53. A particle of mass [tex]$3 \, \text{kg}[tex]$[/tex] is moving in a circular orbit of radius [tex]$[/tex]10 \, \text{m}$[/tex] under the action of a central force whose potential energy is given by [tex]U(r) = 10 r^3 \, \text{joules}[/tex]. The time period of this motion is:
To determine the time period of the motion for a particle of mass \(3 \, \text{kg}\) moving in a circular orbit of radius \(10 \, \text{m}\) under the action of a central force whose potential energy is given by \(U(r) = 10r^3\) joules, we need to follow these steps:
1. Find the force acting on the particle: The force \(F\) is related to the potential energy \(U(r)\) by the relation: [tex]\[
F(r) = - \frac{dU(r)}{dr}
\][/tex] Given \(U(r) = 10r^3\), we find the derivative: [tex]\[
\frac{dU(r)}{dr} = \frac{d}{dr}(10r^3) = 30r^2
\][/tex] Thus, [tex]\[
F(r) = -30r^2
\][/tex]
2. Equate the centripetal force to the force derived from the potential energy: The centripetal force needed to keep the particle in a circular orbit of radius \(r\) is given by: [tex]\[
F_c = m \omega^2 r
\][/tex] where \(m\) is the mass of the particle and \(\omega\) is the angular velocity. The calculated force \(F(r)\) must equal this centripetal force: [tex]\[
m \omega^2 r = 30r^2
\][/tex]
3. Solve for the angular velocity \(\omega\): Plugging in the mass \(m = 3 \, \text{kg}\) and rearranging for \(\omega^2\), [tex]\[
3 \omega^2 r = 30r^2 \implies \omega^2 = \frac{30r^2}{3r} \implies \omega^2 = 10r \implies \omega = \sqrt{10r}
\][/tex]
4. Calculate the time period \(T\): The time period \(T\) of the circular motion is given by: [tex]\[
T = \frac{2\pi}{\omega}
\][/tex] Substituting \(\omega = \sqrt{10r}\) into the formula gives: [tex]\[
T = \frac{2\pi}{\sqrt{10r}} = \frac{2\pi}{\sqrt{10 \cdot 10}} = \frac{2\pi}{\sqrt{100}} = \frac{2\pi}{10} = \frac{\pi}{5}
\][/tex]
Therefore, the time period of the motion is [tex]\(\frac{\pi}{5} \, \text{s}\)[/tex], which corresponds to option (d).
