Question

A stone tied to a string of length $l$ 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 point and has a speed $u$. The magnitude of change in velocity as it reaches its highest point of the circle is

• A. $u-\sqrt{u^2-4gl}$
• B. $\sqrt{2(u^2-gl)}$
• C. $\sqrt{4gl}$
• D. $u+\sqrt{u^2-4gl}$

Answer 1

The correct option is D $u+\sqrt{u^2-4gl}$


From conservation of mechanical energy,
$K_f+U_f=K_i+U_i$
$\frac{m{V}^2}{2}+mg\left(2l\right)=\frac{m{u}^2}{2}$

We have,
$V^2=u^2-4gl$
$V=\sqrt{u^2-4gl}$
Change in velocity
$\overrightarrow{△V}=\overrightarrow{V}-\overrightarrow{u}$
$=-v\hat{i}-(+u\hat{i})$
$=(v+u)(-\hat{i})$
$|\overrightarrow{△V}|=V+u=u+\sqrt{u^2-4gl}$