A turntable is initially at rest.It starts rotating with a constant maximum angular acceleration for a period of 5 seconds.It then starts to slow down with an angular acceleration of -1 rad/s² for another 5 seconds.At the end of the 10 second period the turntable has undergone 20 rotations.What is the maximum angular acceleration the turntable reaches? A)-1 rad/s² B)2 rad/s² C)1 rad/s² D)-3.68 rad/s² E)3.68 rad/s²

Question

Quizplus · Accepted Answer

We can use the following equations of rotational motion:
$	heta = 	heta_i + \omega_i t + \frac{1}{2} \alpha t^2$
$\omega^2 = \omega_i^2 + 2\alpha 	heta$

where $	heta$ is the angular displacement, $\omega$ is the angular velocity, $\alpha$ is the angular acceleration, $t$ is the time, and the subscript $i$ denotes initial values.

Let's use the first equation for the first 5 seconds:

$	heta = 0 + \frac{1}{2} \alpha t^2 = \frac{1}{2} \alpha (5)^2 = \frac{25}{2} \alpha$

Now let's use the second equation for the period of 10 seconds:

$	heta = 20(2\pi) = 40\pi$
$\omega_i = \frac{	heta - \frac{25}{2}\alpha}{5} = \frac{8\pi}{5} - \frac{5}{2}\alpha$
$\omega_f = 0$
$\alpha = -\frac{\omega_i}{5} = -\frac{8\pi}{25} + \frac{1}{2}\alpha$
$\alpha = -\frac{8\pi}{12.5} = -0.64$ rad/s$^2$

Therefore, the maximum angular acceleration is 3.68 rad/s$^2$ in the opposite direction (as given by choice E).

A Turntable Is Initially at Rest