MHT CET · Physics · Rotational Motion
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 the axis is
- A \(\frac{x^2}{2 y}\)
- B \(\frac{y}{2 x}\)
- C \(\frac{x}{2 y}\)
- D \(\frac{y^2}{2 x}\)
Answer & Solution
Correct Answer
(D) \(\frac{y^2}{2 x}\)
Step-by-step Solution
Detailed explanation
The kinetic energy of a rotating body is given by \(x=\frac{I \omega^2}{2}\) and the angular momentum is given by \(\mathrm{y}=\mathrm{I} \omega\), where \(\mathrm{I}\) is the moment of inertia, \(\omega\) the angular velocity.
Therefore, we can write,
\(\mathrm{y}^2=2 \mathrm{I}\left(\frac{\mathrm{I} \omega^2}{2}\right)=2 \mathrm{Ix}\)
On re-writing,
\(I=\frac{y^2}{2 x}\)
Therefore, we can write,
\(\mathrm{y}^2=2 \mathrm{I}\left(\frac{\mathrm{I} \omega^2}{2}\right)=2 \mathrm{Ix}\)
On re-writing,
\(I=\frac{y^2}{2 x}\)
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