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 acceleration and deceleration be one revolution per second square, how long will it take before the two cylinders have equal angular speed ?

Answer 1

A cylinder rotating at an angular speed of 50 rev/s is brought in contact with another identical cylinder in rest. The first and second cylinder has common acceleration and deceleration as $1rad/s^2$ respectively.

Let after 1 sec their angular velocity will be same '$\omega$'.

For the first cylinder

$\omega$ = 50 -at

$\Rightarrow$ t = $\frac{(\omega -50)}{1}$

and for the 2${}^{nd}$ cylinder,

$\Rightarrow \omega ={\alpha }_{2}t$

$\Rightarrow t=\frac{\omega }{I}$

So, $\omega =\left(\frac{\omega -50}{-1}\right)$

$\Rightarrow 2\omega =50$

$\Rightarrow \omega =25\text{rev/s}$

$\Rightarrow t=\frac{25}{1}sec=25\text{sec.}$