Question 158: 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: