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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.

1 Answer

6 votes

Answer:

5.74s

Step-by-step 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^2is 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 = (\Delta \omega)/(\alpha) = (-22.4 - 22.4)/(-7.8) = 5.74s

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