Question

A solid cylinder of mass 2 kg and radius 4 cm is rotating about its axis at the rate of 3 rpm. The torque required to stop after 2$\pi$revolutions is:

• A. $12\times 10^{-4}Nm$
• B. $2\times 10^{6}Nm$
• C. $2\times 10^{-6}Nm$
• D. $2\times 10^{-3}Nm$

Answer 1

Answer: (C)

Work energy theorem .
$W=\frac{1}{2}I\left({\omega }_f^2-{\omega }_i^2\right)\theta =2\pi$
$=2\pi \times 2\pi =4{\pi }^2$
$W_i=3\times \frac{2\pi }{60}\text{rad}/s$
$\Rightarrow -\tau \theta =\frac{1}{2}\times \frac{1}{2}m{r}^2\left(0^2-{\omega }_i^2\right)$
$\Rightarrow -\tau =\frac{\frac{1}{2}\times \frac{1}{2}\times 2\times \left(4\times 10^{-2}\right){\left(-3\times \frac{2\pi }{60}\right)}^2}{4{\pi }^2}$
$\Rightarrow \tau =2\times 10^{-6}Nm$