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# A wheel has moment of inertia $5 \times {10^{ - 3}}\,{\text{kg}}\,{{\text{m}}^2}$ and is making 20 rev/sec. Torque needed to stop it in 10 sec is ……..$\times {10^{ - 2}}\,{\text{N}} - {\text{m}}$:A. $2\pi$B. $2.5\pi$C. $4\pi$D. $4.5\pi$

Last updated date: 13th Jun 2024
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Hint:Convert the angular velocity from rev/sec to rev/s. Determine the angular retardation required to stop the rotating wheel. Recall the formula for the torque acting on the rotating body in terms of its moment of inertia.

Formula used:
1. Angular acceleration, $\alpha = \dfrac{{{\omega _f} - {\omega _0}}}{t}$,
where, ${\omega _f}$ is the final angular velocity and ${\omega _0}$ is the initial angular velocity.
2. Torque, $\tau = I\alpha$
Here, I is the moment of inertia.

We have given that the initial angular velocity of the wheel is 20 rev/sec. Let us convert it into rad/sec as follows,
${\omega _0} = \left( {20\dfrac{{{\text{rev}}}}{{{\text{sec}}}}} \right)\left( {\dfrac{{2{\pi ^c}}}{{1\,{\text{rev}}}}} \right) = 40\pi \,{\text{rad/sec}}$
The angular acceleration or we can say angular retardation required to stop the wheel in 10 seconds is can be calculated as,
$\alpha = \dfrac{{{\omega _0}}}{t}$

Substituting ${\omega _0} = 40\pi \,{\text{rad/sec}}$ and $t = 10\,{\text{s}}$ in the above equation, we get,
$\alpha = \dfrac{{40\pi }}{{10}}$
$\Rightarrow \alpha = 4\pi \,{\text{rad/}}{{\text{s}}^2}$
The torque acting on the rotating wheel to stop it is given by,
$\tau = I\alpha$
Here, I is the moment of inertia.
Substituting $5 \times {10^{ - 3}}\,{\text{kg}}\,{{\text{m}}^2}$ for I and $4\pi \,{\text{rad/}}{{\text{s}}^2}$ for $\alpha$ in the above equation, we get,
$\tau = \left( {5 \times {{10}^{ - 3}}} \right)\left( {4\pi } \right)$
$\therefore \tau = 2\pi \times {10^{ - 2}}\,{\text{N}} - {\text{m}}$

So, the correct answer is option A.

Note:Students should always convert the angular speed from rev/sec into rad/sec. To do so, note that 1 revolution is $360^\circ \,{\text{or 2}}{\pi ^c}$. The angular acceleration is given as, $\alpha = \dfrac{{{\omega _f} - {\omega _0}}}{t}$, where, ${\omega _f}$ is the final angular velocity and ${\omega _0}$ is the initial angular velocity. Since the wheel stops at final time, we have taken the final angular velocity as zero. Note that, if the angular acceleration has negative sign, the body is said to undergo retardation.