MHT CET · Physics · Rotational Motion
A thin uniform rod of length ' \(L\) ' and mass ' \(M\) ', is swinging freely along a horizontal axis passing through its centre. Its maximum angular speed is ' \(\omega\) '. Its centre of mass rises to a maximum height of [g= gravitational acceleration]
- A \(\frac{\omega^2 \mathrm{~L}^2}{12 \mathrm{~g}^2}\)
- B \(\frac{\omega^2 L^2 g}{6}\)
- C \(\frac{\omega^2 g}{12 \mathrm{~L}^2}\)
- D \(\frac{\omega^2 \mathrm{~L}^2}{24 \mathrm{~g}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\omega^2 \mathrm{~L}^2}{24 \mathrm{~g}}\)
Step-by-step Solution
Detailed explanation
The moment of inertia of rod about its centre is \(\frac{\mathrm{ML}^2}{12}\).
Its kinetic energy K.E. \(=\frac{1}{2} \mathrm{I} \omega^2=\frac{1}{2} \times \frac{\mathrm{ML}^2}{12} \times \omega^2\)
\(=\frac{M L^2 \omega^2}{24}\)
If its centre of mass rises by h, then it will gain potential energy.
\(\therefore \quad\) Potential energy, \(\mathrm{Mgh}=\frac{\mathrm{ML}^2 \omega^2}{24}\)
\(\therefore \quad \mathrm{h}=\frac{\mathrm{L}^2 \omega^2}{24 \mathrm{~g}}\).
Its kinetic energy K.E. \(=\frac{1}{2} \mathrm{I} \omega^2=\frac{1}{2} \times \frac{\mathrm{ML}^2}{12} \times \omega^2\)
\(=\frac{M L^2 \omega^2}{24}\)
If its centre of mass rises by h, then it will gain potential energy.
\(\therefore \quad\) Potential energy, \(\mathrm{Mgh}=\frac{\mathrm{ML}^2 \omega^2}{24}\)
\(\therefore \quad \mathrm{h}=\frac{\mathrm{L}^2 \omega^2}{24 \mathrm{~g}}\).
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