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. $\sqrt{5gL}$
• B. $\sqrt{2gL}$
• C. $\sqrt{3gL}$
• D. $\sqrt{7gL}$

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. $T_A-mg=\frac{m{v}_0^2}{L}$
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 $C,mg=\frac{m{v}_C^2}{L}$...(ii)
and total energy, $E=\frac{1}{2}m{v}_C^2+2mgL$
where, $v_C$ is the speed at $C$.
$\Rightarrow v_C^2=gL$ ...(iii)
From Eqs. (ii) and (iii), we get
$E=\frac{1}{2}mgL+2mgL=\frac{5}{2}mgL$ ...(iv)
Equating Eqs. (i) and (iv), we get
$\frac{5}{2}mgL=\frac{m}{2}{v}_0^2$
or $v_0=\sqrt{5gL}$