Question

Two particles of the same mass m are moving in circular orbits because of force given by $F(r)=-\dfrac{16}{r}-{{r}^{3}}$.
The first particle is at a distance of r=1, and the second particle is at r=4. The best estimate for the kinetic energy of the first and second particle is?
A. ${{10}^{-1}}$
B. $6\times {{10}^{-2}}$
C. $6\times {{10}^{2}}$
D. $3\times {{10}^{-3}}$

Answer 1

Hint: A particle moving in a circular orbit is having a centripetal acceleration towards the centre of the circle. The kinetic energy of a particle is a function of mass and velocity of the particle. If we are given a force in a problem associated with the circular motion, it is always good to compare that force term with the equation of centripetal force.

Formula Used:
The centripetal acceleration felt by a body of mass m revolving around a circular path of r is given by, $\text{Centripetal Acceleration}=\dfrac{m{{v}^{2}}}{r}$
The Kinetic Energy of a body of mass m moving with a velocity v is given by, $K.E=\dfrac{1}{2}m{{v}^{2}}$

Complete step-by-step answer:
In the problem, it is given that a force $F(r)=-\dfrac{16}{r}-{{r}^{3}}$ is acting on two particles at distance r=1 and at r=2. The negative sign in the force term indicated that the force is a central force. So, the force acting on a body executing motion is known as centripetal force. It is given by the equation,
$\text{Centripetal Acceleration}=\dfrac{m{{v}^{2}}}{r}$
Where m is the mass of the body.
v is the linear velocity of the body.
r is the radius of the circular orbit.
So, the central force given in the problem provides the necessary centripetal force for the particles revolving around the circular orbit. So, we can write,
$\dfrac{m{{v}^{2}}}{r}=\left| F(r) \right|=\dfrac{16}{r}+{{r}^{3}}$
We can rewrite the equation as,
$\dfrac{m{{v}^{2}}}{r}=\dfrac{16+{{r}^{4}}}{r}$
$\Rightarrow m{{v}^{2}}=16+{{r}^{4}}$ … equation (1)
We know that the kinetic energy of a body of mass m and moving with a velocity v is given by, $K.E=\dfrac{1}{2}m{{v}^{2}}$.
So the equation (1) can be written as,
$\dfrac{1}{2}m{{v}^{2}}=\dfrac{16+{{r}^{4}}}{2}$
So, the kinetic energy associated with the particle at a distance r is given by,
$K.E=\dfrac{16+{{r}^{4}}}{2}$
So the ratio of the kinetic energy of the first particle at r=1 and second particle at r=2 is given by,
$\dfrac{K.{{E}_{1}}}{K.{{E}_{2}}}=\dfrac{16+{{(1)}^{4}}}{16+{{(4)}^{4}}}$
$\dfrac{K.{{E}_{1}}}{K.{{E}_{2}}}=\dfrac{16+1}{16+256}$
$\Rightarrow \dfrac{K.{{E}_{1}}}{K.{{E}_{2}}}=\dfrac{17}{272}$
$\therefore \dfrac{K.{{E}_{1}}}{K.{{E}_{2}}}=6.6\times {{10}^{-2}}\simeq 6\times {{10}^{-2}}$
So the answer to the question is option (B).

Note: A conservative central force acting on a body depends on the radial distance r of the body from the center of force. Also, the central force will be acting along with the radial distance to a particle. A central force will either drive particles towards it or direct the particles away from it. A central force can be expressed mathematically as $F=F(r)\hat{r},\text{ where }\hat{r}$ is a unit vector along the center and the particle.
In a central force system, the particle is moving in a definite plane which is determined by the initial velocity and the position of the particle. The angular momentum of the particle moving under a central force is always conserved.