Question

A disc is rotating with an angular velocity ${\omega }_0$. A constant retarding torque is applied on it to stop the disc. The angular velocity becomes ${\displaystyle \frac{{\omega }_0}{2}}$ after $n$ rotations. How many more rotations will it make before coming to rest?
A. $n$
B. $2n$
C. ${\displaystyle \frac{n}{2}}$
D. ${\displaystyle \frac{n}{3}}$

Answer 1

Hint: In order to solve this question, we should know that a torque produce angular acceleration in the body and since torque is constant then, angular acceleration will also be constant but in negative direction as its retarding torque, here we will use general equation of motion to calculate number of turns a disc will make before it came at rest.

Formula used:
If $\omega ,{\omega }_0,\theta ,\alpha$ be the final angular velocity, initial angular velocity, displacement, angular acceleration then,
${\omega }^2-{{\omega }_0}^2=2\alpha \theta$

Complete step by step answer:
According to the question, we have torque is retarding so, angular acceleration produced by this torque will also be retarding which will be $-\alpha$ and for first n rotations the velocities have,
final velocity $\omega ={\displaystyle \frac{{\omega }_0}{2}}$
initial velocity is ${\omega }_0$
angular acceleration $-\alpha$
Displacement is n rotations so using,
${\omega }^2-{{\omega }_0}^2=2\alpha \theta$ we get,
$\Rightarrow {\displaystyle \frac{{{\omega }_0}^2}{4}}-{{\omega }_0}^2=-2\alpha n$
$\Rightarrow {\displaystyle \frac{3}{4}}{{\omega }_0}^2=2\alpha n\to (i)$

Now let n’ be the further rotations by disc to came at rest at conditions,
final velocity $\omega =0$
initial velocity ${\omega }_{initial}={\displaystyle \frac{{\omega }_0}{2}}$
angular acceleration $-\alpha$
displacement is n’ rotations so using,
${\omega }^2-{{\omega }_0}^2=2\alpha \theta$ we get,
$0-{\displaystyle \frac{{{\omega }_0}^2}{4}}=-2\alpha n^{\prime }$
$\Rightarrow {\displaystyle \frac{{{\omega }_0}^2}{4}}=2\alpha n^{\prime }\to (ii)$
Now, divide first equation by second we get,
${\displaystyle \frac{{\displaystyle \frac{3}{4}}{{\omega }_0}^2}{{\displaystyle \frac{{{\omega }_0}^2}{4}}}}={\displaystyle \frac{2\alpha n}{2\alpha n^{\prime }}}$
$\Rightarrow {\displaystyle \frac{n}{n^{\prime }}}=3$
$\therefore n^{\prime }={\displaystyle \frac{n}{3}}$

Hence, the correct option is D.

Note: It should be remembered that, retarding torque means its applied in such a ways that it opposes the rotational motion of the body and thus produced negative angular acceleration on the body and always check initial and final velocity of the bodies while using equations of motion to solve such problems.