MHT CET · Physics · Rotational Motion
Two bodies have their moments of inertia I and 2I respectively about their axes of rotation. If their kinetic energies of rotation are equal, their angular momenta will be in the ratio
- A \(2: 1\)
- B \(1: 2 \sqrt{2}\)
- C \(1: \sqrt{2}\)
- D \(1: 2\)
Answer & Solution
Correct Answer
(C) \(1: \sqrt{2}\)
Step-by-step Solution
Detailed explanation
\(
\begin{array}{l}
(\mathrm{kE})_{1}=(\mathrm{kE})_{2} \\
\therefore \frac{1}{2} \mathrm{I} \omega_{1}^{2}=\frac{1}{2}(2 \mathrm{I}) \omega_{2}^{2} \\
\therefore \frac{\omega_{1}^{2}}{\omega_{2}^{2}}=2 \\
\therefore \frac{\omega_{1}}{\omega_{2}}=\sqrt{2}
\end{array}
\)
Ratio of angular momenta \(=\frac{I \omega_{1}}{2 I \omega_{2}}=\frac{1}{2} \frac{\omega_{1}}{\omega_{2}}=\frac{1}{\sqrt{2}}\)
\begin{array}{l}
(\mathrm{kE})_{1}=(\mathrm{kE})_{2} \\
\therefore \frac{1}{2} \mathrm{I} \omega_{1}^{2}=\frac{1}{2}(2 \mathrm{I}) \omega_{2}^{2} \\
\therefore \frac{\omega_{1}^{2}}{\omega_{2}^{2}}=2 \\
\therefore \frac{\omega_{1}}{\omega_{2}}=\sqrt{2}
\end{array}
\)
Ratio of angular momenta \(=\frac{I \omega_{1}}{2 I \omega_{2}}=\frac{1}{2} \frac{\omega_{1}}{\omega_{2}}=\frac{1}{\sqrt{2}}\)
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