chapter 12: rotation of rigid bodies - santa rosa …lwillia2/40/40ch12_s14.pdf · chapter 12:...

Chapter 12: Rotation of Rigid Bodies

Center of Mass

Moment of Inertia

Torque

Angular Momentum

Rolling

Statics

Translational vs Rotational

2

/

/

1/ 2

m

x

v dx dt

a dv dt

F ma

p mv

KE mv

Work Fd

P Fv

2

/

/

1/ 2

I

d dt

d dt

I

L I

KE I

Work

P

2

c

t

s r

v r

a r

a r

Fr

L pr

Connection

More…

Center of Mass The geometric ‘center’ or average location of the mass.

Rotational & Translational Motion Objects rotate about their Center of Mass.

The Center of Mass Translates as if it were a point particle.

CMCM

rv

d

dt

The Center of Mass Translates as if it were a point particle

and, if no external forces act on the system, momentum is

then conserved. This means:

EVEN if the bat EXPLODED into a thousand pieces, all the

pieces would move so that the momentum of the CM is

conserved – that is, the CM continues in the parabolic

trajectory!!!! THIS IS VERY VERY IMPORTANT!

Center of Mass

System of Particles Center of Mass

• A projectile is fired into the air and suddenly explodes

• With no explosion, the projectile would follow the dotted line

• After the explosion, the center of mass of the fragments still follows the dotted line, the same parabolic path the projectile would have followed with no explosion!

If no external forces act on the system, then the velocity of the CM doesn’t change!!

Center of Mass: Stability

If the Center of Mass is above the base

of support the object will be stable. If

not, it topples over.

This dancer balances en pointe by having her center of mass directly over her toes, her base of support.

Balance and Stability

Slide 12-88

Center of Mass

CM i i

i

m x

xM

The geometric ‘center’ or average location of the mass.

System of Particles: Extended Body:

CM

1dm

M r r

CM Lecture Problem

1. Identical particles are placed at the 50-cm and 80-cm

marks on a meter stick of negligible mass. This rigid

body is then mounted so as to rotate freely about a

pivot at the 0-cm mark on the meter stick. A) What is

the CM of the system? B) What is the torque acting on

the system? C) What is the rotational inertia of the

system about the end? D) If this body is released from

rest in a horizontal position, what is the angular

acceleration at the release? E) What is the angular

speed of the meter stick as it swings through its lowest

position?

Prelab

A meter stick has a mass of 75.0 grams and has two masses attached to it – 50.0 grams at the 20.0cm mark and 100.0 grams at the 75.0 cm mark.

(a) Find the center of mass of the system - that is, at what mark on the meter stick should the fulcrum be placed so that the system balances?

(b) A fulcrum is then placed at the CM. Sketch and label the system.

(c) Show that the net torque about the cm is zero.

(d) Calculate the rotational inertia of the system about the CM axis. Box answers, make it neat.

Center of Mass The geometric ‘center’ or average location of the mass.

Extended Body:

CM

1dm

M r r

Center of Mass of a Solid Object

Divide a solid object into

many small cells of mass m.

As m 0 and is replaced

by dm, the sums become

Before these can be integrated:

dm must be replaced by

expressions using dx and dy.

Integration limits must be established.

Slide 12-41

Example 12.2 The Center of Mass of a Rod

Slide 12-42

Example 12.2 The Center of Mass of a Rod

Slide 12-43

Extended Body:

Example

CM

1dm

M r r

You must generate an expression for the density and the mass differential, dm, from geometry and by analyzing a strip of the sign. We assume the sign has uniform density. If M is the total mass then the total volume and density is given by:

2( )

1

2

M Mydm dV ytdx ytdx dx

ababt

1,

12

2

MV abt

abt

Where a, b and t are the width, height and thickness of the sign, respectively. Then the mass element for the strip shown is:

CM

1 x x dm

M

3

2

0 0

1 2 2 2 2( )

3 3

aaMy b x

x dx x x dx aM ab ab a a

Newton’s 2nd Law for Rotation

I

The net external torques acting on an object

around an axis is equal to the rotational

inertia times the angular acceleration.

The rotational equation is limited to rotation about a single

principal axis, which in simple cases is an axis of symmetry.

Inertia thing

Acceleration thing

Force thing

Torque: Causes Rotations

The moment arm, d, is

the perpendicular

distance from the axis

of rotation to a line

drawn along the

direction of the force

sinFr Fd

lever arm: sind r

The horizontal component of F (F cos ) has no tendency to produce a rotation

Torque: Causes Rotations

sinFr Fd

The direction convention is: Counterclockwise rotations are positive. Clockwise rotations are negative.

