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

Answer 1

Answer: (D)

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

Answer 2

$\frac{1}{2}mu^{2} - \frac{1}{2}mv^{2} = mgl$

or v² = u² – 2gl

${\overrightarrow{v}}_{i} = u\widehat{i}$

${\overrightarrow{v}}_{f} = \widehat{j}\sqrt{u^{2} - 2gl}$

change in velocity $\Delta\overrightarrow{v} = {\overrightarrow{v}}_{f} - {\overrightarrow{v}}_{i} = \widehat{j}\sqrt{u^{2} - 2gl} - u\widehat{i}$magnitude of $\Delta\overrightarrow{v} = \sqrt{\left( u^{2} - 2gl \right) + u^{2}} = \sqrt{2\left( u^{2} - gl \right)}$.