ExamBro
ExamBro
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]

  1. A \(\frac{\omega^2 \mathrm{~L}^2}{12 \mathrm{~g}^2}\)
  2. B \(\frac{\omega^2 L^2 g}{6}\)
  3. C \(\frac{\omega^2 g}{12 \mathrm{~L}^2}\)
  4. D \(\frac{\omega^2 \mathrm{~L}^2}{24 \mathrm{~g}}\)
Verified Solution

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, \(\mathrm{Mgh}=\frac{\mathrm{ML}^2 \omega^2}{24}\)
\(\therefore \quad \mathrm{h}=\frac{\mathrm{L}^2 \omega^2}{24 \mathrm{~g}}\).