# 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.

Q: 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) $\sqrt{u^2 – g l}$

(b) $u – \sqrt{u^2 – g l}$

(c) $\sqrt{2 g l}$

(d) $\sqrt{2 (u^2 – g l )}$

Ans: (d)

Applying energy conservation

$\frac{1}{2}m u^2 = \frac{1}{2}m v^2 + m gl$

$u^2 = v^2 + 2 gl$

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

Change in velocity $\vec{\Delta v} = v j – u i$

$= \sqrt{v^2 +u^2}$

$= \sqrt{u^2 -2gl + u^2}$

$= \sqrt{2u^2 – 2 g l}$