Question

A stone is tied to a string of length $l$ and is whirled in a vertical circle with the other end of the string as the center. At a certain instant of time, the stone is at its lowest position and has a speed $u$. 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. $\sqrt{2gl}$
• C. $2\sqrt{u^2-2gl}$
• D. $\sqrt{2(u^2-2gl})$

Answer 1

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


The change in velocity is given as, $\overrightarrow{\Delta V}=V_1\hat{i}-V_2\hat{j}$

The magnitude of change in velocity is given as,

$\Delta V=\sqrt{V_1^2+V_2^2}....\left(i\right)$

From the conservation of energy of the stone between the positions $1$ & $2$,

$\frac{1}{2}m{V}_1^2=\frac{1}{2}m{V}_2^2+mgl$

$\Rightarrow V_2=\sqrt{V_1^2-2gl}....\left(ii\right)$

Using equations $\left(i\right)$ and $\left(ii\right)$

$\Delta V=\sqrt{2(V_1^2-gl)}$

According to problem, $V_1=u$

$\therefore \Delta V=\sqrt{2(u^2-gl)}$

Hence, option (d) is correct answer.