Question

A ballet dancer rotates about his own vertical axis on a smooth horizontal floor. The moment of inertia of the dancer decreases by 20 % when he holds himself closer to his axis of rotation. Find the percentage of change in his rotational kinetic energy.
A) Increases by 36 %
B) Decreases by 36 %
C) Increases by 25 %
D) Decreases by 25 %

Answer 1

Hint:When the dancer moves closer to his axis of rotation his angular velocity increases and moment of inertia decreases. Thus his angular momentum remains constant. Angular momentum is defined as the product of the moment of inertia and angular velocity of the rotating body. The dependence of the rotational kinetic energy on the moment of inertia can be found out using the angular momentum conservation theorem.

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

Complete step by step answer.
Step 1: Write down the key points in the question.
The moment of inertia of the dancer is said to reduce by 20 % as he moves closer to his own axis of rotation.
Let $I$ be the initial moment of inertia of the dancer.
Then as the dancer moves closer to his axis of rotation the moment of inertia changes from $I$ to
$0.8I$ .
Step 2: Express the change in the rotational kinetic energy of the dancer in terms of its angular momentum.
The angular momentum of the dancer moving with angular velocity $\omega $ will be $L = I\omega $------- (2).
The rotational kinetic energy of the dancer is given by, $K = \dfrac{1}{2}I{\omega ^2} = \dfrac{1}{2}I\omega \cdot \omega $ ----------- (3).
where $I$ is the moment of inertia of the dancer and $\omega $ is his angular velocity.
From equation (2) we can express the angular velocity as $\omega = \dfrac{L}{I}$ ------ (4).
Substituting equation (4) in (3) we get the rotational kinetic energy in terms of the angular momentum as $K = \dfrac{{{L^2}}}{{2I}}$
Step 3: Based on the conservation of angular momentum find the change in the rotational kinetic energy of the dancer.
The angular momentum conservation theorem states that in the absence of an external torque, the angular momentum of the system is conserved. Here no external torque is present for the rotating dancer and thus the angular momentum is conserved.
Since angular momentum of the dancer is conserved, the rotational kinetic energy can be expressed as $K \propto \dfrac{1}{I}$ -------- (5)
So, the rotational kinetic energy will be inversely proportional to the moment of inertia of the dancer.
We know that the moment of inertia of the dancer changes from $I$ to $0.8I$
Then the corresponding new rotational kinetic energy will be ${K_{new}} \propto \dfrac{1}{{0.8I}} = 1.25K$
The change in the rotational kinetic energy will be $\Delta K = {K_{new}} - K = 1.25K - K = 0.25K$
So, the rotational kinetic energy increases by 25 %.

Hence the correct option is C.

Note: The absence of an external torque refers to the fact that no frictional forces are acting on the dancer since he executes rotations on a smooth horizontal floor. So an increase in his moment of inertia will result in a decrease in his angular velocity and a decrease in the moment of inertia will be followed by an increase in his angular velocity and thus the angular momentum always remains constant.