A record player rotates a record at 45 revolutions per minute. When the record player is switched off, it makes 5 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?

Question

A record player rotates a record at 45 revolutions per minute.  When the record player is switched off, it makes 5 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?

Kabir · Answer

=> Using the rotational analogue of the common equation of motion 2as = v^2 - u^2 , we get:

2 x (Ang. Acc) x (Ang. Distance) = (Final Angular Vel.)^2 - (Initial Angular Vel.)^2

=> Since it comes to a stop, the Final Ang. Vel. is zero. The intial is given as 45 rev per minute, which is 3/4 revs per second. The Ang. Distance is given as 5 complete turns, which is 10π rad.

So, using the above formula:

Ang. Acc. = - (3/4)^2/2(10π)
= - 9/320π