Similar to instantaneous velocity, we can define the Instantaneous Angular Velocity

A change in the Angular Velocity gives

tt

0

lim

ttt f 0

0 Average Angular Acceleration (rads/s2)

Example

A fan takes 2.00 s to reach its operating angular speed of 10.0 rev/s. What is the average angular acceleration (rad/s2)?

Solution:

Given: tf=2.00 s, f=10.0 rev/s

Recognize: t0=0, 0=0, and that f needs to be converted to rads/s

2

2

srad

srad

0

0

srad

srad

srev

4.31

0.10000.2

00.208.62

0.20rev 1

rads 20.10

tt f

f

f

Equations of Rotational Kinematics Just as we have derived a set of equations to describe ``linear’’ or ``translational’’ kinematics, we can also obtain an analogous set of equations for rotational motion

Consider correlation of variables

Translational Rotational

x displacement

v velocity

a acceleration

t time t

)-(2)-(

)-()-(

)-)((2

1

020

200

202

1000

000

ff

ff

fff

fff

tttttt

tt

Replacing each of the translational variables in the translational kinematic equations by the rotational variables, gives the set of rotational kinematic equations

We can use these equations in the same fashion we applied the translational kinematic equations

Example: Problem 8.16A figure skater is spinning with an angular velocity of +15 rad/s. She then comes to a stop over a brief period of time. During this time, her angular displacement is +5.1 rad. Determine (a) her average angular acceleration and (b) the time during which she comes to rest.

Solution:

Given: f=+5.1 rad, 0=+15 rad/s

Infer: 0=0, f=0, t0=0

Find: , tf ?

(a) Use last kinematic equation

2srad

220

20

020

2

22rad) 2(5.1

rad/s) 15(

2

20)(2

f

f

ff

(b) Use first kinematic equation

s 68.0rad/s 15

rad) 1.5(22)0)(0(0

))((

0

021

0021

0

ff

ff

fff

t

ttt

Or use the third kinematic equation

s 68.0rad/s 22-

rad/s 150

)(

20

0

00

f

f

ff

t

ttt

Example: Problem 8.26

At the local swimming hole, a favorite trick is to run horizontally off a cliff that is 8.3 m above the water, tuck into a ``ball,’’ and rotate on the way down to the water. The average angular speed of rotation is 1.6 rev/s. Ignoring air resistance,

Solution:

Given: 0 = f = 1.6 rev/s, y0 = 8.3 m

Also, v0y = 0, t0 = 0, yf = 0

Recognize: two kinds of motion; 2D projectile motion and rotational motion with constant angular velocity.

Method: #revolutions = = t. Therefore, need to find the time of the projectile motion, tf.

determine the number of revolutions while on the way down.

x

y

y0

yf

v0

Consider y-component of projectile motion since we have no information about the x-component.

rev 2.1s) rev/s)(1.3 6.1())((

s 3.1 9.80

m) 3.8(22

02

1v

0021

0

sm

0

221

02

21

0

20000

2

f

ffff

f

ff

fyfyf

ttt

g

yt

gtygty

)-t(ta)-t(tyy

Tangential Velocity and Acceleration

z

r

v

(rad/s) 2

Tt

For one complete revolution, the angular displacement is 2 rad

From our studies of Uniform Circular Motion (Chap. 5), we know that the time for a complete revolution is a period T Therefore the angular frequency can be

written

Also from Chap. 5, we know that the speed for an object in a circular path is

TrT

vr 2

v Tangential velocity (rad/s)

The tangential velocity corresponds to the speed of a point on a rigid body, a distance r from its center, rotating at an angular velocity Each point on the rigid body rotates at the same angular velocity, but its tangential speed depends on its location r

r

vT

r=0