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 speed $u$. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is
A) $\sqrt{u^{2} - 2gL}$
B) $\sqrt{2gL}$
C) $\sqrt{u^{2} - gl}$
D) $\sqrt{2(u^{2} - gL)}$
Explanation
(d) $\frac{1}{2}mu^{2} - \frac{1}{2}mv^{2} = mgL$
$\Rightarrow v = \sqrt{u^{2} - 2gL}$
$\left| \overrightarrow{v} - \overrightarrow{u} \right| = \sqrt{u^{2} + v^{2}} = \sqrt{u^{2} + u^{2} - 2gL} = \sqrt{2(u^{2} - gL)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.