Question

A record player rotates a record at 45 revolutions per minute. When the record player is switched off, it makes 4.0 complete turns at a constant angular acceleration before coming to rest. What was the magnitude of the angular acceleration (in rads/s2) of the record as it slowed down

Answer 1

Answer:

The  angular acceleration is $\alpha = 0.4418 \ rad /s^2$

Explanation:

From the question we are told that

      The  angular speed is $w_f = 45 \ rev / minutes = \frac{45 * 2 * \pi }{60 }= 4.713 \ rad/s$

       The  angular displacement is  $\theta =4 \ rev = 4 * 2 * \pi = 25.14 \ rad$

From the first equation of motion we can define the movement of the record as

      $w_f ^2 = w_o ^2 + 2 * \alpha * \theta$

Given that the record started from rest $w_o = 0$

So

       $4.713^2 = 2 * \alpha * 25.14$

        $\alpha = 0.4418 \ rad /s^2$