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(D) Applying the Law of Conservation of Mechanical Energy between the lowest point and point $P$:$\frac{1}{2} mv_0^2 = mg \ell(1 + \sin 	heta) + \fra

Vedclass · Accepted Answer

$\left(\frac{\sin 	heta}{2+3 \sin 	heta}ight)^{\frac{1}{2}}$

Vedclass · Answer

(D) Applying the Law of Conservation of Mechanical Energy between the lowest point and point $P$:$\frac{1}{2} mv_0^2 = mg \ell(1 + \sin 	heta) + \frac{1}{2} mv_p^2$ ... $(i)$At point $P$,the tension $T_p$ in the string becomes zero. The radial component of the gravitational force provides the necessary centripetal force:$mg \sin 	heta = \frac{mv_p^2}{\ell} \implies mv_p^2 = mg \ell \sin 	heta$ ... $(ii)$Substituting $(ii)$ into $(i)$:$\frac{1}{2} mv_0^2 = mg \ell(1 + \sin 	heta) + \frac{1}{2} mg \ell \sin 	heta$$v_0^2 = 2g \ell(1 + \sin 	heta) + g \ell \sin 	heta = 2g \ell + 3g \ell \sin 	heta$$v_0 = \sqrt{g \ell(2 + 3 \sin 	heta)}$ ... $(iii)$From $(ii)$,$v_p = \sqrt{g \ell \sin 	heta}$.Therefore,the ratio is:$\frac{v_p}{v_0} = \frac{\sqrt{g \ell \sin 	heta}}{\sqrt{g \ell(2 + 3 \sin 	heta)}} = \sqrt{\frac{\sin 	heta}{2 + 3 \sin 	heta}}$

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