Question

An old record completes 0.890 revolutions as it slows down uniformly from 78.0 rpm to 22.8 rpm. What is the magnitude of the angular acceleration of the record, in rad/s²?

Answer 1

Under uniform acceleration,(wf)^2=(wi)^2+2*a*s, where wf is final angular velocity,wi is initial angular velocity ,a is angular acceleration and s is angular displacement.

Here,wf=22.8 rpm=22.8*2π/60 rad/s=2.3876 rad/s, wi=78 rpm=78*2π/60=8.16814 rad/s, s=0.890 revolutions=0.890*2π radians=5.592 radians.

So, (2.3876)^2=(8.16814)^2+2*a*5.592

=>5.7=66.72+11.184a

=>a=(5.7-66.72)/11.184= -5.456 rad/s2.

So, magnitude of angular acceleration is 5.456 rad/s2 and the negative sign denotes that the object is decelerating.