Question

A stone is tied to a string of length ${}^{\prime }{l}^{\prime }$ 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 ${}^{\prime }{u}^{\prime }$. 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. $\sqrt{u^2-2gL}$
• B. $\sqrt{u^2-gL}$
• C. $\sqrt{2gL}$
• D. $\sqrt{2(u^2-gL)}$

Answer 1

The correct option is D $\sqrt{2(u^2-gL)}$

The change in velocity is given as,

$v_c=v\hat{j}-u\hat{i}$

The magnitude of change in velocity is given as,

$\left|v_c\right|=\sqrt{v^2+u^2}$

From energy conservation, it can be written as,

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

$v^2=u^2-2gl$

The magnitude of change in velocity is given as,

$\left|v_c\right|=\sqrt{u^2-2gl+u^2}$

$\left|v_c\right|=\sqrt{2\left(u^2-gl\right)}$