MHT CET · Physics · Oscillations
A thin uniform rod of mass ' \(\mathrm{m}\) ' and length ' \(l\) ' is suspended from one end which can oscillate in a vertical plane about the point of intersection. It is pulled to one side and then released. It passes through the equilibrium position with angular speed ' \(\omega\) '. The kinetic energy while passing through mean position is
- A \(\mathrm{m} l^2 \omega^2\)
- B \(\frac{\mathrm{m} l^2 \omega^2}{4}\)
- C \(\frac{\mathrm{m} l^2 \omega^2}{6}\)
- D \(\frac{\mathrm{m} l^2 \omega^2}{12}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{m} l^2 \omega^2}{6}\)
Step-by-step Solution
Detailed explanation
The kinetic energy of the rod while passing through the mean position will be,
\(
\begin{aligned}
\text { K.E. } & =\frac{1}{2} \mathrm{I} \omega^2 \\
& =\frac{1}{2} \frac{\mathrm{m} l^2}{3} \times \omega^2=\frac{\mathrm{m} l^2 \omega^2}{6}
\end{aligned}
\)
\(
\begin{aligned}
\text { K.E. } & =\frac{1}{2} \mathrm{I} \omega^2 \\
& =\frac{1}{2} \frac{\mathrm{m} l^2}{3} \times \omega^2=\frac{\mathrm{m} l^2 \omega^2}{6}
\end{aligned}
\)
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