If the sum of the torques is zero, the system

is in rotational equilibrium.

Newton’s 1st Law for Rotation

250 3 750girl N m Nm

500 1.5 750boy N m Nm

0

Torque

Is there a difference in torque? (Ignore the mass of the rope)

NO!

In either case, the lever arm is the same!

What is it? 3m

CM Lecture Problem





the CM of the system? B) What is the torque acting

on the system? C) What is the rotational inertia of the





position?


I






Inertia thing

Acceleration thing

Force thing

Torque is a Vector!

= r x F

The direction is given by the right hand rule where the fingers

extend along r and fold into F. The Thumb gives the direction of .

The Vector Product

• The magnitude of C is

AB sin and is equal to the

area of the parallelogram

formed by A and B

• The direction of C is

perpendicular to the plane

formed by A and B

• The best way to determine

this direction is to use the

right-hand rule

ˆ ˆ ˆy z z y x z z x x y y xA B A B A B A B A B A B A B i j k

COMPARE!

CROSS PRODUCT

F and d must be

mutually perpendicular! sin

r F

Fr Fd

sin

L r p

L mvr

cosW F d Fd

DOT PRODUCT

F and d must be

mutually PARALLEL!

CROSS PRODUCT

L and p must be

mutually perpendicular!

Two vectors lying in the xy plane are

given by the equations A = 5i + 2j and

B = 2i – 3j. The value of AxB is

a. 19k

b. –11k

c. –19k

d. 11k

e. 10i – j

ˆ ˆ ˆy z z y x z z x x y y xA B A B A B A B A B A B A B i j k

CM Lecture Problem





the CM of the system? B) What is the torque acting

on the system? C) What is the rotational inertia of the





position?

Net external torques

Find the Net Torque

( 20 )(.5 ) (35 cos30)(1.10 ) N m N m

1 1 2 2 Fd F d

( 20 )(.5 ) (35 )(1.10 sin60) N m N m

OR

23.3 CCW Nm

F and d must be

mutually

perpendicular!

sinFr Fd


I






Inertia thing

Acceleration thing

Force thing

NEW: Rotational Inertia

The resistance of an object to rotate. The further away the mass is from the axis of rotation, the

greater the rotational inertia.

Rotational Inertia

2

i iI m r 2I r dm

System of Particles: Extended Body:

The resistance of an object to rotate.

Moments of Inertia of Various

Rigid Objects

Rotational Inertia

Depends on the axis.

For a system of point particles:

Where r is the distance to the axis of rotation.

2 i i iI m r I

Calculate Rotational Inertia

SI units are kg.m2

CM Lecture Problem







system about the end? D) If this body is released

from rest in a horizontal position, what is the angular



position?

2I r dm

Moment of Inertia of a Rigid

Object: The Rod

Calculate: Moment of

Inertia of a Rigid Object

22 I r dm R dm

The trick is to write dm in terms of the density:

dm dV

2 2 4

0

12

2

R

I r dm r Lr dr LR

2 dV rL drDivide the cylinder into concentric shells with

radius r, thickness dr and length L:

21

2zI MR2But /M R L


I






Inertia thing

Acceleration

thing Force thing

Tangential and Angular

Acceleration

d

dtt

dva

dt

( )

d r

dt

ta r

d

rdt

II

2

0

1 2 2

2 /t t t

I

2

0

1:

2Use t t

A 50 N m torque acts on a wheel of moment of inertia 150 kg m2.

If the wheel starts from rest, how long will it take the wheel to

make a quarter turn (90 degrees)?

2

2 / 23.1

50 /150

radt s

N m kg m

24.3

.314

m kg

R m

) ?

) ?

a

b

1 2

( 90 125 ).314

11

F r F r

N N m

Nm

2

=1

mR2

I

2

11=

1/2 24.3kg(.314m)

Nm

2= 9.2 /rad s

CM Lecture Problem







system about the end? D) If this body is released

from rest in a horizontal position, what is the

angular acceleration at the release? E) What is the

angular speed of the meter stick as it swings through its

lowest position?

