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Q.

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 must be imparted to the mass at A so that it 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

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$\frac{5}{8}$ Let the initial velocity given to the mass at A be u.
Question Image
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 mgl + 2mgl
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
mu2 = 6 mgl or $u = \sqrt{\frac{48}{5} g ℓ}$

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