Chapter 9-10 rotation of rigid bodies

kinematicskinematics dynamicsdynamics

1. model2.Angular quantity

1. model2.Angular quantity

Instantaneouseffect

Instantaneouseffect Time effectTime effectSpace effectSpace effect

theorem of rotation

theorem of rotation

1.Theorem ofangular momentum2.conservation of angular momentum

1.Theorem ofangular momentum2.conservation of angular momentum

1.work 2.kinetic and potential energy3.conservationof energy

1.work 2.kinetic and potential energy3.conservationof energy

Rotational motionRotational motion

Key terms:rigid bodyrotation translationradianrad angular velocityangular accelerationangular speedmoment of inertiarotational kinetic energyparallel-axis theorem

Chapter 9-10 rotation of rigid bodies

Key terms:torqueline of action lever armrotational analog of Newton’s second lawcombined translation and rotationangular momentumconservation of angular momentum

Chapter 9-10 rotation of rigid bodies

Types of Motion

Translation

The linear position of a body changes with respect to a fixed frame, but its angular orientation remains unchanged.

Rotation

The angular orientation of a body changes about a fixed frame of reference, but its linear position remains fixed.

Complex

Simultaneous combination of translation and rotation motions

1. Ideal Model-----Rigid Body

we neglect the deformations and assume that the body has a perfectly definite and unchanging shape and size.

Definite shape and definite size.

The distance between any two points on a rigid body remain unchanging.

A rigid body can be considered as a special collection of particles with mass continuously distributed.

A rigid body in general can have rotational as well as translational motion.

y

xo

rP

2. Angular Quantities

let’s think first about a rigid body that rotates about a fixed axis.

let’s think first about a rigid body that rotates about a fixed axis. (The axis passes through point O and is perpendicular to the x-y plane)

We can use the linear quantities ( r, v, a) as well as the angular quantities ( , , ) to describe the motion of the rigid body.

y

xo

rP

2.1 Angular Coordinate

radr

s

Counterclockwise(+)

s

12

2.2 Angular Velocity

s/raddt

d

rv

2.3 Angular Acceleration

22

2

s/raddt

d

dt

d

If is a constant (=c):t0

20 t2/1t

20

22

rs

2.4 Relating Linear and Angular Kinematics

y

xo

rP

s

v

We can use linear quantities ( r, v, a) or angular quantities ( , , ) to describe the rotational motion. The relationships between them are:

r2

2

n rr

a

rdt

dat

The linear quantities of any point on the rigid body are different, while angular quantities are the same.

3. Theorem of Rotation

Fr

3.1 torque (/moment) of a force

d

o

mi

r

F

x

y

z sinrF FdVector: right-hand rule.

axis//F,r//F,rF:discussion

force times lever-arm, tends to produce rotation.

i

iii

i FrFrFr

2111

Net torque is the vector sum of individual torques:

Example: a force on particle

find the torque on particle if the fixed axis

is in original point.

)( jyixF

Solution:

0)()1(

rrFr

0)()()2(

jyixjyixFr

o

mi

ir

3.2 Rotation Analog of Newton’s Second Law

exF

inf

The net torque on mi to point O:

Proof: A rigid body can be looked as a special collection of particles.

)ji(j

jiiiii frFr

The net torque on the rigid body to point O:

i )ji(j

jiii

iii

i frFr

0fri )ji(j

jii

z

I

i

iii

i Fr

i

iiii

iii rmramr )()( tan, ii

iii

ra

amF

tan,

tan,tan,

i

i2i )mr(

I

i )ji(j

jiii

iii

i frFr 0

)(

i jij

jii fr

i

,taniiFr

“”s are the same for every part of the rigid body.

Rotational analog of Newton’s law for a rigid body.

o

mi

ir

exF

z

i

ii mrI 2

Means external torque, the torque offered by a pair of internal force equals to zero

)1

moment of inertial, measurement of rotational inertial

)2 The angular acceleration of rigid body with fixed axis

0)3 The rigid body keep rest or rotate in uniform angular velocity

Caution:

0

1

, IGiven

I)4

I

i

2iirmI

1) A few point masses: i

2iirmI

2) Continuous distribution of mass: dmrI 2

3.3 Calculation of Moment of Inertial

Moment of inertial depends on the distribution of mass in the system.

Moment of inertia depends on the location and orientation of the axis.

