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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

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}

Answers (1)

best_answer

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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