A flywheel rotating at 12 rev/s is brought to rest in 6 s.The magnitude of the average angular acceleration of the wheel during this process is: A)1/$\pi$ rad/s² B)2 rad/s² C)4 rad/s² D)4$\pi$rad/s² E)72 rad/s²

Question

Quizplus · Accepted Answer

The average angular acceleration ($\alpha$) is given by the change in angular velocity ($\Delta \omega$) over the time period ($\Delta t$), so $\alpha = \Delta \omega / \Delta t$. The initial angular velocity is $12 \, \text{rev/s} \times 2\pi \, \text{rad/rev} = 24\pi \, \text{rad/s}$ (since $1 \, \text{rev} = 2\pi \, \text{radians}$). It comes to rest, so the final angular velocity is $0 \, \text{rad/s}$. The time taken is $6 \, \text{s}$. Thus, $\alpha = (0 - 24\pi) / 6 = -4\pi \, \text{rad/s}^2$. The magnitude of $\alpha$ is $4\pi \, \text{rad/s}^2$, ignoring the negative sign which indicates direction.

A Flywheel Rotating at 12 Rev/s Is Brought to Rest π\piπ

A Flywheel Rotating at 12 Rev/s Is Brought to Rest $\pi$