Question

A particle is moving in a circular path of radius $a$ under the action of an attractive potential $U = - \dfrac{k}{{2{r^2}}}$. Its total energy is:
a) Zero
b) $- \dfrac{{3k}}{{2{a^2}}}$
c) $- \dfrac{k}{{4{a^2}}}$
d) $- \dfrac{k}{{4{a^2}}}$

Answer 1

The total energy of the particle will be equal to the sum of kinetic energy and the potential energy of the particle. As the particle moves, both kinetic energy and potential energy should be calculated.

Complete step by step answer:
Let’s define all the terms given in the equation:
It is given that, Potential energy, $U = - \dfrac{k}{{2{r^2}}}$
We know that, the force of attraction, $F = - \dfrac{{dU}}{{dr}}$
Here, the potential energy, $U$ is given in the question itself.
That is, $U = - \dfrac{k}{{2{r^2}}}$
So, the force of attraction, $F = - \dfrac{{dU}}{{dr}}$
$\Rightarrow F = - \dfrac{d}{{dr}}(\dfrac{{ - k}}{{2{r^2}}})$
Minus sign will get cancelled and taking the magnitude, we get
$\Rightarrow F = \dfrac{k}{2} \times \dfrac{2}{{{r^3}}}$
$\Rightarrow F = \dfrac{k}{{{r^3}}}$
Since the motion is in circular path centripetal force is there, and the force of attraction will be equal to the centripetal force.
We know that the centripetal force is given by,
$centripetal{\text{ }}force = \dfrac{{m{v^2}}}{r}$
Equating the centripetal force and the force of attraction, we get,
$\Rightarrow \dfrac{{m{v^2}}}{r} = \dfrac{k}{{{r^3}}}$
$\Rightarrow m{v^2} = \dfrac{k}{{{r^2}}}$
We know Kinetic energy of the particle is given by the equation,
$K.E. = \dfrac{1}{2}m{v^2}$
Applying the value of $m{v^2} = \dfrac{k}{{{r^2}}}$in the equation for the kinetic energy, we get,
$\Rightarrow K.E. = \dfrac{1}{2} \times \dfrac{k}{{{r^2}}}$
$\Rightarrow K.E. = \dfrac{k}{{2{r^2}}}$
We have already discussed that the total energy of the particle moving will be equal to the sum of the potential energy and the kinetic energy of the particle.
Potential energy is already given in the equation, that is,
$U = - \dfrac{k}{{2{r^2}}}$
And we found the value of kinetic energy, that is,
$K.E. = \dfrac{k}{{2{r^2}}}$
So the total energy will be equal to
$\Rightarrow K.E. + U = \dfrac{{ - k}}{{2{r^2}}} + \dfrac{k}{{2{r^2}}} = 0$
So the final answer is option (A)

Note: Centripetal force is a force on an object directed to the centre of a circular path that keeps the object on the path. Its value is based on three factors: 1) the velocity of the object as it follows the circular path; 2) the object's distance from the centre of the path; and 3) the mass of the object.