table

Again, there are no external torques acting on the student-stool system, so angular momentum will be conserved.Initially: Li = Iii

Finally: Lf = If f f

i

i

f

I

I

i

Ii

f

If

LiLf


ExampleExampleAngular MomentumAngular Momentum

A student sits on a freely turning stool and rotates with constant angular velocity 1. She pulls her arms in, and due to angular momentum conservation her angular velocity increases to 2. In doing this her kinetic energy:

(1) increases (2) decreases (3) stays the same

1 2

I2 I1

L L



I2

LI

21

K2

2 (using L = I)

L is conserved and

I2 < I1 K2 > K1 K increases!

1 2

I2 I1

L L



Since the student has to force her arms to move toward her body, she must be doing positive work! The work/kinetic energy theorem states that this will increase the kinetic energy of the system!

1

I1

2

I2

L L


Angular Momentum of aAngular Momentum of aFreely Moving ParticleFreely Moving Particle

We have defined the angular momentum of a particle about the origin as

This does not demand that the particle is moving in a circle!We will show that this particle has a constant angular

momentum!

y

x

L r p

v



Consider a particle of mass m moving with speed v along the line y = -d. What is its angular momentum as measured from the origin (0,0)?

x

md

y

v



We need to figure out The magnitude of the angular momentum is:

Since and are both in the x-y plane, will be in the z direction (right hand rule):

y

xd

L pdZ

L r p

sin sinL r p rp p r pd

p x dis tance of closest approach

r

p

L

r

p mv



So we see that the direction of is along the z axis, and its magnitude is given by LZ = pd = mvd. L is clearly conserved since d is constant (the distance of closest approach of the particle to the origin) and p is

constant (linear momentum conservation).

y

xd

L

p


Example: Bullet hitting stickExample: Bullet hitting stick

A uniform stick of mass M and length D is pivoted at the center. A bullet of mass m is shot through the stick at a point halfway between the pivot and the end. The initial speed of the bullet is v1, and the final speed is v2.

What is the angular speed F of the stick after the collision? (Ignore gravity)

v1 v2

M

F

initial final

mD

D/4



Conserve angular momentum around the pivot (z) axis! The total angular momentum before the collision is due

only to the bullet (since the stick is not rotating yet).

v1

D

M

initial

D/4m

4

DmvapproachclosestofcedisxpL 1i )tan(



Conserve angular momentum around the pivot (z) axis! The total angular momentum after the collision has

contributions from both the bullet and the stick.

where I is the moment of inertia of the stick about the

pivot.

v2

F

final

D/4

I2 4f F

DL mv



Set Li = Lf and using

v1 v2

M

F

initial final

mD

D/4

I 1

122MD

21 2

1

4 4 12 F

D Dmv mv MD 1 2

3F

mv v

MD

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Physics 2211: Lecture 38