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Average Angular Acceleration (rads/s 2 )

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Similar to instantaneous velocity, we can define the Instantaneous Angular Velocity A change in the Angular Velocity gives. Average Angular Acceleration (rads/s 2 ). Example

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Presentation Transcript
slide1

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

  • A change in the Angular Velocity gives

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)?

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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

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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
  • TranslationalRotational
  • x displacement 
  • v velocity 
  • a acceleration 
  • t time t
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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 16
Example: Problem 8.16

A 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 ?

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(a) Use last kinematic equation

(b) Use first kinematic equation

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Or use the third kinematic equation

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,

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determine the number of revolutions while on the way down.

y

v0

y0

yf

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.

x

tangential velocity and acceleration
Tangential Velocity and Acceleration
  • 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

v

r

z

  • Therefore the angular frequency can be written
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Also from Chap. 5, we know that the speed for an object in a circular path is

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 

vT

  • Each point on the rigid body rotates at the same angular velocity, but its tangential speed depends on its location r

r

r=0

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