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 $1$ rev$/s^2$, then how much time(in s) will it before the two cylinders have equal angular speed?

Answer 1

A cylinder is moving with an angular velocity $50rev/s$ brought in contact with another identical cylinder in rest. The first and second cylinder has common acceleration and deacceleration as $1rad/s^2$ respectively.s

initial velocity $=50rev/sec$

Final velocity $w^1$

$\alpha =1rev/{sec}^2$

Case 1

$w^1=w-\alpha t$

$w^1=50-1\times t$

$=50-t$

Case 2

initial speed $=0$

final speed $w^1$

$w^1=w-\alpha t$

Put value of $w$

$50-t=t\times 1$

$t=25second$