Question

A sphere of mass $2\, kg$ and radius $5\, cm$ is rotating at the rate of $300\, rev / \min$. 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
$\omega^{2}=\omega_{0}^{2}=2 \alpha \theta \ldots$ (i)
where $\omega$ is angular velocity, $\alpha$ the angular acceleration and $\theta$ the angular displacement.
Also, $\omega=2 \pi f \ldots$ (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 $
$\Rightarrow \tau=\frac{2}{5} \times 2 \times\left(5 \times 10^{-2}\right)^{2} \times 12.5$
$\Rightarrow \tau=2.5 \times 10^{-2} \,N-m$