A point particle is held on the axis of a rin | vedclass

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(C) Let $M$ be the mass of the particle.Using the principle of conservation of energy,the total energy at the initial position (at distance $r$ from t

Vedclass · Accepted Answer

$\sqrt {\frac{{2Gm}}{r}\left( {1 - \frac{1}{{\sqrt 2 }}} \right)} $

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(C) Let $M$ be the mass of the particle.Using the principle of conservation of energy,the total energy at the initial position (at distance $r$ from the center) equals the total energy at the center $C$.The gravitational potential $V_p$ at a distance $x$ on the axis of a ring of mass $m$ and radius $r$ is given by $V_p = -\frac{Gm}{\sqrt{r^2 + x^2}}$.Initial potential energy $U_i = M 	imes V_p(r) = -\frac{GMm}{\sqrt{r^2 + r^2}} = -\frac{GMm}{r\sqrt{2}}$.Final potential energy at center $U_f = M 	imes V_p(0) = -\frac{GMm}{r}$.By conservation of energy: $U_i + K_i = U_f + K_f$.$-\frac{GMm}{r\sqrt{2}} + 0 = -\frac{GMm}{r} + \frac{1}{2}MV^2$.$\frac{1}{2}MV^2 = \frac{GMm}{r} - \frac{GMm}{r\sqrt{2}} = \frac{GMm}{r} (1 - \frac{1}{\sqrt{2}})$.$V^2 = \frac{2Gm}{r} (1 - \frac{1}{\sqrt{2}})$.$V = \sqrt{\frac{2Gm}{r} (1 - \frac{1}{\sqrt{2}})}$.

$A$ point particle is held on the axis of a ring of mass $m$ and radius $r$ at a distance $r$ from its centre $C$. When released,it reaches $C$ under the gravitational attraction of the ring. Its speed at $C$ will be

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