Question

A cylinder rotating at an angular speed of $50\text{rev/s}$ is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torque acts on the both the 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 ?

• A. $25\text{sec}$
• B. $50\text{sec}$
• C. $75\text{sec}$
• D. $100\text{sec}$

Answer 1

The correct option is A $25\text{sec}$



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

Let after $t$ seconds, their angular velocity will be same and equal to '$\omega$'.

For the first cylinder $\omega =50-\alpha t=50-t$

$\Rightarrow t=(50-\omega )$

For the $2^{nd}$ cylinder
$\omega =\alpha t$

$\Rightarrow t=\omega$

So, $\omega =(50-\omega )$

$\Rightarrow 2\omega =50\Rightarrow \omega =25\text{rev/s}$

$\Rightarrow t=25\text{sec}$