# A thin uniform rod of length $\dpi{100} l$ and mass $\dpi{100} m$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $\dpi{100} \omega$ . Its centre of mass rises to a maximum height of Option 1) $\frac{1}{3}\frac{l^{2}\omega ^{2}}{g}$ Option 2) $\frac{1}{6}\frac{l\omega }{g}\;$ Option 3) $\frac{1}{2}\frac{l^{2}\omega ^{2}}{g}$ Option 4) $\; \frac{1}{6}\frac{l^{2}\omega^{2}}{g}$

As we learnt in

Kinetic energy of rotation -

$K=\frac{1}{2}Iw^{2}$

- wherein

$I$ = moment of inertia about axis of rotation

$w$ = angular velocity

$\frac{1}{2}\omega^{2}=mgh$

$I=\frac{1}{3}ml^{2}$

$\frac{1}{2}\left ( \frac{1}{3}l^{2} \right )\omega^{2}=mgh$

$h=\frac{\omega^{2}l^{2}}{6g}$

Ans. 4

Option 1)

$\frac{1}{3}\frac{l^{2}\omega ^{2}}{g}$

This is an incorrect option.

Option 2)

$\frac{1}{6}\frac{l\omega }{g}\;$

This is an incorrect option.

Option 3)

$\frac{1}{2}\frac{l^{2}\omega ^{2}}{g}$

This is an incorrect option.

Option 4)

$\; \frac{1}{6}\frac{l^{2}\omega^{2}}{g}$

This is the correct option.

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