Question

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 ${30}^o$ 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 $V_c$ and $V_a$ be the minimum tangential velocity in clockwise and anticlockwise directions respectively, given to the bob in order to hit the peg $B$ then ratio $V_c:V_a$ will be equal to

• A. $1:1$
• B. $1:\sqrt{2}$
• C. $1:2$
• D. $1:4$

Answer 1

The correct option is C $1:2$

For anti-clockwise motion, speed at the highest point should be $=gR$

Conserving energy at $\left(1\right)$ and $\left(2\right)$

$\frac{1}{2}m{v_a}^2=mg\frac{R}{2}+\frac{1}{2}m\left(gR\right)$

$\Rightarrow {v_a}^2=gR+gR=2gR$

$\Rightarrow v_a=\sqrt{2gR}$

For clockwise motion, the bob must have atleast that much speed initially,so that the string must not become loose any where until it reaches the ped $B$

At the initial position:

$T+mg\cos {60}^{\circ }=\frac{m{v_c}^2}{R}$

$v_c$ being the initial speed in clockwise direction.

For $V_{c_{min}}:$Put $T=0$

$\Rightarrow v_c=\sqrt{\frac{gR}{2}}$

$\Rightarrow \frac{v_c}{v_a}=\frac{\sqrt{\frac{gR}{2}}}{\sqrt{2gR}}=\frac{1}{2}$

$\Rightarrow v_c:v_a=1:2$