Example: An engineer is designing a one-piece machine part consisting of three heavy connectors linked by light molded struts. The connectors can be considered as massive particles connected by massless rods.

a) What is the moment of inertia of this body about an axis through point A, perpendicular to the plane of the diagram?

b) What is the moment of inertia about an axis coinciding with rod BC?

(see page 277)

i

2iirmI 0.30m

0.40m

0.50m

mA=0.30kg

mB=0.10kg

mC=0.20kg

B

AC

Solution:

i

2iia rmI

a)

22 )m40.0)(kg20.0()m50.0)(kg10.0( 2m.kg057.0

i

2iib rmI 2)m40.0)(kg30.0( 2m.kg048.0

Solution:

a)

b)

0.30m

0.40m

0.50m

mA=0.30kg

mB=0.10kg

mC=0.20kg

B

AC

Example: A slender uniform rod with mass M and length L. It might be a baton held by a twirler in a marching band. Compute its moment of inertia about an axis through center of mass point O.

dmrI 2dmrI 2

dmx 2

2

L

2

L2dxx

L

M

2ML12

1

Cdx

dmx

xo

Solution:

22cm MRdmRI

Uniform thin loop (M ,R):

Uniform thin disc(M, R)

cI2r 2R

M

rdr2

2MR2

1

R

0

R

r dr

2cmp MdII

Icm: about axis through CM

M: total mass

d : distance between CM and point p.

2) Parallel-Axis Theorem: (p281)

CM

Icm

O

Ip

p d

0M

xdmxc o

M

ydmyc

2cmp MdII

dmbyaxdmrI p ])()[( 222

dmbaydmbxdmadmyxI p )(22)( 2222

o

Example: If the rod rotates about the axis through one end, Use parallel-axis theorem:

2cm MdII

22 ML4

1ML

12

1

2ML3

1

3.3 Application of rotational analog of Newton’s second law

ICaution:

1) The left is vector sum of external torque about a fixed axis, and the direction is parallel to axis.

2) “ I ” is the Moment of inertial

3) “” means the angular acceleration to fixed axis, same direction with torque

Example: In judo, a weaker and smaller fighter who understands physics can defeat a stronger and larger fighter who doesn’t. The fact is demonstrated by the basic ‘ hip throw ’, in which a fighter rotates his opponent around his hip.

IFd 1

ImgddF 21

FF

Solution:

Example: The massive shield door at a neutron test facility at Lawrence Livermore Laboratory is the world’s heaviest hinged door M=44000kg, I=8.7*104 kgm2, and face width 2.4m find the smallest force to move the door from rest through an angle of 900 in 30s.

y

O

Example: We wrap a light, flexible cable around a solid cylinder with mass M and radius R. The cylinder rotates with negligible friction about a stationary horizontal axis. We tie the free end of the cable to an object of mass m and release the object with no initial velocity. Find the acceleration of the object of mass m and the angular acceleration of the cylinder

M

R

m

h

N

MgTT

mg

Solution:

For m: )1(maTmg For M: )2()MR

2

1(TR 2

)3(Ra

Example: An Atwood’s machine. There is no slipping between the cord and the surface of the wheel. Find the linear accelerations of blocks A and B, the angular acceleration of the wheel C, and the tension in each side of the cord

a) If weight of cylinder is negligible ;b) If weight of cylinder is not negligible.

C

mAmB

R

m2gm1g

T2T1

T’2 T’

1

T

y

OSolution:

a) The weight of cylinder is negligible …

b)If weight of cylinder is not negligible:

For m1: )1(amTgm y1111

For m2: )2(amTgm y2222

For m1: )1(amTgm y1111 For m2: )2(amTgm y2222

For M: )3()MR2

1(RTRT 2'

1'2

)4(aa y2y1 )5(Ra y1

From (1),(2),(3),(4),(5) :

g2Mmm

mmaa

21

21y2y1

g2Mmm

)2Mm2(mT,g

2Mmm

)2Mm2(mT

21

122

21

211

C

m1m2

R

m2gm1g

T2T1

T’2 T’

1

T

y

O

IL

mg cos2

d

dI

dt

dI

Lmg cos

2

0

2

cos2

dIdL

mg

2

2

1)sin1(

2 I

Lmg

Solution:

Example: A grindstone 1.0 m in diameter, of mass 50 kg, is rotating at 900 rev.min-1. A tool is pressed against the rim with a normal force of 200N, and the grindstone comes to rest in 10s. Find the coefficient of friction between the tool and the grindstone. Neglect friction in the bearings.

