Answer :
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).
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).
