The potential energy of a particle of mass at a distance from a fixed point O is given by , where is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius about the point O. If is the speed of the partic

Question

The potential energy of a particle of mass $m$ at a distance $r$ from a fixed point O is given by $V (r) = \frac{k r^{2}}{2}$ , where $k$ is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius $R$ about the point O. If $v$ is the speed of the particle and $L$ is the magnitude of its angular momentum about O, which of the following statements is (are) true? [2018]

Smart Achievers · Accepted Answer

(2, 3)

Given : potential energy, $r$

$Applying,$ $k$ $= \frac{d V}{d r} = \frac{d}{d r} [\frac{k r^{2}}{2}] = k r$

$For r = R, F = k R$

$Also F = \frac{m v^{2}}{R}$ $(∵ particle is moving in circular orbit)$

$or, \frac{m v^{2}}{R} = k R \Rightarrow v = \sqrt{\frac{k}{m}} R$

$And angular momentum, L = m v R = m (\sqrt{\frac{k}{m}} R) R = \sqrt{k m} R^{2}$

Solution

1 $v = \sqrt{\frac{k}{2 m}} R$

2 $v = \sqrt{\frac{k}{m}} R$

3 $L = \sqrt{m k} R^{2}$

4 $L = \sqrt{\frac{m k}{2}} R^{2}$

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Solution

1 v=k2mR

2 v=kmR

3 L=mkR2

4 L=mk2R2

Ans. (2, 3) Given : potential energy, V(r)=Kr22 Applying, |F|=dVdr=ddr[kr22]=kr For r=R, F=kR Also F=mv2R (∵particle is moving in circular orbit) or, mv2R=kR⇒v=kmR And angular momentum, L=mvR=m(kmR)R=kmR2

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1 $v = \sqrt{\frac{k}{2 m}} R$

2 $v = \sqrt{\frac{k}{m}} R$

3 $L = \sqrt{m k} R^{2}$

4 $L = \sqrt{\frac{m k}{2}} R^{2}$