Question

A sphere of mass $2kg$ and radius $5cm$ is rotating at the rate of $300rev/\min$. Then the torque required to stop it in $2\pi$ revolutions is

• A. 1. $6\times 10^{5}N-m$
• B. $1.6\times 10^{-4}N-m$
• C. $2.5\times 10^{-3}N-m$
• D. $2.5\times 10^{-2}N-m$

Answer 1

Answer: (D)

From equation of rotational motion, we have
${\omega }^2={\omega }_0^2=2\alpha \theta \dots$ (i)
where $\omega$ is angular velocity, $\alpha$ the angular acceleration and $\theta$ the angular displacement.
Also, $\omega =2\pi f\dots$ (ii)
Where $f$ is number of revolutions.
From Eqs. (i) and (ii), we get
For $\omega =0$ (sphere stop),
$\theta =2\pi \times 2\pi =4{\pi }^2$
$\therefore n=300$ rev / min
$=\frac{300}{60}=5$ rev / s
$\therefore 0=(2\pi \times 5{)}^2-2\alpha (2\pi {)}^2$
$\Rightarrow \alpha =\frac{100{\pi }^2}{8{\pi }^2}=12.5$ rad / s
Torque applied $\tau =I\alpha$
where $I$ is moment of inertia, a the angular acceleration.
For a sphere, $I=\frac{2}{5}m{R}^2$
$\therefore \tau =\frac{2}{5}m{R}^2\alpha$
$\Rightarrow \tau =\frac{2}{5}\times 2\times {\left(5\times 10^{-2}\right)}^2\times 12.5$
$\Rightarrow \tau =2.5\times 10^{-2}N-m$