The potential energy of a particle of mass m at a distance r from a fixed point O is given by $V (r) = k \frac{r^{2}}{2}$ .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?

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

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

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

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

answer is B, C.

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Detailed Solution

$V = \frac{k r^{2}}{2}$

$F = - \frac{d V}{d r} = - k r$

At r=R,

$k R = \frac{m v^{2}}{R}$ (Centripetal force)

$\begin{array}{l} v = \sqrt{\frac{k R^{2}}{m}} = \sqrt{\frac{k}{m}} R \\ Angular Momentum = L = m v R = m \sqrt{\frac{k}{m}} R^{2} = \sqrt{m k} R^{2} \end{array}$

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