chapters 10, 11 rotation and angular momentum

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

0

lim

ttt if

ifavg

Angular acceleration

• Average angular acceleration

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

dt

d

tt

0

lim

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:

Chapter 10Problem 6

A rotating wheel requires 3.00 s to rotate through 37.0 revolutions. Its angular speed at the end of the 3.00-s interval is 98.0 rad/s. What is the constant angular acceleration of the wheel?

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 (recall Ch. 4)

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 (Ch. 4)

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 9 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

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

Mdmrrcom /

0/ˆˆ Mydmjxdmi

dmrI 2


dmbyax ])()[( 22

dmyx )( 22 dmba )( 22

R dmR )( 2 dmh )( 2

CMI 2Mh

2MhII CM


2MhII CM

Chapter 10Problem 22

Rigid rods of negligible mass lying along the y axis connect three particles. The system rotates about the x axis with an angular speed of 2.00 rad/s. Find (a) the moment of inertia about the x axis and the total rotational kinetic energy and (b) the tangential speed of each particle and the total kinetic energy. (c) Compare the answers for kinetic energy in parts (a) and (b).

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


An electric motor turns a flywheel through a drive belt that joins a pulley on the motor and a pulley that is rigidly attached to the flywheel. The flywheel is a solid disk with a mass of 80.0 kg and a diameter of 1.25 m. It turns on a frictionless axle. Its pulley has much smaller mass and a radius of 0.230 m. The tension in the upper (taut) segment of the belt is 135 N, and the flywheel has a clockwise angular acceleration of 1.67 rad/s2. Find the tension in the lower (slack) segment of the belt.

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 P (point of momentary contact between 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


A cylinder of mass 10.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed of 10.0 m/s. Determine (a) the translational kinetic energy of its center of mass, (b) the rotational kinetic energy about its center of mass, and (c) its total energy.

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

Fnet,x = M aCM,x

fs – M g sin θ = M aCM,x

R fs = ICM α

α = – aCM,x / R

fs = – ICM aCM,x / R2

21

sin

/ MRI

g a

CMCM,x

Rolling down a ramp

21

sin

/ MRI

g a

CMCM,x

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


A particle of mass m moves in a circle of radius R at a constant speed. The motion begins at point Q at time t = 0. Determine the angular momentum of the particle about point P as a function of time.

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

• Δmi is described by

im

ir

ip

Angular momentum of a rigid body

))(( iiiiz vmrL

i

izz LL i

iii vmr ))((

i

iii rmr )(2

)( i

ii rm zI

zz 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 projectile of mass m moves to the right with a speed v. The projectile strikes and sticks to the end of a stationary rod of mass M, length d, pivoted about a frictionless axle through its center. (a) Find the angular speed of the system right after the collision. (b) Determine the fractional loss in mechanical energy due to the collision.

Answers to the even-numbered problems

Chapter 10

Problem 4− 226 rad/s2


Chapter 10

Problem 16(a) 54.3 rev;(b) 12.1 rev/s


Chapter 10

Problem 2611mL2/12


Chapter 10

Problem 32168 N m clockwise⋅


Chapter 10

Problem 34(a) 1.03 s;(b) 10.3 rev


Chapter 10

Problem 48276 J


Chapter 11

Problem 4(a) 168°; (b) 11.9° principal value;(c) Only the first is unambiguous.


Chapter 11

Problem 12(− 22.0 kg m⋅ 2/s)ˆk

chapters 10, 11 rotation and angular momentum

Documents

rotation axis

body angular position

angular speed

angular variables

fixed axis rigid body

constant angular acceleration

angular velocityrotation

angular momentumrotation