Solution: t,I 0 ,Rfk kk Nf

s10t,0

)s/rad(3060/2900

t

0

)s/rad(3t/)( 20t

2MR2

1I 59.ok

Example: A wheel rotates about a fixed axis from rest. A constant torque,=20N.m, acts on it for 10 seconds. =100rev/min at t=10s. Then it takes 100 seconds for the wheel to come to stop again. Find the moment of inertial of the wheel with respect to the axis.

Solution: =I, = o+ t

20=I 1 ， 1= /t1 (for: o=0) 20- r=I1 ， 1= /t1 (for: o=0) (1)

When release external torque: - r=I2 , 2=- /t2 (2)

t1=10s , t2=100s , =(100×2)/60=10.5rad/s,

From (1) 、 (2): I=17.3kg.m2 。

acted by external torque:

r

iv

Pmi

4. Energy in Rotational Motion

O

The kinetic energy of mi :

2iii vm

2

1K 22

iirm2

1 iii rv

The rigid body : 2

i

2ii

i

2ii

ii )rm(

2

1vm

2

1K

i

iirmI 2

2I2

1K

4.1 Rotational Kinetic Energy

Example: Each of the helicopter rotor blades is 5.2m long, m=240kg. The rotor is rotating at 350rev/min.

Find 1) the rotational inertial of the rotor assemble about the axis of rotation.

2)what is the total kinetic energy of rotation.

4.2 Gravitational Potential Energy

i

ii gymU i

cmii MyymcmMgy

The gravitational potential energy is the same as though all the mass were concentrated at the center of mass of the rigid body.

4.3 Work and Power in Rotational Motion

sdFdW

2

1

dW

Work of the torque

Suppose a force F acts on the rigid body. The rigid body rotates through an infinitesimal angle d about a fixed axis. The work dW done by the force while point P moves distance ds is:

F

dsr

d

P

dsinFr d

work done by a torque

2

1

dW

CAUTION:

2

1

ddt

dI

dt

d

I

21

22 I

2

1I

2

1dI

2

1

2

1

dW

a work done by the forces doesn’t equal to the work done by the torques.

21

22 2

1

2

1 mmrdFW

For a particle:

dt

d

dt

dWP

Power of a torque:

Compare with the work-energy theorem for collection of particles:

21

22 2

1

2

1mvmvWtot

21

22 2

1

2

1 IIWtot

vFp

4.4 Conservation of Mechanical Energy

If the work done by all forces other than conservative force equals zero. The total mechanical energy of the rigid body is conservative.

CEWW innonext 0

y

O

Example: We wrap a light, flexible cable around a solid cylinder with mass M and radius R. The cylinder rotates with negligible friction about a stationary horizontal axis. We tie the free end of the cable to an object of mass m and release the object with no initial velocity. Find the speed of the falling object and the cylinder just as the object strikes the floor.

See page 280, example 9-9.

M

R

m

h

N

MgTT

mg

Solution:

No work is done by friction, so

mgh 2mv2

1)MR

2

1( 2 2)

R

v(

2

1

)1(2

1sin

222 I

Lmg

Lmg

Solution:

)2(3

1 2mlI

We get

Example: A uniform spherical shell of mass M and radius R rotates about a vertical axis on frictionless bearings. A massless cord passes around the equator of the shell, over a pulley .

Find the speed of the object when it falls a distance h from rest.

r

v

R

vps

mghE (1)

(2)

From the conservation law of energy

222

21

21

21

khIMvMgh

rvmrI ，221

mM

khMghv

21

2 2

Solution:

1)

Example: the given is as follows,find the velocity of the object and the acceleration of M 。

k

h

M

222

21

21

21

khIMvMgh

Example: the given is as follows,find the velocity of the object and the acceleration of M 。

Solution:

2)

Make a derivation

khR

aImamg

khvImvamgv

2

2R

Im

khmga

We get:

k

h

M

Alternative solution:

)2()( 21 IRTT

)3(2 khT

)4(Ra

We get:

2R

Im

khmga

k

h

M

)1(1 maTmg

Example: The power output of an automobile engine is advertised to be 200 hp at 6000 rpm. What is the corresponding torque?

