A bob of mass m is suspended at a point O by a light string of length l and left to perform vertical motion (circular) as shown in figure. Initially, by applying horizontal velocity $$v_o$$ at the point ' A ', the string becomes slack when, the bob reaches at the point ' D '. The ratio of the kinetic energy of the bob at the points B and C is

________.

The string becomes slack at the highest point D,means the tension is zero. The centripetal force is provided entirely by gravity at this point.

$$ \frac{m v_D^2}{l} = mg $$

$$ v_D^2 = gl $$

$$ KE_D = \frac{1}{2} m v_D^2 = \frac{1}{2} mgl $$

Let the center O be the reference level for zero potential energy. The height of point D from O is $$l$$, so its potential energy is:

$$ U_D = mgl $$

According to the law of conservation of mechanical energy, the total energy $$E$$ at any point is constant.

$$ E = K_D + U_D $$

$$ E = \frac{1}{2} mgl + mgl = \frac{3}{2} mgl $$

Now, let us find the kinetic energy at point B ($$K_B$$).

The vertical depth of point B below O is $$l \cos 60^\circ$$.

$$ U_B = -mgl \cos 60^\circ = -mgl \left(\frac{1}{2}\right) = -\frac{1}{2} mgl $$

Using energy conservation at point B:

$$ E = K_B + U_B $$

$$ \frac{3}{2} mgl = K_B - \frac{1}{2} mgl $$

$$ K_B = \frac{3}{2} mgl + \frac{1}{2} mgl $$

$$ K_B = 2mgl $$

Next, let us find the kinetic energy at point C ($$K_C$$).

The vertical height of point C above O is $$l \cos 60^\circ$$.

$$ U_C = mgl \cos 60^\circ = mgl \left(\frac{1}{2}\right) = \frac{1}{2} mgl $$

Using energy conservation at point C:

$$ E = K_C + U_C $$

$$ \frac{3}{2} mgl = K_C + \frac{1}{2} mgl $$

$$ K_C = \frac{3}{2} mgl - \frac{1}{2} mgl $$

$$ K_C = mgl $$

Finally, we find the ratio of the kinetic energy at point B to that at point C.

$$ \text{Ratio} = \frac{K_B}{K_C} $$

$$ \text{Ratio} = \frac{2mgl}{mgl} $$

$$ \text{Ratio} = 2 $$