Laws of Motion 5 Question 2

2. A particle is moving in a circular path of radius $a$ under the action of an attractive potential energy $U=-\frac{k}{2 r^{2}}$. Its total energy is

(2018 Main)

(a) $-\frac{3}{2} \cdot \frac{k}{a^{2}}$

(b) $-\frac{k}{4 a^{2}}$

(c) $\frac{k}{2 a^{2}}$

(d) zero

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Answer:

Correct Answer: 2. (d)

Solution:

  1. Given, $U=-\frac{k}{2 r^{2}} \Rightarrow F _r=-\frac{d U}{d r}=-\frac{k}{r^{3}}$

Since, the particle moves in a circular path of radius ’ $a$ ‘, the required centripetal force is provided by the above force.

Hence, $\quad \frac{m v^{2}}{a}=\frac{k}{a^{3}} \Rightarrow m v^{2}=\frac{k}{a^{2}}$

Kinetic energy, $K=\frac{1}{2} m v^{2}=\frac{k}{2 a^{2}}$

Total energy $=K+U=-\frac{k}{2 a^{2}}+\frac{k}{2 a^{2}}=0$



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