1 HP = 746 WSolution:

Whp

WhphpP 51049.1)

1

746(200200

sradrev /628min/6000

mNsrad

smNP

237

/628

/1049.1 5

5. Angular Momentum

i

iivmP

0

Coordinate(x) --- Angular Coordinate()

Distance(s) --- Angular Displacement()

Velocity(v) --- Angular Velocity()

Acceleration(a) --- Angular Acceleration()

Momentum(P) --- ?

v

m

M, L

?v '

5.1 Angular Momentum of a particle

r

mvm

o

x

y

z

d

vmrprL

Unit: kg m2/s

sinrmvL

Direction: determined by right-hand rule for vector products.

Magnitude: mvd

L

o

m

r

F

x

y

z

d

Fr Compare with torque

Caution: The Value of L and both depends on the choice of origin O.

vmrprL

O

R

vm

L=mvR

P

O

mvd

To point O: L=mvd

To point P: L=0

Angular momentum of collection of particles:

i

iii

iii

i vmrprLL

r

mvm

o

x

y

z

d

L

5.2 Angular Momentum of a rigid body

The angular momentum of mi to point O:

iiii vmrL The angular momentum of the rigid body to the fixed axis (z-axis):

ii rv 2iirm

)rm(LLi

2ii

iiz

i

2iirmI

zz IL

iv

ir

im

O

xy

z

CAUTION: Giving angular momentum you must always specify about axis

Frdt

Ld

5.3 The Theorem of Angular Momentum

vmrprL

for one single particle:

dt

Ld

The rate of change of angular momentum of a particle equals the torque of the net Force acting on It.

dt

)pr(d

pdt

rd

dt

pdr

,dt

pdF

0

,

vmv

vdt

rd

122

1

2

1LLdLdt

L

L

t

t

dt

)L(di

i

ii

for collection of particles:

i )ji(j

jiii

iii

i frFr

ofri )ji(j

jii

i

iii

i Frdt

Ld

for rigid body:

dt

dLzz zz IL

dt

Id z )( zI

1122

2

1

2

1

)( IIIddtL

L

t

t

dt

Ld

Lddt

2

1

t

t 2

1

L

L

12 LL

If the torque of all the external forces equals 0:

,0i

i 12 LL

------conservation of the angular momentum for collection of particles or one single particle.

1

t

t

P

P 2 PPpddtF2

1

2

1

Compare with conservation of momentum:

,0i

i 12 LL

,0FFi

i �

12 PP�

Example: the disk rotate on the table, angular velocity and friction coefficient is , how much time it will take for the disk to stop its motion

r dr

Solution :

drgrrdmgd 22

RMgdR

3

2

0

12 IIdt

g

RIt

It

4

3

Theorem of angular momentum

When the net external torque acting on a system is zero, the total angular momentum of the system is constant.

5.4 Conservation of Angular Momentum

,0ex ,LL 12

2211 II

Example: The position of a object with mass m is given by r=acost i+bsint j (a, b, and are all constant), find the angular moment of the object and the net torque acting on it.

Solution: vmrL

jtsinbitcosar

jtcosbitsinadt

rdv

kji0tsinbtcosa 0tsinbmtsinam

L

kabm

amrFr

2

2

dt

rda

r2

0)( 2 rr

Example: A small block on a frictionless horizontal surface has a mass of m. it is attached to a massless cord passing through a hole in the surface. The block is originally revolving at a distance of r from the hole with an angular speed 0. Pull the cord from below slowly, shorten the radius of the circle to r/2. How much work was done in pulling the cord?

See page 322 10-33, 10-76, 10-87

F

0

ro

m

Solution: zeroisOtonet

21 LL:Lofonconservati

W

040

2mr 2)2r(m

22)2

r(m

2

1 20

2mr2

1 20

2mr2

3

Example: see page 314 example 10-16

s/m400v

1.00m

0.50m

Solution: Consider the door and bullet as a system, net torque about the axis through O equals zero, so angular momentum is conserved.

2

lmvL:Linitial 1

)II(L:Lfinal rodbul2

21 LL

momentumofonconservati

,)2

l(mI 2

bul ,Ml3

1I 2rod

s/rad40.0

:so

Example: see page 327 10-75

Solution:

A BC

S S

Consider A and B as a system, when remove the accelerating torque from A, angular momentum is conserved.

