Question

A fly-wheel of mass $25\, kg$ has a radius of $0.2\, m$. It is making $240\, rpm .$ What is the torque necessary to bring to rest in $20\, s$?

• A. $2 \pi\, Nm$
• B. $0.4 \pi\, Nm$
• C. $\frac{2}{\pi} Nm$
• D. $4 \pi \,Nm$

Answer 1

Answer: (B)

$\alpha=\frac{2 \pi n}{t} =\frac{2 \pi \times \frac{240}{60}}{20}$
$=\frac{2 \pi \times 4}{20}$
$\alpha =\frac{2 \pi}{5}$
Torque, $\tau =I \alpha$
$\tau =M R^{2} \alpha$
$=25 \times(0.04) \times \frac{2 \pi}{5}$
$=0.4 \pi\, Nm$