chapters 10, 11 rotation and angular momentum. rotation of a rigid body we consider rotational...

Chapters 10, 11

Rotation and angular momentum

Rotation of a rigid body

• We consider rotational motion of a rigid body about a fixed axis

• Rigid body rotates with all its parts locked together and without any change in its shape

• Fixed axis: it does not move during the rotation

• This axis is called axis of rotation

• Reference line is introduced

Angular position

• Reference line is fixed in the body, is perpendicular to the rotation axis, intersects the rotation axis, and rotates with the body

• Angular position – the angle (in radians or degrees) of the reference line relative to a fixed direction (zero angular position)

Angular displacement

• Angular displacement – the change in angular position.

• Angular displacement is considered positive in the CCW direction and holds for the rigid body as a whole and every part within that body

if

Angular velocity

• Average angular velocity

• Instantaneous angular velocity – the rate of change in angular position

dt

d

tt

0lim

ttt if

ifavg

Angular acceleration

• Average angular acceleration

• Instantaneous angular acceleration – the rate of change in angular velocity

dt

d

tt

0lim

ttt if

ifavg

Rotation with constant angular acceleration

• Similarly to the case of 1D motion with a constant acceleration we can derive a set of formulas:

Relating the linear and angular variables: position

• For a point on a reference line at a distance r from the rotation axis:

• θ is measured in radians

rs

Relating the linear and angular variables: speed

• ω is measured in rad/s

• Period

rs dt

dsv

dt

dr

r

v

rT

2

dt

rd )(

2

Relating the linear and angular variables: acceleration

• α is measured in rad/s2

• Centripetal acceleration

dt

dvat

dt

dr

r

r

vac

2

r

r 2)(

dt

rd )(

r2

Rotational kinetic energy

• We consider a system of particles participating in rotational motion

• Kinetic energy of this system is

• Then

i

iivmK

2

2

i

iivmK

2

2

i

iii rm

2

)( 2 i

ii rm 22

)(2

Moment of inertia

• From the previous slide

• Defining moment of inertia (rotational inertia) as

• We obtain for rotational kinetic energy

2

2IK

i

ii rmK 22

)(2

i

ii rmI 2)(

Moment of inertia: rigid body

• For a rigid body with volume V and density ρ(V) we generalize the definition of a rotational inertia:

• This integral can be calculated for different shapes and density distributions

• For a constant density and the rotation axis going through the center of mass the rotational inertia for 8 common body shapes is given in Table 10-2 (next slide)

volume

dVrI 2 dmr 2


• The rotational inertia of a rigid body depends on the position and orientation of the axis of rotation relative to the body

Chapter 10Problem 25

Four equal masses m are located at the corners of a square of side L, connected by essentially massless rods. Find the rotational inertia of this system about an axis (a) that coincides with one side and (b) that bisects two opposite sides.

Parallel-axis theorem

• Rotational inertia of a rigid body with the rotation axis, which is

perpendicular to the xy plane and

going through point P:

• Let us choose a reference frame, in which the center of mass coincides with the origin

volumevolume

dmrdVrI 22

dmrI 2


dmbyax ])()[( 22

dmyx )( 22 dmba )( 22

ydmbxdma 22

MdmyjxirCM /)ˆˆ(

0/ˆˆ Mydmjxdmi

dmrI 2


dmbyax ])()[( 22

dmyx )( 22 dmba )( 22

R dmR )( 2 dmh )( 2

CMI 2Mh

2MhII CM


2MhII CM


A uniform rectangular flat plate has mass M and dimensions a by b. Use the parallel-axis theorem in conjunction with Table 10.2 to show that its rotational inertia about the side of length b is Ma2/3.

Torque

• We apply a force at point P to a rigid body that is

free to rotate about an axis passing through O

• Only the tangential component Ft = F sin φ of the

force will be able to cause rotation

Torque

• The ability to rotate will also depend on how far from the rotation axis the force is applied

• Torque (turning action of a force):

• SI unit: N*m (don’t confuse with J)

))(sin())(( rFrFt

Torque

• Torque:

• Moment arm: r┴= r sinφ

• Torque can be redefined as:

force times moment arm

τ = F r┴

))(sin())(( rFrFt )sin)(( rF

Newton’s Second Law for rotation

• Consider a particle rotating under the influence of a force

• For tangential components

• Similar derivation for rigid body

rFt

I

rmat rrm )( )( 2mr I

Newton’s Second Law for rotation

I

i

i


A 2.4-kg block rests on a slope and is attached by a string of negligible mass to a solid drum of mass 0.85 kg and radius 5.0 cm, as shown in the figure. When released, the block accelerates down the slope at 1.6 m/s2. Find the coefficient of friction between block and slope.

