Question

A particle is rotated in a vertical circle by connecting it to a string of length l and keeping the other end of the string fixed. The minimum speed of the particle when the string is horizontal for which the particle will complete the circle is
(a) $\sqrt{gl}$
(b) $\sqrt{2gl}$
(c) $\sqrt{3gl}$
(d) $\sqrt{5gl}$

Answer 1

(c) $\sqrt{3gl}$

Suppose that one end of an extensible string is attached to a mass m, while the other end is fixed. The mass moves with a velocity v in a vertical circle of radius R. At some instant, the string makes an angle θ with the vertical as shown in the figure.



For a complete circle, the minimum velocity at L must be $v_{\mathrm{L}}=\sqrt{5gl}$.

Applying the law of conservation of energy, we have:

Total energy at M = total energy at L
$$\begin{gathered} \mathrm{i}.\mathrm{e}.,\frac{1}{2}m{v_{\mathrm{M}}}^2+mgl=\frac{1}{2}m{v_{\mathrm{L}}}^2 \\ \Rightarrow \frac{1}{2}m{v_{\mathrm{M}}}^2=\frac{1}{2}m{v_{\mathrm{L}}}^2-mgl \\ \mathrm{Using}{v}_L\ge \sqrt{5gl},\mathrm{we}\mathrm{have}: \\ \frac{1}{2}m{v_{\mathrm{M}}}^2\ge \frac{1}{2}m\left(5gl\right)-mgl \\ \therefore v_{\mathrm{M}}=\sqrt{3gl} \end{gathered}$$