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A particle of mass $9\,kg$ is moving under the action of a central force whose potential energy is given by $U = \dfrac{{10}}{r}$. For what energy it will orbit a circle of radius $10\,m$? Calculate the time period of this motion.

Last updated date: 21st Jul 2024
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Hint: The net force which acts on an object to keep it moving in a circular path is called centripetal force. Newton’s first law says that an object will continue moving along a straight line path until an external force acts on it. The external force in this case is the centripetal force.

Complete step by step answer:
Given that:
$U = \dfrac{{10}}{r}$
$\Rightarrow E = \dfrac{{10}}{{10}} = 1\,J$
The centripetal force:
$|{\text{f}}| = \left| {\dfrac{{ - du}}{{dr}}} \right| \\
\Rightarrow |{\text{f}}|= + \dfrac{{10}}{{{r^2}}}$
Now, centripetal force = centrifugal force
$\dfrac{{10}}{{{r^2}}} = \dfrac{{m{v^2}}}{r}$
$\Rightarrow {v^2} = \dfrac{{10}}{{10 \times 9}} \\
\Rightarrow {v^2} = \dfrac{1}{9}$
Adding square root on both sides:
$v = \dfrac{1}{3}\,m/s$
The time period:
$T = \dfrac{{2\pi r}}{v} \\
\Rightarrow T = \dfrac{{2\pi \times 10 \times 3}}{1} \\
\therefore T = 60\pi {\text{ sec}}$

Therefore, the time period of this motion is $60\pi {\text{ sec}}$.

Note: The force which is needed to keep an object moving in a curved path that is directed inward towards the center of rotation is called centripetal force whereas the apparent force that is felt by an object which is moving in a curved path that acts outwardly away from the center is called as centrifugal force. The centrifugal force is equal in both the magnitude and dimensions with the centripetal force.