Question

Continuing with the previous proble, if at t=0 the turntable begins to slow down, how many complete revolutions will the bug have made if it takes 30 seconds for the turntable to come to a complete stop? (Assume the turntable is initially rotating at 10 rev/min).

Answer 1

N = 10 rpm

Angular speed = 2pi*N/60

             = 2pi*10 / 60 = 1.05 rad/s

angular acceleration = (w - wo) / t

                  = (0 - 1.05) / 30

                  = -0.035 rad/s2

using the equation of angular motion we hav e

w^2 -wo^2 = 2

0 - 1.05^2 = 2*-0.035*

= 15.8 rad

= 15.8 / 2pi = 2.5 revolutions