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A weightless rod of length $2 l$ carries two equal masses ‘m’, one secured at lower end A and the other at the middle of the rod at B. The rod can rotate in vertical plane about a fixed horizontal axis passing through C. The horizontal velocity that must be imparted to the mass at A so that is just completes the vertical circle is $\sqrt{\frac{P}{5} g l}$ . Then the value of $P$ is _____?

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answer is 48.

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Detailed Solution

Let the initial velocity given to the mass at A be u.

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Then the velocity of mass at b is u/2
As the system moves from initial the final position increase in potential energy is $= 4 m g l + 2 m g l$
Decrease in kinetic energy $= \frac{1}{2} m u^{2} + \frac{1}{2} m {(\frac{u}{2})}^{2} = \frac{5}{8} m u^{2}$

From conservation of energy $\frac{5}{8} m u^{2} = 6 m g l$ or $u = \sqrt{\frac{48}{5} g l}$

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