Question

A light rod of length L can revolve in a vertical circle around point O. The rod carries two equal masses of mass m each such that one mass is connected at the end of the rod and the second mass is fixed at the middle of the rod. u is the velocity imparted to the end P to deflect the rod to the horizontal position. Again mass m in the middle of the rod is removed and mass at end P is doubled. Now v is the velocity imparted to end P to deflect it to the horizontal position. Then v/u is

• A. (6/5)^1/2
• B. 1
• C. 2.5
• D. (5/6)^1/2

Answer 1

Answer: (D)

(5/6)^1/2

Answer 2

Using conservation of energy

In first case, $\frac{1}{2}mu^{2} + \frac{1}{2}m\left( \frac{u}{2} \right)^{2}$

= mgL + mg $\frac{L}{2}$

or u = $\left( \frac{12}{5}gL \right)$

In second case

$\frac{1}{2}$ (2) v² = 2 mgL v = (2gL)^1/2

∴ $\frac{u}{v} = \frac{(12/5gL)^{1/2}}{\lbrack 2gL\rbrack^{1/2}} = \left\lbrack \frac{6}{5} \right\rbrack^{1/2}$

or v = $\left\lbrack \frac{6}{5} \right\rbrack^{1/2}$