A bicycle tire is spinning counterclockwise at $3.80 \mathrm{rad} / \mathrm{s}$. During a time period $\Delta t=2.50 \mathrm{~s}$, the tire is stopped and spun in the opposite (clockwise) direction, also at $3.80 \mathrm{rad} / \mathrm{s}$. Calculate the change in the tire's angular velocity $\Delta \omega$ and the tire's average angular acceleration $\alpha_{\mathrm{av}}$. (Indicate the direction with the signs of your answers.)

HINT
(a) the change in the tire's angular velocity $\Delta \omega$ (in rad/s) $\mathrm{rad} / \mathrm{s}$
(b) the tire's average angular acceleration $\alpha_{\mathrm{av}}$ (in $\mathrm{rad} / \mathrm{s}^{2}$ ) $\mathrm{rad} / \mathrm{s}^{2}$
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Step 1 ：The initial angular velocity of the tire is \(3.80 \, \mathrm{rad/s}\) in the counterclockwise direction. This is considered as positive.

Step 2 ：The tire is then spun in the opposite (clockwise) direction at the same speed, so the final angular velocity is \(-3.80 \, \mathrm{rad/s}\).

Step 3 ：The change in angular velocity \(\Delta \omega\) is given by the final angular velocity minus the initial angular velocity. So, \(\Delta \omega = -3.80 \, \mathrm{rad/s} - 3.80 \, \mathrm{rad/s} = -7.60 \, \mathrm{rad/s}\).

Step 4 ：The average angular acceleration \(\alpha_{\mathrm{av}}\) is given by the change in angular velocity divided by the change in time. The change in time \(\Delta t\) is given as \(2.50 \, \mathrm{s}\). So, \(\alpha_{\mathrm{av}} = \frac{-7.60 \, \mathrm{rad/s}}{2.50 \, \mathrm{s}} = -3.04 \, \mathrm{rad/s^2}\).

Step 5 ：Final Answer: The change in the tire's angular velocity \(\Delta \omega\) is \(\boxed{-7.60 \, \mathrm{rad/s}}\) and the tire's average angular acceleration \(\alpha_{\mathrm{av}}\) is \(\boxed{-3.04 \, \mathrm{rad/s^2}}\).