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