Question

Moment of inertia of the wheel about the axis of rotation is 3 MKS units. Its kinetic energy will be 600 J, if period of rotation is in (seconds)
$\text{A}\text{. }\pi $
$\text{B}\text{. }\dfrac{\pi }{5}$
$\text{C}\text{. }\dfrac{\pi }{10}$
$\text{D}\text{. 2}\pi $

Answer 1

Hint: When a body is pure rotational motion with angular velocity $\omega $, it kinetic energy is given as $K=\dfrac{1}{2}I{{\omega }^{2}}$ and is time period is equal to $T=\dfrac{2\pi }{\omega }$. Use these two formulas to find the time period of the wheel.

Formula used:
$K=\dfrac{1}{2}I{{\omega }^{2}}$
$T=\dfrac{2\pi }{\omega }$

Complete step-by-step answer:
Moment of inertia of the body about an axis of rotation is the measure of the tendency to resist any change in the rotational motion of the body. It is similar to mass in a pure translational motion of a body.
When a body is in a pure translational motion and moving with speed with v, it possesses some energy associated with its motion called the translational kinetic energy of the body. Similarly, when a body is in a pure rotational motion and moving with angular velocity $\omega $ about an axis of rotation, it possesses an energy called rotational kinetic energy of the rotating body.
The value of the rotational kinetic energy of the body is given as $K=\dfrac{1}{2}I{{\omega }^{2}}$ ….. (i).
When the body is rotating about a fixed axis, we define the time period (T) of its rotation. The value of time period is $T=\dfrac{2\pi }{\omega }$ ….. (ii).
It is asked to find the time period of the rotating wheel.
For that let us first find the angular velocity of the wheel.
It is given that the moment of inertia of the wheel about the axis of rotation is 3 MKS units. The MKS unit if moment of inertia is $kg{{m}^{2}}$. Hence, I = 3$kg{{m}^{2}}$.
It is also given that the kinetic energy of the wheel is 600 J. Hence, K = 600 J.
Substitute the values of I and K in equation (i).
 $\Rightarrow 600=\dfrac{1}{2}\times 3{{\omega }^{2}}$
$\Rightarrow {{\omega }^{2}}=600\times \dfrac{2}{3}$
$\Rightarrow {{\omega }^{2}}=400{{s}^{-2}}$
$\Rightarrow \omega =20{{s}^{-1}}$
Substitute the value of $\omega $ in equation (ii).
$\Rightarrow T=\dfrac{2\pi }{20}=\dfrac{\pi }{10}s$
Therefore, the time period of the wheel is $\dfrac{\pi }{10}$ seconds.
Hence, the correct option is C.

Note: We can always take help of the knowledge of translation mechanics to solve the questions involving rotational motion. Just we should know which quantities of rotational motion are analogous to the quantities in transitional motion.
If we know this, then the formulas used in translational motion can help us to solve questions.
We know that translational kinetic energy of a body is $K=\dfrac{1}{2}m{{v}^{2}}$
Moment of inertia is analogous to mass (m).
Angular velocity ($\omega $) is analogous to velocity (v).
Therefore, we can write the rotational kinetic energy as $K=\dfrac{1}{2}I{{\omega }^{2}}$.