Superposition of Inertia

The Parallel Axis Thm

The theorem states I = ICM + MD 2 I is about any axis parallel to the axis through the center of mass of the object ICM is about the axis through the center of mass D is the distance from the center of mass axis to the arbitrary axis

iI ISuperposition: Inertia ADD

Moment of Inertia for a Rod Rotating

Around One End

The moment of inertia of the

rod about its center is

The position of the CM is

D=½ L

Therefore,

21

12CMI ML

2

CM

2

2 21 1

12 2 3

I I MD

LI ML M ML

It is easier to rotate a rod about its center than about an end.

Rotational Inertia: Parallel Axis Theorem

A uniform rod (mass = 2.0 kg, length = 0.60 m) is free to

rotate about a frictionless pivot at one end. The rod is

released from rest in the horizontal position. What is the

magnitude of the angular acceleration of the rod at the

instant it is 60° below the horizontal?

a. 15 rad/s2

b. 12 rad/s2

c. 18 rad/s2

d. 29 rad/s2

e. 23 rad/s2

© 2013 Pearson Education, Inc.

Rotational Energy

A rotating object has kinetic energy because all particles in the object are in motion.

The kinetic energy due to rotation is called rotational kinetic energy.

Adding up the individual kinetic energies, and using vi = ri :

Slide 12-47

© 2013 Pearson Education, Inc.

Define the object’s moment of inertia:

The units of moment of inertia are kg m2. Moment of inertia depends on the axis of

rotation. Mass farther from the rotation axis contributes more

to the moment of inertia than mass nearer the axis. This is not a new form of energy, merely the familiar

kinetic energy of motion written in a new way.

Then the rotational kinetic energy is simply

Rotational Energy

Slide 12-48

CM Lecture Problem










speed of the meter stick as it swings through its

lowest position?

Rotational Inertia

Which reaches the bottom first?

(Same mass and radius)

Why Solid Cylinder?

2CMI mr

21

2CMI mr

PROVE IT!

Rolling

The red curve shows the path moved by a point on the rim of the object. This path is called a cycloid

The green line shows the path of the center of mass of the object which moves in linear motion.

Fig. 10.28, p.317

All points on a rolling object move in a direction perpendicular to an axis through the instantaneous point of contact P. In other words, all points rotate about P. The center of mass of the object moves with a velocity vCM, and the point P ’ moves with a velocity 2vCM. (Why 2?)

Rolling without Slipping

For pure rolling motion, (no slipping) as the cylinder rotates through an angle , its center moves a linear distance s = R with speed vCM. At any instant, the point on the rim located at point P is at rest relative to the surface since no slipping occurs.

Fig. 10.29, p.318

The motion of a rolling object can be modeled as a combination of pure translation and pure rotation.

Translation: CM

moves with vCM.

Rotation: All points

rotate about P with

angular speed .

Rolling Without Slipping

Friction

If an object rolls without slipping

then the frictional force that causes

the rotation is a static force.

If no slipping occurs, the point of

contact is momentarily at rest and

thus friction is static and does no

work on the object. No energy is

dissipated and Mechanical Energy is

Conserved. If the object slips, then

friction is not static, does work on

the object and dissipates energy.

21

2 PK I

2 21 1

2 2 CM CMK I mv


Kinetic Energy

m 2

CMUse Parallel-axis Th : PI I MR

2 2

CM

1( )

2 K I MR

The total kinetic energy of a rolling object is the sum of the rotational

kinetic energy about the center of mass and the translational kinetic

energy of the center of mass.

2 21 1

2 2 CM CMMgh Mv I

i fE E

2 21 1( )

2 2 CM

CM CM

vMv I

R


Conservation of Energy

2 2 21 1 2( )( )

2 2 5 CM

CM

vMv MR

R

2 21 1

2 5 CM CMMv Mv 27

10 CMMv

10

7CMv gh

21

2Mgh Mv

i fE E



10

7CMv gh

Compare to Slipping Only:

2CMv gh

Some of the gravitational

potential energy goes into

rolling the sphere so the

translational velocity of the

cm is less when rolling!

What about little ball big ball?

21

2Mgh Mv

i fE E



10

7CMv gh

Compare to Slipping Only:

2CMv gh

What about the

acceleration down the

incline?

Rolling down:

Does change while in flight?

Rolling Down the Incline

Rotational Inertia

Which reaches the bottom first?

(Same mass and radius)

Why Solid Cylinder?

2CMI mr

21

2CMI mr

PROVE IT!

