Question

A wheel is rotating at the rate of 1000rpm and its kinetic energy is ${10^6}{\text{J}}$. Determine the moment of inertia of the wheel about its axis of rotation.
A) $180{\text{ kg}}{{\text{m}}^2}$
B) $1800{\text{ kg}}{{\text{m}}^2}$
C) $18{\text{ kg}}{{\text{m}}^2}$
D) $0 \cdot 180{\text{ kg}}{{\text{m}}^2}$

Answer 1

Hint: Here the angular velocity and kinetic energy of the wheel are given. The moment of inertia of a body is considered as the mass analogous in rotational mechanics. And so just as the translational kinetic energy depends on the mass of a body and its linear velocity, here the rotational kinetic energy depends on the moment of inertia of the wheel and its angular velocity.

Formula used:
The kinetic energy of a rotating body is given by, $K = \dfrac{1}{2}I{\omega ^2}$ where $I$ is the moment of inertia of the body and $\omega $ is the angular velocity of the body.

Complete step by step answer:
Step 1: Listing the given parameters of the wheel.
The problem at hand involves a wheel which rotates about its axis of rotation.
The angular velocity of the rotating wheel is mentioned to be $\omega = 1000{\text{rpm}}$.
The kinetic energy of the rotating wheel is mentioned to be $K = {10^6}{\text{J}}$.
We have to obtain the moment of inertia of the rotating wheel.
Let $I$ be the moment of inertia of the given wheel about its axis of rotation.
Step 2: Expressing the relation for the kinetic energy of the rotating wheel to obtain an expression for the moment of inertia of the wheel.
The kinetic energy of the given wheel can be expressed as $K = \dfrac{1}{2}I{\omega ^2}$ --------- (1)
We can express equation (1) in terms of the moment of inertia of the wheel as
 $I = \dfrac{{2K}}{{{\omega ^2}}}$ -------- (2).
The angular velocity of the wheel, $\omega = 1000{\text{rpm}}$ is expressed in units of rotations per minute. This has to be converted to radian per second before substituting its value in equation (2).
We know that $1{\text{ rotation}} = 2\pi {\text{rad}}$ and $1{\text{min}} = 60{\text{s}}$.
So the angular velocity in radian per second will be given by, $\omega \left( {{\text{rad}}{{\text{s}}^{ - 1}}} \right) = \dfrac{{\omega \left( {{\text{rpm}}} \right) \times 2\pi }}{{60}}$
$ = \dfrac{{1000 \times 2\pi }}{{60}}$
$ = \dfrac{{100\pi }}{3}$
Now substituting for $\omega = \dfrac{{100\pi }}{3}{\text{rad}}{{\text{s}}^{ - 1}}$ and $K = {10^6}{\text{J}}$ in equation (2) we get,
$\Rightarrow I = \dfrac{{2 \times {{10}^6} \times {3^2}}}{{{{100}^2}{\pi ^2}}} = 182 \cdot 5{\text{kg}}{{\text{m}}^2}$
Thus the moment of inertia of the wheel is $I \approx 180{\text{kg}}{{\text{m}}^2}$ .

Therefore, the correct option is A.

Note:
Here the wheel is mentioned to be rotating. So the given kinetic energy is indeed its rotational kinetic energy. The wheel will possess an additional translational kinetic energy if it was said to be rolling. While substituting the values of physical quantities in any equation, each quantity must be expressed in their respective S.I. units. Here the angular velocity was not expressed in its S.I. unit so we had to convert it to its S.I. unit to get the correct result.