Question

A stone tied to a string of length L is whirled in a vertical circle with the other end of string at the centre. At a centre instant of time the stone is at it lowest position and has a speed u. The magnitude of change in velocity as it reaches a position where string is horizontal is

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

Answer 1

Answer: (D)

$\sqrt{2(u^2-gL)}$

At the lowest position the string has only Kinetic energy which is the total energy.
$T.E=K.E=\frac{1}{2}m{u}^2$
when the string is in horizontal position it has both kinetic and potential energy. But the total energy remains constant. $P.E=mgL$ , $K.E=\frac{1}{2}m{v}^2$
As the total energy remains const. $\frac{1}{2}m{u}^2=mgL+\frac{1}{2}m{v}^2$
$⟹v^2=u^2-2gL$
$v=\sqrt{u^2-2gL}$
here initial velocity is in horizontal direction($\hat{i}$)
and final velocity is in vertical direction($\hat{j}$)
hence change in velocity will be $u\hat{i}-v\hat{j}$
$\therefore$ Magnitude of change in velocity is $|v|=\sqrt{u^2+u^2-2gL}=\sqrt{2(u^2-gL)}$