
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.
Answer
408.3k+ views
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.
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.
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