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Average Angular Acceleration (rads/s 2 )PowerPoint Presentation

Average Angular Acceleration (rads/s 2 )

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Presentation Transcript

- 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)?

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

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

- 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

- Replacing each of the translational variables in the translational kinematic equations by the rotational variables, gives the set of rotational kinematic equations (for constant )

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

Example Problem translational kinematic equations by the rotational variables, gives the set of rotational kinematic equations (for constant

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 ?

(a) Use last kinematic equation translational kinematic equations by the rotational variables, gives the set of rotational kinematic equations (for constant

(b) Use first kinematic equation

Or use the third kinematic equation translational kinematic equations by the rotational variables, gives the set of rotational kinematic equations (for constant

Example Problem

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,

determine the number of revolutions while on the way down. translational kinematic equations by the rotational variables, gives the set of rotational kinematic equations (for constant

y

v0

y0

yf

Solution:

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

Also, vy0 = 0, t0 = 0, y0 = 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

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

Tangential Velocity and Acceleration information about the x-component.

- For one complete revolution, the angular displacement is 2 rad
- From our studies of Uniform Circular Motion (Chap. 6), we know that the time for a complete revolution is a period T

v

r

z

- Therefore the angular velocity (frequency) can be written

- Also from Chap. 12, 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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