Q: A bob is attached to one end of a string other end of which is fixed at peg A. The bob is taken to a position where string makes an angle of 30° with the horizontal. On the circular path of the bob in vertical plane there is a peg B’ at a symmetrical position with respect to the position of release as shown in the figure. If Vc and Va be the minimum tangential velocity in clockwise and anticlockwise directions respectively, given to the bob in order to hit the peg B’ then ratio V_{c} : V_{a} is equal

(A) 1 : 1

(B) 1 : √2

(C) 1 : 2

(D) 1 : 4

Ans: (C)

Sol: For anti-clockwise motion, speed at the highest point should be $\sqrt{gR}$ . Conserving energy at (1) & (2) :

$\displaystyle \frac{1}{2}m v_a^2 = m g\frac{R}{2} + \frac{1}{2}m g R$

$\displaystyle v_a^2 = g R + g R $

$\displaystyle v_a = \sqrt{2 g R} $

For clock-wise motion, the bob must have atleast that much speed initially, so that the string must not

become loose any where until it reaches the peg B. At the initial position :

$T + m g cos60^o = \frac{m v_c^2}{R}$

V_{C} being the initial speed in clockwise direction. For V_{C min} : Put T = 0 ;

$\displaystyle V_c = \sqrt{\frac{gR}{2}} $

$\displaystyle \frac{V_c}{V_a} \frac{\frac{gR}{2}}{\sqrt{2gR}} = \frac{1}{2}$

V_{c} : V_{a} = 1 : 2