Question

A wheel of mass 2 kg and radius 20 cm has a moment of inertia = I, is rotating on its axis with an angular velocity of w rad/s. If the radius of the wheel is reduced to 10 cm and mass is 1 kg, and the wheel rotates with the same angular velocity, then the rotational kinetic energy of the wheel will be:
A). $\dfrac{1}{8}$th the kinetic energy of the wheel of mass 2 kg
B). $\dfrac{1}{2}$ of the kinetic energy of the wheel of mass 2 kg
C). $\dfrac{1}{4}$th the kinetic energy of the wheel of mass 2 kg
D). Kinetic energy doesn’t change

Answer 1

Hint: The moment of inertia depends on the mass of the body and square of the distance of that mass from the axis of rotation. Farther the mass from the axis of rotation, higher is the moment of inertia.

Formula Used:
The moment of inertia of a body about an axis is given by the following formula:
$I = \sum {{m_i}r_i^2} = M{r^2}{\text{ }}.........{\text{(A)}}$
where ${m_i}$ is the ith mass in the body and ${r_i}$ is the distance of that mass from the axis of rotation; M is the mass of the whole body and r is its distance from the axis of rotation.
Rotational kinetic energy is given as:
$K = \dfrac{1}{2}I{\omega ^2}{\text{ }}.........{\text{(B)}}$
where $\omega$ is the angular velocity of rotation.

Complete step by step solution:
We are given a wheel which has mass m = 2 kg and radius r = 20 cm = 0.2 m which is rotating with angular velocity $\omega$ rad/s. Therefore, we can write expressions for the moment of inertia and kinetic energy of rotation as follows using equations (A) and (B).
I = m{r^2} = 2 \times {\left( {0.2} \right)^2} = 0.08{\text{kg - }}{{\text{m}}^2}{\text{ }}.............{\text{(i)}} \\\
$\Rightarrow K = \dfrac{1}{2}I{\omega ^2}{\text{ }}..............{\text{(ii)}} \\\$
The new wheel has mass m’ = 1 kg and radius r’ = 10 cm = 0.1 m while the angular velocity is the same as before. Therefore, we have
I' = m'r{'^2} = 1 \times {\left( {0.1} \right)^2} = 0.01{\text{kg - }}{{\text{m}}^2}{\text{ }}....................{\text{(iii)}} \\\
$\Rightarrow K' = \dfrac{1}{2}I'{\omega ^2}{\text{ }}................{\text{(iv)}} \\\$
We see that there is decrease in the moment of inertia due to decrease in mass of the body and distance of the mass from the axis of rotation.
Dividing equation (iv) by (ii), we get
\dfrac{{K'}}{K} = \dfrac{{\dfrac{1}{2}I'{\omega ^2}}}{{\dfrac{1}{2}I{\omega ^2}}} = \dfrac{{I'}}{I} = \dfrac{{0.01}}{{0.08}} = \dfrac{1}{8} \\\
$\Rightarrow K' = \dfrac{K}{8} \\\$
Therefore the correct answer is option A.

Note: The expression for kinetic energy for rotational motion is analogous to the expression for kinetic energy in translational motion. Moment of inertia is analogous to mass and the angular velocity is analogous to translational velocity.