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Chapter 8: Rotational Motion

• Rotation (rigid body) versus translation (point particle)• Rotation concepts and variables• Rotational kinematic quantities

Angular position and displacement Angular velocity Angular acceleration

• Rotation kinematics formulas for constant angular acceleration

“Radian”

radians) (in r slength arc r

srad

Example: r = 10 cm, = 100 radians s = 1000 cm = 10 m.

Definition:• 2 radians = 360 degree o

oo

.π

π

radian 357180

2

3601

s

r’

“radian” : more convenient unit for angle than degree

( )2

2

in rads r r

2

Rigid body: A “rigid” object, for which the position of each point relative to all other points in the body does not change.

Rigid body can still have translational and rotational motion.

Rigid body

Example: Solid: Rigid body Liquid: Not rigid body

• By convention, is measured CCW from the x-axis

• It keeps increasing past 2, can be negative, etc.

• Each point of the body moves around the axis in a circle with some specific radius

x

y

rigid bodyrotation axis “o” fixed to body

parallel to z-axis

Reference line rotates with body

Angular position of rotating rigid body

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Angular displacement: • Net change in the angular coordinate

rad.) in angle (aninitalfinal

Arc length: s• Measures distance covered by a point as it moves

through (constant r) yReference line rotating with body

x

s = r

o

fr r

arc) circular a along distance (a rs

x

y

rigid bodyrotation axis “o” fixed to body

parallel to z-axis

Reference line rotates with body

Angular displacement of rotating rigid body

Rigid body rotation: angular & tangential velocity

Tangential velocity vT:• Rate at which a point sweeps out arc length along

circular path

Tv r

Angular velocity :• Rate of change of the angular displacement

dt

d

t

tLim

t instave

0

• Units: radians/sec. Positive in Counter-Clock-Wise sense

• Frequency f = # of complete revolutions/unit time

• f = 1/T T = period (time for 1 complete revolution

/2f /T2f 2x

vT

r

For any point, r is the perpendiculardistance to the rotation axis

s r sr

t t

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1.1. The period of a rotating wheel is 12.57 seconds. The radius of the wheel is 3 meters. It’s angular speed is closest to:

iClicker Quiz

A. 79 rpmB. 0.5 rad/sC. 2.0 rad/sD. .08 rev/sE. 6.28 rev/s

1.2. A point on the rim of the same wheel has a tangential speed closest to:

A. 12.57 rev/sB. 0.8 rev/sC. 0.24 m/sD. 1.5 m/sE. 6.28 m/s

/T2f 2

rvT

rs

A ladybug sits at the outer edge of a merry-go-round, and a gentleman bug sits halfway between her and the axis of rotation. The merry-go-round makes a complete revolutiononce each second. The gentleman bug’s angular velocity is

A. half the ladybug’s.B. the same as the ladybug’s.C. twice the ladybug’s.D. impossible to determine

GL

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A ladybug sits at the outer edge of a merry-go-round, and a gentleman bug sits halfway between her and the axis of rotation. The merry-go-round makes a complete revolutiononce each second. The gentleman bug’s velocity is

A. half the ladybug’s.B. the same as the ladybug’s.C. twice the ladybug’s.D. impossible to determine

GL

Rigid body rotation: angular acceleration

Angular acceleration • Rate of change of the angular velocity inst

0 =

tavet

dLim

t dt

• Units:

• CCW considered positive

• for CONSTANT tf 0

2rad/s

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1D and Angular Kinematics Equations(Same mathematical forms)

dt

dva

dt

dxv

1D motion withconstant acceleration a

x, v, a

Angular motion withconstant angular acceleration

dt

d

dt

d

variables

Definitions

KinematicEquations

atv)t(vf 0

22

100 attvx)t(xf

]xx[av)t(v ff 020

2 2

t)t(f 0

22

100 tt)t(f

][)t( ff 020

2 2

At t = 0, a wheel rotating about a fixed axis at a constant angular acceleration has an angular velocity of 2.0 rad/s. Two seconds later it has turned through 5.0 complete revolutions. Find the angular acceleration of this wheel?

Example: Wheel rotating and accelerating

t)t(f 0

22

100 tt)t(f

][)t( ff 020

2 2

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Rigid body rotation: radial and tangential accelerationCentripetal (radial) acceleration ac or ar• Radial acceleration component, points toward rotation axis

22

) (use v r T

Tr

va r

r r rF ma

x

vT

rac

aT

Tangential acceleration aT:• Tangential acceleration component

• Proportional to angular acceleration α and also to radius r

• Units: length / time 2

r Ta tangential TF ma

Rotation variables: angular vs. linear

T

rΔθ

v rω

s

22 =r T

r

va

r

Ta r

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Rolling Motion (Without Slipping)

vcenter

s R

sR

t t

centerv R

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Rotational Dynamics

• Moment of inertia – rotational analog of mass• Torque – rotational analog of force

We want something like “F=ma” for rotational motion…..

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Something like mass for rotational motion: Moment of Inertia, I

Kinetic energy of ladybug and gentlemanbug

GL

This image cannot currently be displayed.

2 2L L G GI m r m r

2 2 21 1 2 2 3 3 ...I m r m r m r Generally, 21

2K IKinetic energy:

Example: Find moment of inertia for a crossed dumbbell

•Four identical balls as shown: m = 2 kg

•Connected by massless rods: length d = 1 m.

m

m

m m

d

d

d dAB C

d 2

A) Choose rotation axis perpendicular to figure through point “A”

B) Now choose axis perpendicular to figure through point “B”

C) Let rotation axis pass through points “B” and “C”

Rotational inertia I depends on axis chosen

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Calculation of Moment of inertia for continuous mass distributions requires “Integration, a kind of calculus”.We will just use the result.

Axis

Moments of Inertia of Various Rigid Objects

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What makes things rotate?What changes angular velocity?What causes angular acceleration?

“Torque”

Now we want to define “torque, τ”, so that “τ = I α”.

F

rp

r

FT

axis

m T TF ma m r Newton’s Law along tangential direction

2 TrF m r I

Multiplying “r”, so that we have “I” on right side

So, let’s define torque as TrF I Then we get

sinTF F Since

sinT prF rF r F

sinpr r and

(Torque τ could be either positive (Counter-Clockwise) or negative(Clockwise)

Line of action

Moment arm

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net I

1 2 3 ...net For multiple forces

F

rp

r

FT

axis

m

sinT prF rF r F

If r = 0, torque is zero.

If theta = 0 or 180 degree, the torque is zero.

m1=100 kg adult, m2=10 kg baby.Distance to fulcrum point is 1 m and 11 m respectively.The seesaw starts at horizontal position from rest. Which direction will it rotates?

(a) Counter-Clockwise(b) Clockwise(c) No rotation(d) Not enough information

m1 m2

Example: Find the net torque, moment of inertia, and initial angular acceleration. Choose axis of rotation through fulcrum point.