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. The string becomes slack only on reaching the topmost point $C$ as shown in figure. Then, the speed of bob $\left(v_{0}\right)$ at point $A$ is

• A. $\sqrt{5 g L}$
• B. $\sqrt{2 g L}$
• C. $\sqrt{3 g L}$
• D. $\sqrt{7 g L}$

Answer 1

Answer: (A)

The total mechanical energy $E$ of the system is conserved. We take the potential energy of the system to be zero at the lowest point $A$. Thus, at $A$,
$E=\frac{1}{2} m v_{0}^{2}$ ...(i)
The resultant force at $A$ provides the necessary centripetal force.
i.e. $T_{A}-m g=\frac{m v_{0}^{2}}{L}$
where, $T_{A}$ is the tension in the string at $A$.
At the highest point $C$, the string becomes slack, as the tension in the string $\left(T_{C}\right)$ becomes zero.
Thus, at $C, m g=\frac{m v_{C}^{2}}{L}$...(ii)
and total energy, $E=\frac{1}{2} m v_{C}^{2}+2 m g L$
where, $v_{C}$ is the speed at $C$.
$\Rightarrow v_{C}^{2}=g L$ ...(iii)
From Eqs. (ii) and (iii), we get
$E=\frac{1}{2} m g L+2 m g L=\frac{5}{2} m g L$ ...(iv)
Equating Eqs. (i) and (iv), we get
$\frac{5}{2} m g L=\frac{m}{2} v_{0}^{2}$
or $v_{0}=\sqrt{5 g L}$