Question

You have a grindstone with a mass of 90.0 kg, 0.340-m radius, and is turning at 90.0 rpm. You
press a steel ax against it, bringing it to a stop in 36.0 s. (a) What is the change in kinetic energy of
the grindstone? (b) If the tangential force due to the friction between the ax and the grindstone is
4.00 N, how many rotations will the grindstone make as it comes to a stop?

Answer 1

calculate I of solid disc

omegai = 90 rev/min * 2 pi rad/rev * 1 min/60 s
= 9.42 radians/second

Ke = (1/2) I omegai^2 at start, zero at finish

Torque = 4 R = 4*.34 newton meters

alpha = Torque/I

omega = omegai - alpha t
omega = 0 at finish
so
t = omegai/alpha
which is time to stop
then
d = total radians turned to stop
d = omegai t - .5 alpha t^2
divide that by 2 pi radians/revolution to get number of revolutions