Physics - Circular Motion Question with Solution | TestHub

Suppose the string becomes slack when the particle reaches the point P (figure). Suppose the string OP makes an angle $\theta$ with the upward vertical. The only force acting on the particle at P is its weight mg . The radial component of the force is $\mathrm{mg}\cos \theta$. As the particle moves on the circle upto P ,

$$\mathrm{mg}\cos \theta =\mathrm{m}\left(\frac{{\mathrm{v}}^2}{\ell }\right)$$

or, ${\mathrm{v}}^2=\mathrm{g}\ell \cos \theta$ where v is it speed at P . Using conservation of energy,

$$\begin{array}{rlr}& \frac{1}{2}m{v}_0^2=\frac{1}{2}m{v}^2+mg\ell (1+\cos \theta )& \text{ or, }{v}^2=v_0^2-2g\ell (1+\cos \theta )\end{array}$$

From (i) and (ii), ${\mathrm{v}}_0^2-2\mathrm{g}\ell (1+\cos \theta )$ or, ${\mathrm{v}}_0^2=\mathrm{g}\ell (2+3\cos \theta )$ Now onwards the particle goes in a parabola under the action of gravity. As it passes through the point of suspension O , the equations for horizontal and vertical motions give,

$$\ell \sin \theta =(\mathrm{v}\cos \theta )\mathrm{t}$$

and $-\ell \cos \theta =(\mathrm{v}\sin \theta )\mathrm{t}-\frac{1}{2}{\mathrm{gt}}^2$ or, $-\ell \cos \theta =(\mathrm{v}\sin \theta )\left(\frac{\ell \sin \theta }{\mathrm{v}\cos \theta }\right)-\frac{1}{2}\mathrm{g}{\left(\frac{\ell \sin \theta }{\mathrm{v}\cos \theta }\right)}^2$ or, $-{\cos }^2\theta ={\sin }^2\theta -\frac{1}{2}\mathrm{g}\frac{\ell {\sin }^2\theta }{{\mathrm{v}}^2\cos \theta }$ or, $-{\cos }^2\theta =1-{\cos }^2\theta -\frac{1}{2}\frac{\mathrm{g}\ell {\sin }^2\theta }{\mathrm{g}\ell {\cos }^2\theta }[\mathrm{From}(\mathrm{i})]$ or, $1=\frac{1}{2}{\tan }^2\theta$ or, $\tan \theta =\sqrt{2}$, From (iii), ${\mathrm{v}}_0=[\mathrm{g}\ell (2+\sqrt{3}){]}^{1/2}=3.16\mathrm{m}/\mathrm{s}$