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 $\frac{n}{3}$

Applying equation for rotational motion
For ${\omega }_0$ to ${\omega }_0/2$ ,

${\omega }_f^2={\omega }_i^2-2\alpha \theta$
$\Rightarrow {\left(\frac{{\omega }_o}{2}\right)}^2={\omega }_o^2-2\alpha \theta$
$\Rightarrow \theta =\frac{3{\omega }_o^2}{8\alpha }$
So, number of rotations,
$n=\frac{\frac{3{\omega }_o^2}{8\alpha }}{2\pi }=\frac{3{\omega }_o^2}{16\pi \alpha }$
$\Rightarrow \alpha =\frac{3{\omega }_o^2}{16\pi n}$ ..............(1)

Now, for ${\omega }_0/2$ to $0$ (rest),
${\omega }_f^2={\omega }_i^2-2\alpha {\theta }^{\prime }$
$\Rightarrow 0={\left(\frac{{\omega }_o}{2}\right)}^2-2\alpha {\theta }^{{}^{\prime }}$
$\Rightarrow 2\alpha {\theta }^{{}^{\prime }}=\frac{{\omega }_o^2}{4}$
$\Rightarrow 2\times \frac{3{\omega }_o^2}{16\pi n}\times {\theta }^{{}^{\prime }}=\frac{{\omega }_o^2}{4}$
[from (1)]
$\Rightarrow {\theta }^{{}^{\prime }}=\frac{2\pi n}{3}$
So, number of rotations,
$n^{{}^{\prime }}=\frac{\frac{2\pi n}{3}}{2\pi }=\frac{n}{3}$