Question

A body is rotating about its own axis. Its rotational kinetic energy is '\(x\)' and its angular momentum is '\(y\)'. Hence its moment of inertia about its own axis is

• A. \(\frac{x}{2y}\)
• B. \(\frac{y}{2x}\)
• C. \(\frac{x^2}{2y}\)
• D. \(\frac{y^2}{2x}\)

Hint:

Combine K = (1/2) I w^2 and L = I w to eliminate w.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Rotational kinetic energy is \(x = \frac12I\omega^2\) and angular momentum is \(y = I\omega\).

Step 2: Eliminate omega:
From \(y = I\omega\), \(\omega = \frac yI\). Substitute into the energy:
\[ x = \frac12I\cdot\frac{y^2}{I^2} = \frac{y^2}{2I} \]

Step 3: Solve for I:
\[ I = \frac{y^2}{2x} \]

Step 4: Why the other options are wrong.
\(\frac{x}{2y}\) and \(\frac{y}{2x}\) fail dimensional checks: \(I\) has units of kg m\(^2\), while \(\frac{y}{x}\) has units of seconds. \(\frac{x^2}{2y}\) also has wrong units.

Final Answer:
The moment of inertia is \(\frac{y^2}{2x}\), option (D). \[ \boxed{\frac{y^2}{2x}} \]