Q: A stone tied to a string of length L is whirled in a vertical circle with the other end of the string at 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 its velocity as it reaches a position, where the string is horizontal, is

(a) $\large \sqrt{u^2 – 2 g l}$

(b) $\large \sqrt{2 g l}$

(c) $\large \sqrt{u^2 – g l}$

(d)$\large \sqrt{2(u^2 – g l)}$

Ans: (d)

Sol: Applying conservation of energy between two positions , one is its lowest position & other is when string is horizontal .

$\large \frac{1}{2}mu^2 = m g L + \frac{1}{2}mv^2$

$\large v^2 = u^2 – 2 g L$

Change in velocity = Final velocity – Initial velocity

$\large \vec{\Delta v} = v\hat{j} – u\hat{i}$

$\large |\Delta v| = \sqrt{v^2 + u^2}$

$\large |\Delta v| = \sqrt{u^2 – 2 g L + u^2}$

$\large |\Delta v| = \sqrt{2u^2 – 2 g L }$

$\large |\Delta v| = \sqrt{2(u^2 – g L ) }$