Rotational work

• Work

• Power

• Work – kinetic energy theorem

dsFdW t rdFt d f

i

dW

dt

dWP

WII

K if 22

22

dt

d

Corresponding relations for translational and rotational motion

Smooth rolling

• Smooth rolling – object is rolling without slipping or bouncing on the surface

• Center of mass is moving at speed vCM

• Point of momentary contact between the two

surfaces is moving at speed vCM

s = θR

ds/dt = d(θR)/dt = R dθ/dt

vCM = ds/dt = ωR

Rolling: translation and rotation combined

• Rotation – all points on the wheel move with the

same angular speed ω

• Translation – all point on the wheel move with the

same linear speed vCM

Rolling: translation and rotation combined

22

22CMCM MvI

K

Rolling: pure rotation

• Rolling can be viewed as a pure rotation around the

axis P moving with the linear speed vcom

• The speed of the top of the rolling wheel will be

vtop = (ω)(2R)

= 2(ωR) = 2vCM

Friction and rolling

• Smooth rolling is an idealized mathematical description of a complicated process

• In a uniform smooth rolling, P is at rest, so there’s no tendency to slide and hence no friction force

• In case of an accelerated smooth rolling

aCM = α R

fs opposes tendency to slide

Rolling down a ramp

22

22ff

i

IMvMgh

2

22

22 MR

Ivvgh ff

i

IMR

MRghv i

f

2

22


What fraction of a solid disk’s kinetic energy is rotational if it’s rolling without slipping?

Vector product of two vectors

• The result of the vector (cross) multiplication of two vectors is a vector

• The magnitude of this vector is

• Angle φ is the smaller of the two angles between and

cba

sinabc

b

a

Vector product of two vectors

• Vector is perpendicular to the plane that contains vectors and and its direction is determined by the right-hand rule

• Because of the right-hand rule, the order of multiplication is important (commutative law does not apply)

• For unit vectors

)( baab

c

b

a

ii ˆˆ 0 kkjj ˆˆˆˆ

ji ˆˆ k̂ ikj ˆˆˆ jik ˆˆˆ

Vector product in unit vector notation

)ˆˆˆ()ˆˆˆ( kbjbibkajaiaba zyxzyx

ibia xxˆˆ

jbia yxˆˆ

kabbajabba

iabbaba

yxyxxzxz

zyzy

ˆ)(ˆ)(

ˆ)(

)ˆˆ( iiba xx 0

)ˆˆ( jiba yx kba yxˆ

Torque revisited

• Using vector product, we can redefine torque (vector) as:

Fr

Fr

Fr

sinrF Fr sin

Angular momentum

• Angular momentum of a particle of mass m and

velocity with respect to the origin O is defined as

• SI unit: kg*m2/s

)( vrmprL

v

Newton’s Second Law in angular form

)( vrmprL

v

dt

rd

dt

vdrm

dt

Ld

vvarm

arm

amr

netFr

i

iFr

i

i net

netdt

Ld

Angular momentum of a system of particles

n

nLL

n

n

dt

Ld

dt

Ld

n

nnet ,net

netdt

Ld

Angular momentum of a rigid body

• A rigid body (a collection of elementary masses

Δmi) rotates about a fixed axis with constant angular

speed ω

• For sufficiently symmetric objects:

IL

Conservation of angular momentum

• From the Newton’s Second Law

• If the net torque acting on a system is zero, then

• If no net external torque acts on a system of particles, the total angular momentum of the system is conserved (constant)

• This rule applies independently to all components

netdt

Ld

0dt

Ld

constL

constLxxnet 0,


constIL

iiI ffI


constL

More corresponding relations for translational and rotational motion


A skater has rotational inertia 4.2 kg·m2 with his fists held to his chest and 5.7 kg·m2 with his arms outstretched. The skater is spinning at 3.0 rev/s while holding a 2.5-kg weight in each outstretched hand; the weights are 76 cm from his rotation axis. If he pulls his hands in to his chest, so they’re essentially on his rotation axis, how fast will he be spinning?

Questions?

Answers to the even-numbered problems

Chapter 10

Problem 240.072 N m⋅


Chapter 10

Problem 302.58 × 1019 N m⋅


Chapter 10

Problem 40hollow


Chapter 11

Problem 1669 rad/s; 19° west of north


Chapter 11

Problem 18(a) 8.1 N m kˆ⋅(b) 15 N m kˆ⋅


Chapter 11

Problem 241.7 × 10-2 J s⋅


Chapter 11

Problem 26(a) 1.09 rad/s(b) 386 J


Chapter 11

Problem 30along the x-axis or 120° clockwise from the x-axis


Chapter 11

Problem 4226.6°

chapters 10, 11 rotation and angular momentum. rotation of a rigid body we consider rotational...

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