A particle of mass $m$ moves under the action of some force along a circular orbit of radius $R$ with constant velocity. The potential energy $U$ of the particle in the field of this force depends only on the distance to the center of the orbit as follows:
a) $U(R) = kR (k > 0)$;
b) $U(R)=kR^{2} \: (k>0)$.
Find the velocity of the particle in cases a) and b).
Decision:
A particle of mass $m$ moves along a circle of radius $R$ with constant velocity $v$ if a force $F$, constant in magnitude and directed to the center of the circle at any time, acts on it. This force produces a centripetal acceleration
$\frac{v^{2}}{R}$,
and by Newton's second law.
$F = \frac{mv^{2}}{R}$. (1)
Formula (1) allows us to determine the velocity of the particle and, if the force $F$ as a function of $R$ is known:
$v = \sqrt {RF(R)/m}$. (2)
Let us find the relationship between the function $F(R)$ and the quantity $U(R)$.
Let us mentally move a particle with some constant velocity (so that its kinetic energy remains unchanged) in a field with potential energy $U(R)$ along an arbitrary radius. To do this, the particle must be acted upon by a force $F(R)$, external to the "particle-field" system, equal in magnitude to the desired force and directed in the opposite direction, i.e. from the center. The total force will be equal to zero, and therefore the motion will be uniform. The force $F(R)$ on the path $\Delta R$ from a point with radial coordinate $R$ to a point with radial coordinate $R + \Delta R$ does work $\Delta A \approx F(R) \Delta R$. This approximation is more accurate the smaller the value of $\Delta R$ is. All the work $\Delta A$ done by the external force, goes to replenish only the potential energy of the system "particle-field" (kinetic energy is constant!). As a result of this work, the potential energy of the system increases by the value $\Delta U = \Delta A \approx $F(R)$ \Delta R$. Hence roughly $F(R) \approx \frac{ \Delta U }{ \Delta R}$. The smaller $\Delta R$ is, the more accurate this equality will be. In the limit, when $\Delta R \rightarrow 0$ we get the exact equality:
$F(R)=\lim_{\Delta R \rightarrow 0} \frac{\Delta U}{ \Delta R} = \frac{dU}{dR}$. (3)
So, the desired function $F(R)$ is equal to the derivative of the potential energy $U(R)$ over $R$.
Using the general formula (3), in case (a) we obtain
$F(R) = \frac{dU}{dR} = k$, (4)
i.e., $F$ is a constant, independent of $R$. Substituting expression (4) into (2), we find:
$v = \sqrt{ \frac{kR}{m}}$.
In case b):
$F(R) = \frac{dU}{dR}=2kR$. (5)
Substituting expression (5) into (2), we have:
$v = R \sqrt{ \frac{2k}{m}}$.
a) $U(R) = kR (k > 0)$;
b) $U(R)=kR^{2} \: (k>0)$.
Find the velocity of the particle in cases a) and b).
Decision:
A particle of mass $m$ moves along a circle of radius $R$ with constant velocity $v$ if a force $F$, constant in magnitude and directed to the center of the circle at any time, acts on it. This force produces a centripetal acceleration
$\frac{v^{2}}{R}$,
and by Newton's second law.
$F = \frac{mv^{2}}{R}$. (1)
Formula (1) allows us to determine the velocity of the particle and, if the force $F$ as a function of $R$ is known:
$v = \sqrt {RF(R)/m}$. (2)
Let us find the relationship between the function $F(R)$ and the quantity $U(R)$.
Let us mentally move a particle with some constant velocity (so that its kinetic energy remains unchanged) in a field with potential energy $U(R)$ along an arbitrary radius. To do this, the particle must be acted upon by a force $F(R)$, external to the "particle-field" system, equal in magnitude to the desired force and directed in the opposite direction, i.e. from the center. The total force will be equal to zero, and therefore the motion will be uniform. The force $F(R)$ on the path $\Delta R$ from a point with radial coordinate $R$ to a point with radial coordinate $R + \Delta R$ does work $\Delta A \approx F(R) \Delta R$. This approximation is more accurate the smaller the value of $\Delta R$ is. All the work $\Delta A$ done by the external force, goes to replenish only the potential energy of the system "particle-field" (kinetic energy is constant!). As a result of this work, the potential energy of the system increases by the value $\Delta U = \Delta A \approx $F(R)$ \Delta R$. Hence roughly $F(R) \approx \frac{ \Delta U }{ \Delta R}$. The smaller $\Delta R$ is, the more accurate this equality will be. In the limit, when $\Delta R \rightarrow 0$ we get the exact equality:
$F(R)=\lim_{\Delta R \rightarrow 0} \frac{\Delta U}{ \Delta R} = \frac{dU}{dR}$. (3)
So, the desired function $F(R)$ is equal to the derivative of the potential energy $U(R)$ over $R$.
Using the general formula (3), in case (a) we obtain
$F(R) = \frac{dU}{dR} = k$, (4)
i.e., $F$ is a constant, independent of $R$. Substituting expression (4) into (2), we find:
$v = \sqrt{ \frac{kR}{m}}$.
In case b):
$F(R) = \frac{dU}{dR}=2kR$. (5)
Substituting expression (5) into (2), we have:
$v = R \sqrt{ \frac{2k}{m}}$.