A weightless rod of length l carries two equal masses m one fixed at the end and other in the middle of the rod. The rod can revolve in a vertical plane about A . Then horizontal velocity which must b

Question

A weightless rod of length l carries two equal masses m one fixed at the end and other in the middle of the rod. The rod can revolve in a vertical plane about A . Then horizontal velocity which must be imparted to end C of the rod to deflect it to the horizontal position is

TestHub · Accepted Answer

Loss in kinetic energy = gain in potential energy

$∴ \frac{1}{2} I ω^{2} = m g \frac{l}{2} + m g l$ ...(i)

Here, $I = \frac{m l^{2}}{4} + m l^{2} = \frac{5}{4} m l^{2}$

From (i),

$\frac{1}{2} (\frac{5}{4} m l^{2}) ω^{2} = \frac{m g l}{2} + m g l = \frac{3}{2} m g l$

$m l^{2} ω^{2} = \frac{12}{5} m g l$

$∴ v_{c} = \sqrt{(\frac{12 g l}{5})}$

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