Question

A stone is tied to a string of length $l$ and is whirled in a vertical circle with the other end of the string as the centre. At a certain instant of time, the stone is at its lowest position and has a speed $u$. The magnitude of the change in velocity as it reaches a position, where the string is horizontal, ($g$ being acceleration due to gravity) is

• A. $\sqrt{2(u^2-gl)}$
• B. $\sqrt{u^2-gl}$
• C. $u-\sqrt{u^2-2gl}$
• D. $\sqrt{2gl}$

Answer 1

The correct option is A $\sqrt{2(u^2-gl)}$


In this problem, there are no external and non-conservative forces, so we can apply conservation of mechanical energy between initial and final position.

$U_f+K_f=U_i+K_i$

Taking the lowest point of the circle as a reference point,

$\frac{1}{2}m{v}^2+mgh=\frac{1}{2}m{u}^2+mg\left(0\right)$

Here, $h=l$ for the given interval

$\therefore v=\sqrt{u^2-2gl}$

Now, the magnitude of change in velocity vector is,

$|\overrightarrow{\Delta v}|=|\overrightarrow{v}-\overrightarrow{u}|$


$|\overrightarrow{\Delta v}|=\sqrt{v^2+u^2-2uv\cos {90}^o}$

$|\overrightarrow{\Delta v}|=\sqrt{v^2+u^2}=\sqrt{(u^2-2gl)+u^2}$

$\therefore |\overrightarrow{\Delta v}|=\sqrt{2(u^2-gL)}$

Hence, answer will be option $\left(A\right)$.