Question

A bob of mass m is suspended by a light string of length L. It is imparted a horizontal velocity $v_0$ at the lowest point A such that it completes a semi-circular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point, C. This is shown in Fig. 6.6. Obtain an expression for (i) $v_0$; (ii) the speeds at points B and C; (iii) the ratio of the kinetic energies $\left(K_B/K_C\right)$ at B and C. Comment on the nature of the trajectory of the bob after it reaches the point C.

Answer 1

$\left(i\right)$Let the particle at the top most point be $um/s$

As the particle is in vertical circular motion;

So the centripetal acceleration is being provided by mg alone as string has slacked so, $mg=\frac{m{u}^2}{L}$

So, $u=\sqrt{gL}$ at the top most point (when string slacks).......Speed of C

Now appliying the energy conservation between A and C

We have,

$\frac{m{v}_0^2}{2}+0=\frac{m{u}^2}{2}+mg\left(2L\right)$

so , $\frac{m{v}_0^2}{2}=\frac{m{u}^2}{2}+mg\left(2L\right)$

so, $v_0=\sqrt{5gL}$......Speed of A

$\left(ii\right)$For point C

Applying conservation of energy

$\frac{m{v}_0^2}{2}+0=\frac{m{u}_B^2}{2}+mg\left(L\right)$

$\frac{m{v}_0^2}{2}=\frac{m{u}_B^2}{2}+mg\left(L\right)$

so, $u_B=\sqrt{3gL}$......Speed of B

$\left(iii\right)$Ratio of kinetic energy will be

$\frac{K{E}_B}{K{E}_C}=\frac{\frac{m{u}_B^2}{2}}{\frac{m{u}^2}{2}}=\frac{u_B^2}{u^2}=\frac{2gL}{gL}=\frac{3}{1}$

The object will complete the circular path and its trajectory will be circular as string has slacked at the highest point.