Question

A circular disc is rotating about its own axis at uniform angular velocity$\omega$. The disc is subjected to uniform angular retardation by which its angular velocity is decreased to $\dfrac{\omega }{2}$during $120\;\;$rotations. The number of rotations further made by it before coming to rest is:
A. $120\;\;$
B. $60\;\;$
C. $40$
D. $20$

Answer 1

Given that the disc is rotating about its axis. Uniform angular retardation is acting on the rotating disc. Retardation means the final velocity of the body is reduced when compared to the initial velocity. Retardation can also be termed as deceleration or negative acceleration. Therefore it is obvious that this retardation will reduce the rotations of the disc. Let us see how many rotations that disc will make after the retardation.

Complete step by step solution:
Given that the initial angular velocity . After the angular retardation, the disc’s angular velocity is reduced by $\dfrac{\omega }{2}$. Also initially the number of rotations of the disc is $120\;\;$. We need to find the number of rotations of the disc after the retardation.
We know the third equation of motion,
${v^2} = {u^2} + 2as.$
Here $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration, $s$ is the displacement.
The above equation is for linear motion, for rotational motion the above equation becomes,
${\omega _2}^2 = {\omega _1}^2 + 2\alpha \theta$
Here ${\omega _2}$ is the said to be final angular velocity, ${\omega _1}$ is said to be the initial angular velocity, $\alpha$ is the angular acceleration, $\theta$ is the angular displacement.
We can rearrange the above equation to find the angular acceleration as,
$\alpha = \dfrac{{{\omega _1}^2 - {\omega _2}^2}}{{2{\theta _1}}}$
Here it is said that the disc is subjected to uniform retardation.
Therefore $\alpha$ here is the angular retardation which is constant.
Similarly, we can write for the above equation after the angular retardation.
$\alpha = \dfrac{{{\omega _2}^2 - {\omega _3}^2}}{{2{\theta _2}}}$
Equating both the equations.
$\dfrac{{{\omega _1}^2 - {\omega _2}^2}}{{2{\theta _1}}} = \dfrac{{{\omega _2}^2 - {\omega _3}^2}}{{2{\theta _2}}}$
Here, ${\omega _3}$ is the case in which the disc comes to rest. Therefore ${\omega _3} = 0$
Also ${\theta _1} = {120^0}$
${\omega _2} = \dfrac{{{\omega _1}}}{2}$
Therefore we can say that ${\omega _1} = \omega$and ${\omega _2} = \dfrac{\omega }{2}$. Substituting all the values we get,
$\dfrac{{{\omega ^2} - {{(\dfrac{\omega }{2})}^2}}}{{2{\theta _1}}} = \dfrac{{{{(\dfrac{\omega }{2})}^2}}}{{2{\theta _2}}}$
Solving the above equation we get,
${\theta _2} = \dfrac{{{\theta _1}}}{3}$
${\theta _2} = \dfrac{{120}}{3}$
${\theta _2} = 40$
Therefore the disc will make $40$ rotations further made by it before coming to rest.
Therefore the correct option is C.

Note:
One should know the difference between retardation and acceleration. Retardation means the reduction of the velocity of the body which means the final velocity is less than the initial velocity. Acceleration means an increase in velocity of the body which means the final velocity is greater than the initial velocity.