Question

A cylinder rotating at an angular speed of $50$ rev/s is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torques act on the two cylinders, accelerating the stationary one and decelerating the moving one. If the common magnitude of the angular acceleration and deceleration be $1rev/s^2$, then how much time (in s) will it take before the two cylinder have equal angular speed.

Answer 1

Given: Initial velocity $=50rev/s,\alpha =1rev/s^2$

Let final velocity $={\omega }^{2}rev/sec$

Let time taken $=t$

For first cylinder

${\omega }^2-\omega -\alpha -t\Rightarrow {\omega }^2=50-t$

For second cylinder

${\omega }^2=0-\alpha t\Rightarrow 50-t=-\alpha t\Rightarrow 50=2t\Rightarrow t=25sec$