Question

a uniform rod of mass m and length l is hinged at upper end the rod is free to rotate in vertical plane a ball of mass m moving horizontally with velocity v colloides at lower end of rod perpendicularly to it and stick to it the minimum velocity of ball such that combined system just complete the vertical circle will be

Answer 1

The Minimum velocity of the ball such that combined system just complete the vertical circle is $2\sqrt{\frac{gl}{3}}$

Explanation:

We know that for a rod of mass m and length l is

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

Initially the angular velocity of the rod is zero

If after impact the combined angular velocity of the rod and the ball is $\omega$

Then,

Initial Kinetic Energy of the ball

$=\frac{1}{2}mv^2$

This will comvert into rotational kinetic energy of the system (Rod + Ball) after impact

Therefore,

the kinetic energy of the system (Rod + Ball) after impact

$=\frac{1}{2}I_{Sys}\omega^2$

Where, $I_{Sys}$ is the moment of inertia of rod + ball system

$I_{Sys}=\frac{1}{3}ml^2+ml^2$

$\implies I_{Sys}=\frac{4}{3}ml^2$

From the conservation of energy

The kinteic energy before the impact and after the impact will be same

Hence,

$\frac{1}{2}mv^2=\frac{1}{2}I_{Sys}\omega^2$

$\implies mv^2=\frac{4}{3}ml^2\times \omega^2$

$\implies \omega^2=\frac{3v^2}{4l^2}$

$\implies \omega=\sqrt{\frac{3}{4}}\frac{v}{l}$

Thus, the linear velocity of the system

$V=\omega\times l$

$\implies V=\sqrt{\frac{3}{4}}v$

For the system to just complete a full circle

$V=\sqrt{gl}$

$\implies \sqrt{\frac{3}{4}}v=\sqrt{gl}$

$\implies v^2=\frac{4}{3}gl$

$\implies v=2\sqrt{\frac{gl}{3}}$

Hope this answer is helpful.

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