Kinematics of Rotation of Rigid Bodies

Angle of rotationAngular displacement Δθ = θ – θ0

Δθ > 0 if rotation is counterclockwiseΔθ < 0 if rotation is clockwise

z

Angle of rotation θ is a dimensionless quantity, but it is a vector

zi

rs

RadiuslengthArcradiansin )( If θ is in radians, then

s=rθ and s’=r’θ with the same θ.

r

s

r’s’

00

0 3.5723601360221

radrad

rrrevolution

Example:A total eclipseof the sun

Average angular velocityAngular Velocity and Acceleration

rpmrevors

radttt zz

min

,0

0

Instantaneous angular velocity

zz idtd

t

,lim0t

z z

ωz>0 ωz<0

Average angular acceleration

20

0 ,srad

ttt zzzz

z

Instantaneous angular acceleration

zzzz

z idtd

dtd

t

,lim 2

2

0t

Instantaneous angular speed is a scalar .||

Kinematic Equations of Linear and Angular Motionwith Constant Acceleration

(1)

(3)

(4)

(2)

Rotational motion Quantity Linear motionθ Displacement x

ω0z Initial velocity v0x

ωz Final velocity vx

αz Acceleration ax

t Time t

Relations between Angular and Tangential Kinematic Quantities

Ice-skating stunt“crack-the-whip”

dtdr

dtdv

dtdr

dtds

rs

Centripetal and Tangential Accelerations in Rotational Kinematics

Exam Example 22: Throwing a Discus (example 9.4)

Rotational Kinetic Energy and Moment of InertiaKinetic energy of one particle 222

21

21 mrmvK T

Rotational kinetic energy is the kinetic energy of the entire rigid body rotating with the angular speed ω

222

21

21 IKrmK R

iiiR

Definition of the moment of inertia i V

ii dVrrmI 22

Total mechanical energy

UIMvE cmcm 22

21

21

Translationalkinetic energy

Rotational kinetic energy cmr

r

'rcmv

0cm

Parallel-Axis Theorem 2cmcm MrII

Proof:

i

iicmi i

cmii

iicmi rmrrmrmrrmI '2)(')'( 222 =0

Potentialenergy

ω

vh

Rotation about an Axis Shifted from a Center of Mass Position

Exam Example 23: Blocks descending over a massive pulley (problem 9.75)

Rm1

m2

gm

2

gm1

1NF

2T

1T

2T

1T

x

y

F

ay

ω

0

Δy

Data: m1, m2, μk, I, R, Δy, v0y=0Find: (a) vy; (b) t, ay; (c) ω,α; (d) T1, T2

kf

Solution: (a) Work-energy theoremWnc= ΔK + ΔU, ΔU = - m2gΔy,Wnc = - μk m1g Δy , since FN1 = m1g,ΔK=K=(m1+m2)vy

2/2 + Iω2/2 = (m1+m2+I/R2)vy2/2 since vy = Rω

221

1212

2221 /

)(2)(21

RImmmmygvmmygv

RImm k

yky

(b) Kinematics with constant acceleration: t = 2Δy/vy , ay = vy2/(2Δy)

(c) ω = vy/R , α = ay/R = vy2/(2ΔyR)

(d) Newton’s second law for each block: T1x + fkx = m1ay → T1x= m1 (ay + μk g) ,

T2y + m2g = m2ay → T2y = - m2 (g – ay)

Moments of Inertia of Various BodiesScaling law and order of magnitude I ~ ML2

22

2

31

2121 MLLMMLI

Moment of Inertia CalculationsCylinder rotating about axis of symmetry

2122

21

22

41

42

32

242

2)2(2

1

2

1

RRRRLRRL

drrLrLdrrIR

R

R

R

The total mass of the cylinder is

)( 21

22 RRLVM

Result: )(21 2

221 RRMI