Question

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

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

Answer 1

The correct option is A $2\times {10}^{-6}Nm$

$\theta =2\pi \times 2\pi =4{\pi }^2{\omega }_0=3rpm=3\left(\frac{2\pi }{60}\right)rad/s{\omega }^2={{\omega }_0}^2-2\alpha \theta \Rightarrow 0={\left(\frac{3\times 2\pi }{60}\right)}^2-2\alpha {\left(4\pi \right)}^2\therefore \alpha =\frac{1}{800}rad/s^2\tau =\frac{m{R}^2}{2}\alpha =\frac{2}{2}{\left(\frac{4}{100}\right)}^2\times \left(\frac{1}{800}\right)=2\times {10}^{-6}Nm$