A bob of mass $m$ is suspended at a point $O$ by a light string of length $l$ and left to perform vertical circular motion as shown in the figure. Initially, by applying horizontal velocity $v_0$ at the point $A$, the string becomes slack when the bob reaches the point $D$. The ratio of the kinetic energy of the bob at the points $B$ and $C$ is
Answer: (B) $2$
The string just becomes slack at the top $D$, so tension is zero there and gravity alone provides the centripetal force:
$$\frac{mv_D^2}{l} = mg \Rightarrow v_D^2 = gl$$
Energy conservation from $A$ (height $0$) to $D$ (height $2l$): $v_A^2 = v_D^2 + 4gl = 5gl$.
Heights above $A$:
- $B$ is $60^\circ$ from the lowest point: $h_B = l(1-\cos60^\circ) = l/2$
- $C$ is $60^\circ$ from the highest point: $h_C = l(1+\cos60^\circ) = 3l/2$
$$v_B^2 = 5gl - gl = 4gl, \qquad v_C^2 = 5gl - 3gl = 2gl$$
$$\frac{K_B}{K_C} = \frac{4gl}{2gl} = 2$$
Solution by Sreeraj P, M.Sc Physics