Question

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

Answer 1

The correct option is D

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

Now, since the two velocity vectors shown in Fig. 8.296 are mutually perpendicular, the magnitude of change of velocity will be given by
$|\Delta \overrightarrow{v}|=\sqrt{u^2+v^2}$

Substituting the value of $v^2$ from Eq. (i),
$|\Delta \overrightarrow{v}|=\sqrt{u^2+u^2-2gL}=\sqrt{2(u^2-gL)}$