Question 148: Work, Energy and Power & Circular Motion

Question

A particle is moving in a circular path of radius $$a$$ under the action of an attractive potential $$U =  - {k \over {2{r^2}}}$$ Its total energy is:

Accepted Answer

We know, Total energy = Kinetic energy + Potential energy
Potential energy given as $$U =  - {k \over {2{r^2}}}$$
We need to find Kinetic Energy.
As Force acting on the particle (F) = $$ - {{dU} \over {dr}}$$
$$ \Rightarrow F =  - {d \over {dr}}\left( {{{ - k} \over {2{r^2}}}} ight)$$
$$= {k \over 2} 	imes \left( { - 2} ight) 	imes {r^{ - 3}}$$
$$ =  - {k \over {{r^3}}}$$
Because of this force particle is having circular motion so it will provide possible centripetal force.
$$\left| F ight| = {{m{v^2}} \over r}$$
$$ \Rightarrow {{m{v^2}} \over r} = {k \over {{r^3}}}$$
$$ \Rightarrow $$ $$m{v^2} = {k \over {{r^2}}}$$
We know kinetic energy of particle, K = $${1 \over 2}m{v^2}$$ = $${k \over {2{r^2}}}$$
As Total energy = Kinetic energy + Potential energy
So Total energy = $${k \over {2{r^2}}}$$ $$ - {k \over {2{r^2}}}$$ = 0

Answer

$$ - {3 \over 2}{k \over {{a^2}}}$$

Answer

Zero

Answer

$$ - {k \over {4{a^2}}}$$

Answer

$$ {k \over {2{a^2}}}$$

Work, Energy and Power & Circular Motion

A particle is moving in a circular path of radius $a under the action of an attractive potential U = - {k \over {2{r^2}}}$ Its total energy is:

Choose the correct answer: