Angular position and displacement Angular velocity...

1

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:

• 2p radians = 360 degree

ooo

.π

π

radian 357180

2

3601

s

r ’

“radian” : more convenient unit for angle than degree

( )2

2

in rads r r

p

p

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 2p, can be

negative, etc.

• Each point of the body moves around

the axis in a circle with some specific

radius x

y

rigid body rotation axis “o” fixed to body

parallel to z-axis

Reference line rotates with body

Angular position of rotating rigid body

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) y Reference line rotating with body

x

s = r

o

f r r

arc) circular a along distance (a rs

x

y

rigid body rotation 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

tinstave

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 ppp 2x

vT

t

r

For any point, r is the perpendicular

distance to the rotation axis

s r s

rt t

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 rpm

B. 0.5 rad/s

C. 2.0 rad/s

D. .08 rev/s

E. 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/s

B. 0.8 rev/s

C. 0.24 m/s

D. 1.5 m/s

E. 6.28 m/s

/T2f pp 2

rvT

rs

Rigid body rotation: angular acceleration

Angular acceleration a:

• Rate of change of the angular velocity inst0

=t

avet

dLim

t dt

a a

• Units:

• CCW considered positive

• for CONSTANT a: tf a 0

2rad/s

1D and Angular Kinematics Equations (Same mathematical forms)

dt

dva

dt

dxv

1D motion with

constant acceleration a

x(t), v(t), a(t) (t), (t), a(t)

Angular motion with

constant angular acceleration a

dt

d

dt

da

variables

Definitions

Kinematic

Equations

atv)t(vf 0

2

2

100 attvx)t(xf

]xx[av)t(v ff 020

22

t)t(f a 0

2

2

100 tt)t(f a

][)t( ff 020

22 a

Rotational variables are vectors, having direction

The angular displacement, speed, and acceleration

( , , a )

are vectors with direction.

The directions are given by the right-hand rule:

Fingers of right hand curl along the angular direction (See Fig.)

Then, the direction of thumb is the direction of the angular quantity.

at t = 0:

Example:

A grindstone is rotating with constant angular

acceleration about a fixed axis in space.

Initial conditions

Positive directions:

right hand rule

rad/s 4.6 - rad/s 0.35 2 a 0

When is = 0 again in addition to t=0?

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 a 0

2

2

100 tt)t(f a

][)t( ff 020

22 a

Rigid body rotation: radial and tangential acceleration

Centripetal (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

,a

r ac

aT

Tangential acceleration aT:

• Tangential acceleration component

• Proportional to angular acceleration α and also to radius r

• Units: length / time 2

r Ta a tangential TF ma

Rotation variables: angular vs. linear

T

rΔθ

v rω

s

22 =r T

r

va

r

Ta ra

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 revolution

once 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

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 revolution

once 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

17

Rotational Dynamics

• Moment of inertia – rotational analog of mass

• Torque – rotational analog of force

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

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

Kinetic energy of ladybug and gentlemanbug

G L

( )

2 2 2 2

2 2 2 2 2 2 2 2

1 1 1 1( ) ( )

2 2 2 2

1 1 1 1

2 2 2 2

L L G G L L G G

L L G G L L G G

K m v m v m r m r

m r m r m r m r I

2 2

L L G GI m r m r 2 2 2

1 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 d

A B 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

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

Moments of Inertia of Various Rigid Objects

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

F

rp

r

FT

axis

m T TF ma m r a

Newton’s Law along tangential direction

2 TrF m r Ia a

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

So, let’s define torque as TrFt

It aThen we get

sinTF F Since

sinT prF rF r Ft

sinpr r and

(could be positive or negative)

net It a

1 2 3 ...nett t t t

For multiple forces

F

rp

r

FT

axis

m

sinT prF rF r Ft

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.

PP10606-49*: When she is launched from a springboard, a diver's angular speed

about her center of mass changes from zero to 6.20 rad/s in 220 ms. Her rotational

inertia about her center of mass is constant at 12.0 kg·m2. During the launch, what

are the magnitudes of (a) her average angular acceleration and (b) the average

external torque on her from the board?

Example: second law for rotation

I totnet at

F=5 N

5 N tangential force is applied at the edge of a uniform disk

of radius 2 m and mass of 8 kg.

Find angular acceleration.

Formula: 21

2I MR

Axis of rotation

1m

F=5 N

5 N tangential force is applied as in the figure, on a uniform disk

of radius 2 m and mass of 8 kg.

Find angular acceleration.

Formula: 21

2I MR

Axis of rotation

F=5 N

5 N force is applied at 1 m from the center of a uniform disk of

radius 2 m and mass of 8 kg.

Find torque.

Axis of rotation

45 degree

Torque on extended object by gravitational force

Assume that the total gravitational force effectively

acts at the center of mass.

Gravitational potential energy of extended object

M g H, where H is the height of the center of mass and M

is the total mass.

Axis of rotation

Horizontal uniform rod

of length L & mass M

iClicker Q

Find the torque by gravitational force.

A. LMg

B. (L/2)Mg

C. 2LMg

D. (3/2)LMg

E. None of the above

Find the angular acceleration. 2

,

1

3end rodI ML

Axis of rotation

Horizontal uniform rod of length L & mass M is released from rest.

Example of energy conservation

Find its angular speed at the lowest point, assuming no friction

between axis of rotation and the rod.

Initial

Final

A thin uniform rod (length = 1.2 m, mass = 2.0 kg) is

pivoted about a horizontal, frictionless pin through one

end of the rod. (The moment of inertia of the rod about

this axis is ML2/3.) The rod is released when it makes an

angle of 37° with the horizontal. What is the angular

acceleration of the rod at the instant it is released?

Example

Example: Torque and Angular Acceleration of a Wheel

r

a

mg

•Cord wrapped around disk, hanging weight • Cord does not slip or stretch • Let m = 1.2 kg, M = 2.5 kg, r =0.2 m • Find acceleration of mass m, find angular acceleration a for disk, tension, and torque on the disk

2

2

1MrI Formula:

Rolling : motion with translation and rotation about center of mass

vcm

total rot cmK K K

21

2rot cmK I

21

2cm cmK Mv

gravity cmU Mgh

mech totE K U

For rolling without slipping

s R

cmv R

mech non conservativeE W

Example: Use energy conservation to find the speed of the bowling ball as it rolls w/o slipping to the bottom of the ramp

Rotation accelerates if there is friction between the sphere and the ramp Friction force produces the net torque

and angular acceleration.

There is no mechanical energy change because the contact point is always at rest relative to the surface, so no work is done against friction

Formula: For a solid sphere

2

5

2MRIcm

Hint:

Given: h=2m

iClicker Q: A solid sphere and a spherical shell of the same radius r and same mass M roll to the bottom of a ramp without slipping from the same height h. True or false? : “The two have the same speed at the bottom.”

A) True

B) False. Shell is faster.

C) False. Solid sphere is faster.

D) Not enough information.

I_(cm, spherical shell) = (2/3) MR^2

I_(cm, solid sphere)=(2/5) MR^2