MHT CET · Physics · Rotational Motion
A thin uniform rod of length ' \(L\) ' and mass ' \(M\) ' is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is ' \(\omega\) '. Its centre of mass rises to a maximum height of ( \(g=\) acceleration due gravity)
- A \(\frac{\mathrm{L}^2 \omega^2}{2 \mathrm{~g}}\)
- B \(\frac{\mathrm{L} \omega}{6 \mathrm{~g}}\)
- C \(\frac{\mathrm{L} \omega}{2 \mathrm{~g}}\)
- D \(\frac{\mathrm{L}^2 \omega^2}{6 \mathrm{~g}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{L}^2 \omega^2}{6 \mathrm{~g}}\)
Step-by-step Solution
Detailed explanation
Kinetic Energy = Potential Energy \( \frac{1}{2} I \omega^2 = Mg h \)
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