Question

A particle is moving in a circle of radius r under the
action of a force F=αr² which is directed towards
centre of the circle .Total mechanical energy (kinetic
energy potential energy) of the particle is (take
potential energy=O for r=0):​

Answer 1

Answer:

Total mechanical energy is $\dfrac 12 m r^2\left(\dfrac{d\theta}{dt} \right)^2 -\alpha \dfrac{r^3}{3}$.

where, $\theta$ is the angle of the position of the particle at an instant of time with respect to the horizontal axis passing through the center of the circle.

Explanation:

Given:

• Force on the particle, $\rm F=\alpha r^2$.

Finding Potential Energy:

The potential energy $\rm U$ of the particle is related with the force $\rm F$ as

$\rm F=-\dfrac{dU}{dr}$

Therefore, the potential energy of the particle is given by

$\rm U =-\int F\ dr=-\int \alpha r^2\ dr = -\alpha \dfrac{r^3}{3} -C.$

where, C is the constant of integration.

Given that potential energy=0 for r=0,

Using this condition,

$\rm (U(r))_{r=0} = - \alpha \dfrac{0^3}{3} -C=0.\\\Rightarrow C=0.\\\\\text{which gives,}\\U(r) = -\alpha \dfrac{r^3}{3}.$

Finding Kinetic Energy:

The kinetic energy of a particle at a point is given by

$\rm T=\dfrac 12 mv^2$

where,

• $m$ is the mass of the particle.

• $v$ is the velocity of the particle at that point.

The particle is moving in circle, consider a point on the circle where the particle is located at some instant t. The coordinates of the point be $\rm (x,\ y)$ such that the center of the circle is at origin.

In polar coordinates,

$\rm x=r\cos\theta\\y=r\sin\theta$

$\theta$ is the angle to the point measured by the horizontal axis passes through the center of the circle.

Therefore, the velocities of the particle in x and y direction are respectively given as

$\rm v_x = \dfrac{dx}{dt} = \dfrac{d}{dt} (r\cos\theta) = -r\sin\theta\ \dfrac{d\theta}{dt}.\\v_y = \dfrac{dy}{dt} = \dfrac{d}{dt} (r\sin\theta) = r\cos\theta\ \dfrac{d\theta}{dt}.$

Therefore, the total kinetic energy of the particle is given by

$\rm T=\dfrac 12 m \left(v_x^2+v_y^2 \right)=\dfrac 12 m\left[\left(-r\sin\theta\ \dfrac{d\theta}{dt} \right)^2+\left(r\cos\theta\ \dfrac{d\theta}{dt} \right)^2\right]\\=\dfrac 12 m \left[r^2\sin^2\theta\ \left(\dfrac{d\theta}{dt} \right)^2+r^2\cos^2\theta\ \left(\dfrac{d\theta}{dt} \right)^2\right]\\=\dfrac 12 m r^2\left(\sin^\theta+\cos^2\theta\right )\left(\dfrac{d\theta}{dt} \right)^2\\=\dfrac 12 m r^2\left(\dfrac{d\theta}{dt} \right)^2$

Total Mechanical Energy:

The total mechanical energy of the particle is given by

$\rm E = T+U\\=\dfrac 12 m r^2\left(\dfrac{d\theta}{dt} \right)^2+ \left(-\alpha \dfrac{r^3}{3}\right)\\=\dfrac 12 m r^2\left(\dfrac{d\theta}{dt} \right)^2 -\alpha \dfrac{r^3}{3}.$