Total Energy

2 2

0

1 1

2 2mgh mv I

trans rotE KE KE PE

0

2

2

/

mghv

m I r

2 21 1( )

2 2

vmv I

r

2

2

/

mghv

m I r

0v gh

0

4

3v gh

General equation

for the total final

velocity at the

end of the ramp:

Solid Disk has the

greatest velocity at the

bottom of the ramp!

Note: the velocity is

independent of the radius!

Total Kinetic Energy A 1.0-kg wheel in the form of a

solid disk rolls without slipping

along a horizontal surface with a

speed of 6.0 m/s. What is the total

kinetic energy of the wheel?

2 21 1

2 2mv I total trans rotK K K

2 2 21 1 1( )( )

2 2 2

vmv mr

r

2 21 1

2 4mv mv

23(1 )(6 / )

4kg m s23

4mv 27J

What I to use?

A tennis ball starts from rest and

rolls without slipping down the hill.

Treat the ball as a solid sphere. Find

the range x.

Rolling Ball Problem

2 2 21 1 2( )( )

2 2 5

vmv Mr

r

2 21 1

2 2 CM CMmgh mv I

10

7v gh

cosxR v t v t

210 sin

2v t gt

2 sinvt

g

cosxR v t v t 2 2sin cosv

g

(10 / 7 )2sin cosgh

g

10 sin 2

7

hR

g

= .25m

21

2y yy v t a t

Does change while in flight?

Angular Momentum

L I

finalL I

If no NET external Torques act on a

system then

Angular Momentum is Conserved.

IinitialL

Angular Momentum

L I

Angular Momentum

IL

A Skater spins with an angular speed of 2 rev/s. If she

brings her arms in and decreases her rotational inertia by

a factor of 5, what is her new angular speed in rev/s?

f f fL I 0 0 0L I

0 0 0 f f fL LI I

00f

f

I

I

2 // 5

Irev s

I

10 /rev s

Angular Momentum for a

Point Particle

r v L I

2 vmr mvr

r

L mvr

Single mass, a distance r from the axis of rotation.

68.37 10

625.1 10

8450 /

?

r x mP

r x mA

v m sP

vA

What is the velocity

of the satellite at apogee?

68.37 10

625.1 10

8450 /

?

r x mP

r x mA

v m sP

vA

Is the angular momentum of the system

CONSERVED?

A PL L

68.37 10

625.1 10

8450 /

?

r x mP

r x mA

v m sP

vA

A A P Pmr v mr v

PA P

A

rv v

r /2820m s

Angular Momentum is a Vector

Angular Speed is a Vector

When a rigid object rotates about an axis, the angular

momentum L is in the same direction as the angular velocity ,

according to the expression L = I, both directions given by the

RIGHT HAND RULE.

Angular Momentum is a Vector

The angular momentum L of a particle of mass m and linear momentum p located at

the vector position r is a vector given by L = r p. The value of L depends on the

origin about which it is measured and is a vector perpendicular to both r and p.

Only the perpendicular

component of p

contributes to L.

sin

L r p

L mvr

Torque Changes Angular Momentum

ext

L

t

S and L must be measured about the same origin

This is valid for any origin fixed in an inertial frame

Sext= dL/dt

Looks like

SFext= dp/dt

A particle whose mass is 2 kg moves in

the xy plane with a constant speed of

3 m/s along the direction r = i + j. What is

its angular momentum (in kg · m2/s)

relative to the origin?

a. 0 k

b. 6k

c. –6k

d. 6 k

e. –6 k

p

= 0

Only the

perpendicular

component of p

contributes to L!

sinL mvr

Torque Changes Angular Momentum

ext

L

t

The torque is perpendicular to both the applied force and the

lever arm, but parallel to the angular speed, angular acceleration

and angular momentum.

S and L must be measured about the same origin

This is valid for any origin fixed in an inertial frame

Sext= dL/dt

= r x F = dL/dt

COMPARE!

CROSS PRODUCT

F and d must be

mutually perpendicular! sin

r F

Fr Fd

sin

L r p

L mvr

cosW F d Fd

DOT PRODUCT

F and d must be

mutually PARALLEL!

CROSS PRODUCT

L and p must be

mutually perpendicular!

Colliding Probems

sin

L r p

L mvr

A particle of mass m = 0.10 kg and speed v0 = 5.0

m/s collides and sticks to the end of a uniform solid

cylinder of mass M = 1.0 kg and radius R = 20 cm.

