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 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 velocity as it reaches a position where the string is horizontal (g being acceleration due to gravity) is

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

Answer 1

Answer: (A)

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

Answer 2

Key Idea : When stone reaches a position where string is horizontal, it attains the energy partially as kinetic and partially as potential. When stone is at its lowest position, it has only kinetic energy, given by $K=\frac{1}{2} m u^{2}$ At the horizontal position, it has energy $E = U + K = \frac{1}{2} mu'^2 + mgl$ According to conservation of energy, $K =E$ $\therefore \frac{1}{2} m u^{2} =\frac{1}{2} m u^{\prime 2}+m g l$ or $\frac{1}{2} m u^{\prime 2}=\frac{1}{2} m u^{2}-m g l$ or $u^{\prime 2} =u^{2}-2 g l$ or $u^{\prime} =\sqrt{u^{2}-2 g l} \ldots$(i) So, the magnitude of change in velocity $|\Delta \vec{ u }| =|\overrightarrow{ u }|=\sqrt{u^{\prime 2}+u^{2}+2 u^{\prime} u \cos 90^{\circ}}$ $|\Delta \overrightarrow{ u }| =\sqrt{u^{\prime 2}+u^{2}}$ $=\sqrt{2\left(u^{2}-g L\right)}$ [from E (i)]