Question

A spinning wheel on a fireworks display is initially rotating in a counterclockwise direction. The wheel has an angular acceleration of -7.80 rad/s2. Because of this acceleration, the angular velocity of the wheel changes from its initial value to a final value of -22.4 rad/s. While this change occurs, the angular displacement of the wheel is zero. (Note the similarity to that of a ball being thrown vertically upward, coming to a momentary halt, and then falling downward to its initial position.) Find the time required for the change in the angular velocity to occur.

Answer 1

Answer:

5.74s

Explanation:

We can first solve for the initial angular velocity using the following formula

$\omega^2 - \omega_0^2 = 2\alpha\theta$

Where $\omega = -22.4rad/s$ is the final angular velocity, $\alpha = -22.4 rad/s^2$is the angular acceleration and $\theta = 0$ is the angular displacement

$22.4^2 - \omega_0^2 = 2*(-7.8)*0$

$\omega_0^2 = 22.4^2$

$\omega_0 = 22.4rad/s$

So for the wheel to get from 22.4 to -22.4 with angular acceleration of -7.8 then the time it takes must be

$t = \frac{\Delta \omega}{\alpha} = \frac{-22.4 - 22.4}{-7.8} = 5.74s$