Rotation Ques 93
- The potential energy of mass $m$ at a distance $r$ from a fixed point $O$ is given by $V(r)=k r^{2} / 2$, 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?
(a) $v=\sqrt{\frac{k}{2 m}} R$
(b) $v=\sqrt{\frac{k}{m}} R$
(c) $L=\sqrt{m k} R^{2}$
(d) $L=\sqrt{\frac{m k}{2}} R^{2}$
(2018 Adv.)
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Answer:
Correct Answer: 93.(b,c)
Solution:
- $V=\frac{K r^{2}}{2}$
$$ F=-\frac{d V}{d r}=-K r(\text { towards centre }) F=-\frac{d V}{d r} $$
$$ \begin{aligned} k R & =\frac{m v^{2}}{R}(\text { Centripetal force }) \\ v & =\sqrt{\frac{k R^{2}}{m}}=\sqrt{\frac{k}{m}} R \\ L & =m v R=\sqrt{m k} R^{2} \end{aligned} $$
BETA
