Question

A wheel having a moment of inertia $2\mathrm{kg}-{\mathrm{m}}^2$ about its vertical axis rotates at the rate of $60\mathrm{rpm}$ about this axis. The torque which can stop the wheel's rotation in one minute would be

• A. $\frac{2\mathrm{\pi }}{13}\mathrm{N}-\mathrm{m}$
• B. $\frac{\mathrm{\pi }}{14}\mathrm{N}-\mathrm{m}$
• C. $\frac{\mathrm{\pi }}{15}\mathrm{N}-\mathrm{m}$
• D. $\frac{\mathrm{\pi }}{20}\mathrm{N}-\mathrm{m}$

Answer 1

The correct option is C

$\frac{\mathrm{\pi }}{15}\mathrm{N}-\mathrm{m}$

Step 1: Given parameters

Moment of inertia, $\mathrm{I}$ = $2\mathrm{kg}-{\mathrm{m}}^2$.

Rotating rate, ${\mathrm{\omega }}_0$ = $60\mathrm{rpm}$

${\mathrm{\omega }}_0=\frac{60}{60}\times 2\mathrm{\pi }\frac{\mathrm{rad}}{\mathrm{s}}$

Time, $\mathrm{t}=60\mathrm{s}$

Step 2: Formula used

The torque required to stop the wheel's rotation is given as follows:

$$\begin{gathered} \mathrm{\tau }=\mathrm{l\alpha } \\ =\mathrm{l}\left(\frac{{\mathrm{\omega }}_0-\mathrm{\omega }}{\mathrm{t}}\right)\left[\because \mathrm{\alpha }=\left(\frac{{\mathrm{\omega }}_0-\mathrm{\omega }}{\mathrm{t}}\right)\right] \end{gathered}$$

Where, $\mathrm{I}$ is the moment of inertia,$\mathrm{\tau }$ is the torque, $\mathrm{t}$ is the time, ${\mathrm{\omega }}_0$ is the initial angular velocity, $\mathrm{\omega }$ is the final angular velocity, $\mathrm{\alpha }$ is the retardation

As the final angular velocity is zero,

$$\begin{gathered} \mathrm{\alpha }=\left(\frac{{\mathrm{\omega }}_0-0}{\mathrm{t}}\right) \\ \Rightarrow \mathrm{\alpha }=\frac{{\mathrm{\omega }}_0}{\mathrm{t}} \end{gathered}$$

Therefore, the above formula becomes,

$$\begin{gathered} \mathrm{\tau }=\mathrm{I}\times \mathrm{\alpha } \\ \mathrm{\tau }=\mathrm{I}\times \frac{{\mathrm{\omega }}_0}{\mathrm{t}}...\left(1\right) \end{gathered}$$

Step 3: Calculating the value of torque

Substitute the given values in equation (1).

$$\begin{gathered} \mathrm{\tau }=\mathrm{I}\times \frac{{\mathrm{\omega }}_0}{\mathrm{t}} \\ \mathrm{\tau }=\frac{2\times 2\mathrm{\pi }\times 60}{60\times 60} \end{gathered}$$

$\mathrm{\tau }=\frac{\mathrm{\pi }}{15}\mathrm{N}-\mathrm{m}$

Hence, the value of torque is $\frac{\mathrm{\pi }}{15}\mathrm{N}-\mathrm{m}$.

Hence, option (C) is the correct answer.