Question

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

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

Answer 1

The correct option is D $2.5\times {10}^{-2}N-m$

$M=2kg,R=5cm=5\times {10}^{-2}m$,
${\omega }_0=300rev/min=\frac{300\times 2\pi }{60}=10\pi rad/s$
$\theta =2\pi rev=2\pi \times 2pi=4{\pi }^{2}rad$
$\alpha =?\omega =0$
$\alpha =\frac{{\omega }^2-{\omega }_0^2}{2\theta }=-\frac{100{\pi }^2}{2\times 4{\pi }^2}=-\frac{25}{2}rad/s^2$
$\tau =I\alpha$
$=\frac{2}{5}M{R}^2\alpha =\frac{2}{5}\times 2\times (5\times {10}^{-2}{)}^2\times \frac{25}{2}$
$=2.5\times {10}^{-2}Nm$