Question

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

• A. $\frac{n}{4}$
• B. $\frac{n}{5}$
• C. $\frac{n}{2}$
• D. $\frac{n}{3}$

Answer 1

The correct option is D $\frac{n}{3}$

For first case,
Initial angular velocity, ${\omega }_i={\omega }_o$
Final angular velocity, ${\omega }_f=\frac{{\omega }_o}{2}$
Since, torque is constant. Therefore,
$\alpha =$ constant and also negative in nature as torque is retarding the motion.
We can apply,
$\Rightarrow {\omega }_f^2={\omega }_i^2-2\alpha \Delta \theta$
$\Rightarrow {\left(\frac{{\omega }_o}{2}\right)}^2={\omega }_o^2-2\alpha \Delta \theta$
$\Rightarrow \Delta \theta =\frac{3{\omega }_o^2}{8\alpha }$
So, number of rotations,
$n=\frac{\frac{3{\omega }_o^2}{8\alpha }}{2\pi }=\frac{3{\omega }_o^2}{16\pi \alpha }$
[one rotation $=2\pi$]
$\alpha =\frac{3{\omega }_o^2}{16\pi n}$ ..............(1)
Now,
For second case,
Initial angular velocity, ${\omega }_i=\frac{{\omega }_o}{2}$
Final angular velocity, ${\omega }_f=0$
We can apply,
$\Rightarrow {\omega }_f^2={\omega }_i^2-2\alpha \Delta \theta$
$\Rightarrow 0={\left(\frac{{\omega }_o}{2}\right)}^2-2\alpha \Delta {\theta }^{{}^{\prime }}$
$\Rightarrow 2\alpha \Delta {\theta }^{{}^{\prime }}=\frac{{\omega }_o^2}{4}$
$\Rightarrow 2\times \frac{3{\omega }_o^2}{16\pi n}\times \Delta {\theta }^{{}^{\prime }}=\frac{{\omega }_o^2}{4}$
[from (1)]
$\Rightarrow \Delta {\theta }^{{}^{\prime }}=\frac{2\pi n}{3}$
So, number of rotations,
$n^{{}^{\prime }}=\frac{\frac{2\pi n}{3}}{2\pi }=\frac{n}{3}$