Question

The potential energy of a particle of mass $ m $ at a distance $ r $ from a fixed point $ O $ is given by $ V(r) = \dfrac{{k{r^2}}}{2} $ , where $ k $ is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius $ R $ about the point $ O $ . If $ v $ is the speed of the particle and $ L $ is the magnitude of its angular momentum about O, which of the following is (are) true?
(A) $ v = \sqrt {\dfrac{k}{{2m}}} R $
(B) $ v = \sqrt {\dfrac{k}{m}} R $
(C) $ L = \sqrt {mk} {R^2} $
(D) $ L = \sqrt {\dfrac{{mk}}{2}} {R^2} $

Answer 1

Hint
Find the expression of force on the particle from the given expression of the potential energy of the particle. From the expression of the force, we can easily obtain the speed and angular momentum of the particle.
Formula Used: The formulas used to solve this problem are
- $ F = - \dfrac{{dV}}{{dr}} $ ; Where $ F $ is the force acting on the particle, and $ V $ is the potential energy of the particle
- $ {F_c} = \dfrac{{m{v^2}}}{R} $ ; Where $ {F_c} $ is the centripetal force acting on a particle of mass $ m $ moving in a circle of radius $ r $
- $ L = mvR $ ; Where $ L $ is the angular momentum of a particle of mass $ m $ , $ v $ is the speed of the particle

Complete step by step answer
It is given that the potential energy of the particle, $ V(r) = \dfrac{{k{r^2}}}{2} $
We know that the force acting on a particle in a conservative field is given by
  $ F = - \dfrac{{dV}}{{dr}} $
Substituting $ V = \dfrac{{k{r^2}}}{2} $
 $ F = - \dfrac{d}{{dr}}\left( {\dfrac{{k{r^2}}}{2}} \right) $
As $ \dfrac{k}{2} $ is a constant, it can be taken out
 $ F = - \dfrac{k}{2}\dfrac{d}{{dr}}\left( {{r^2}} \right) $
 $ F = - \dfrac{k}{2}(2r) $
Cancelling $ 2 $ , we get
 $ F = - kr $
As given the question, $ k $ is a positive constant. So the force is opposite to the direction of $ r $ . This means that the force is towards the point $ O $ , the centre of the circle. As the radius of the circle is $ R $ , so for obtaining the magnitude of the force, we make the substitution $ r = R $ in the above expression.
 $ F = kR $ (1)
Here we have omitted the negative sign as we are concerned with the magnitude of the force. As stated above, this force is towards the centre of the circle in which the particle is revolving. So, from Newton’s second law of motion, this force must be equal to the centripetal force.
As we know centripetal force, $ {F_c} = \dfrac{{m{v^2}}}{R} $ (2)
Equating (1) and (2), we get
 $ \dfrac{{m{v^2}}}{R} = kR $
Multiplying with $ \dfrac{R}{m} $ on both the sides
 $ {v^2} = \dfrac{{k{R^2}}}{m} $
Finally, taking the square root
 $ v = \sqrt {\dfrac{k}{m}} R $ (3)
So, the velocity of the particle is $ \sqrt {\dfrac{k}{m}} R $
 $ \therefore $ option B correct.
We also know that the angular momentum, $ L = mvR $
Substituting $ v $ from (3), we get
 $ L = m\left( {\sqrt {\dfrac{k}{m}} R} \right)R $
 $ L = m\dfrac{{\sqrt k }}{{\sqrt m }}{R^2} $
On simplifying, we get
 $ L = \sqrt m \sqrt k {R^2} $
 $ L = \sqrt {mk} {R^2} $
So, the angular momentum of the particle about O is $ \sqrt {mk} {R^2} $
 $ \therefore $ option C is also correct.
Hence, both the options B and C are correct.

Note
The relation $ F = - \dfrac{{dV}}{{dr}} $ is valid for a particle in a conservative force field only. So, before using the relation for a particle, always confirm whether the particle is in a conservative field or not. If any non-conservative force is applied to the particle, then this relation is not valid.