If the cylinder is initially at rest and is pivoted about

a frictionless axle through its center, what is the

final angular velocity (in rad/s) of the system after

the collision?

a. 8.1

b. 2.0

c. 6.1

d. 4.2

e. 10

sin

L r p

L mvr

System of Particles Center of Mass

tot CMp vM

CMCM

( / )

v

rv

i i i i

i i

d m r M md

dt dt M

CM toti i i

i i

M m v v p p

tot,final tot,initial CM Mp p v

CM

1 v vi i

i

mM

CM tot

1

Mv p

Note: If the velocity of the CM is zero, the total momentum is zero!

If the sum of the net external torques is zero,

the system is in rotational equilibrium.

Chapter 12: Statics

Newton’s 1st Law: Conditions for Equilibrium

0

If the sum of the net external forces is zero,

the system is in translational equilibrium.

SF = 0

The ladder is 8m long and

weighs 355 N. The weight of the

firefighter is 875N and he stands

2.30m from the center of mass

of the ladder. If the wall is

frictionless, find the minimum

coefficient of friction of the

floor so that the ladder doesn’t

slip.

The Ladder Problem

What Keeps a Spinning Gyroscope

from Falling Down?

The “Couple” are

equal and opposite

forces so

SF = 0.

Since the normal

force is through the

axis of rotation P

(the support) it

produces no torque.

Only the weight of

the CM produces a

torque with lever

arm r that changes

the angular

momentum, L.

Precession Keeps it from Falling!

L

t

r L is in the direction of !

This causes the Precession!

P

Spinning Wheel What happens if you rotate the wheel while sitting in a spin stool?

The stool will spin in the direction of the torque.

Conservation of Angular Momentum!

The Gyroscopic Effect Once you spin a gyroscope, its axle wants to keep pointing

in the same direction. If you mount the gyroscope in a set

of gimbals so that it can continue pointing in the same

direction, it will. This is the basis of the gyro-compass

which is used in navigation systems.

What is the best gyroscope on

Earth?

Earth’s Precession

The Earth’s

precession is due to

the gravitational tidal

forces of the Moon

and Sun applying

torque as they attempt

to pull the Equatorial

Bulge into the plane

of Earth’s orbit.

The Earth’s

Precession causes the

position of the North

Pole to change over a

period of 26,000

years.

TOTAL Angular Momentum Orbital and Spin Angular Momentum

The TOTAL angular momentum includes both the angular

momentum due to revolutions (orbits) and rotations (spin).

It is the TOTAL angular momentum that is conserved.

Earth- Moon System:

Total Angular Momentum is Conserved!

•Earth Rotation Slowing due to friction of ocean on bottom

•.0023 s per century: 900 Million yrs ago, Earth day was 18 hrs!

•Decrease of Earth’s spin angular momentum, increases the

orbital angular momentum of the Moon by increasing the

distance, r, in order to keep L conserved!

•Earth is slowing down and Moon is moving further away!

Bouncing laser

beams off the

Moon

demonstrates

that it slowly

moving away

from the Earth

~.25 cm/month

Angular Momentum as a

Fundamental Quantity • The concept of angular momentum is also valid on

a submicroscopic scale

• Angular momentum has been used in the development of modern theories of atomic, molecular and nuclear physics

• In these systems, the angular momentum has been found to be a fundamental quantity

– Fundamental here means that it is an intrinsic property of these objects

– It is a part of their nature

Orbital Angular Momentum

& its Z-component are Quantized

h( 1) ( 0,1,2,... 1)

2L l l l n

h ( ,.. 1,0,1,2,... )

2Z l lL m m l l

Angular momentum magnitude is quantized:

Angular momentum direction is quantized:

Note: for n = 1, l = 0. This means that the ground

state angular momentum in hydrogen is zero, not

h/2, as Bohr assumed. What does it mean for

L = 0 in this model?? Standing Wave

Intrinsic Spin is Quantized

h 1( 1) ( )

2 2S s s s

h 1 1 ( , )

2 2 2Z s sS m m

Fine Line Splitting: B field due to intrinsic spin interacts with B

field due to orbital motion and produces additional energy states.

Spin magnitude is quantized:

Spin direction is quantized:

Cosmic Rotations: Conservation

of Angular Momentum

Does the Universe have NET

Angular Momentum? According the General Relativity,

The Universe is ISOTROPIC – it looks the

same in every direction. Net Angular

Momentum would define a direction.

Isotropic Universe

chapter 12: rotation of rigid bodies - santa rosa …lwillia2/40/40ch12_s14.pdf · chapter 12:...

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