Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Get the answers you need quickly and accurately from a dedicated community of experts on our Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
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).
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.