0AI )II( BA The thermal energy equals the lose of the total rotational kinetic energy:

Q 20AI2

1 2BA )II(

2

1

BA I3

1I,J2400Q

J3200I2

1K

:so

20A0A

20

220 )(

21

21

21

llkmvmv

v

O

l0

l

v0

d

We get ： v =4m/s ， =300

Solution:

sin00 mvllmv

Example ： a block with spring was in a frictionless surface, k=100N/m, a spring was fixed in o point ， m=1kg, the natural length of the spring l0=0.2m, the velocity of the block is v0=5m/s, perpendicular to the spring 。 When the spring rotate 900, the length of l=0.5m find the velocity of the block.

RMm

GmvR

MmGmv

321

21 22

0

Conservation of the angular momentum

v

v0

MR

OmA

CO’

3R

Solution:

)43(3sin 2

0

20

GMRv

Rv

We get ：

sin30 RmvRmv

Conservation of the energy

Example: a rocket’s mass is m, velocity is v0, oc=3R 。

Find the angle between v and v0 。 (the earth mass is M 、radius=R)

R

Example: A man of mass m runs around the edge of a horizontal turntable that is mounted on a frictionless vertical axis through its center. The velocity of the man, relative to the disk, is v. The turntable is rotating in the opposite direction with an angular velocity of , relative to the earth. The radius of the turntable is R, and its mass is M. Find the final angular velocity of the system if the man comes to rest, relative to the turntable.

Solution: Consider the man and turntable as a system, angular momentum is conserved.

discI 'mandisc )II( Rm

The equation is wrong, because conservation of L is only valid in inertial reference frame

R

vME

R

discI 'mandisc )II(

MEmanI

,mRI 2man ,MR

2

1I 2disc

2/Mm

2/MmR/mv'

EMME D D

Summary:

amF

I

zz

yy

xx

maF

maF

maF

)

dt

dva(,maF

)R

va(,maF

ttt

2

nnn

dmrI

rmI

2

i

2ii

2cmp MdII

Instantaneous Effect

(One particle)

(Rigid body)

rdFWb

atot

inex WW

)UU(W 12con

non,inex WW

Accumulation Effect of Space

(One particle)

(particles)

,0WWif non,inex

dWb

atot (Rigid body)

2cm I

2

1mgyE

21

22 mv

2

1mv

2

1

12 KK i

21ii

i

22ii vm

2

1vm

2

1

)UK()UK( 1122 12 EE

12 EE

21

22 I

2

1I

2

1

时间积累

dtFJ2

1

t

t

(One particle)

12

t

tPPdtFJ

2

1

(particles)

,0Fif

(Rigid body)dt

Ld

(One particle)

dt

Ld

(particles)dt

dLzz

2

1

t

tdt矩冲量

=,0if

12 PP

12 vmvm

i

1iii

2ii vmvm

12 PP

i

1ii

2i LL

221112 IIorLL

12 LL

6* Rigid-body Rotation about a Moving Axis

6.1 Combined translation and rotation: Energy Relations

cmcmv

mi

ri

cmv'

iv

iv

For mi: 'icmi vvv

)vv()vv(m2

1K '

icm'icmii

)vvv2v(m2

1 2'i

'icm

2cmi

For mi: )vvv2v(m2

1K 2'

i'icm

2cmii

For body:

)vm2

1(vvm

)vm2

1(KK

2'ii

'icmi

2cmii

0)vm(vvvm 'iicm

'icmi

2cm

2cm

2'ii

2cmi

I2

1Mv

2

1

)vm2

1()vm

2

1(K

1

2

3

4

R

2. Rolling without slipping

cmv

cmv

cmv

cmv

cmv

cm'1 vv

'2v

'3v

'4v

2v

3v

4v

0v1

Rvcm2

121 I

2

1Mv

2

1K

22cm )MRI(

2

10

2cm

2cm Mv

2

1I

2

1

Center of mass: cmi aMF

Rotation about axis through

CM:

cmI

Example13: See page 303 example 10-6.

Use two ways to find the speed vcm of solid cylinder after it has dropped a distance h.

h

Solution1: 2cm

2cm I

2

1Mv

2

1Mgh

Solution2:

Rvcm

T

Mg

)1(MaTMg cm

)2(ITR cm 2cm MR

2

1I

From (1),(2): mg3

1T,g

3

2acm

gh3

4vha2v cmcm

2cm

Ld

7* Gyroscopes and precession

d

O

p

IL

mg

r

dsinLdL

sinIdt

d mgp

dtdL mg

gmrmg