JEE MAIN - Physics (2018 (Offline) - No. 17)
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:
$$ - {3 \over 2}{k \over {{a^2}}}$$
Zero
$$ - {k \over {4{a^2}}}$$
$$ {k \over {2{a^2}}}$$
Explanation
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) = $$ - {{dU} \over {dr}}$$
$$ \Rightarrow F = - {d \over {dr}}\left( {{{ - k} \over {2{r^2}}}} \right)$$
$$= {k \over 2} \times \left( { - 2} \right) \times {r^{ - 3}}$$
$$ = - {k \over {{r^3}}}$$
Because of this force particle is having circular motion so it will provide possible centripetal force.
$$\left| F \right| = {{m{v^2}} \over r}$$
$$ \Rightarrow {{m{v^2}} \over r} = {k \over {{r^3}}}$$
$$ \Rightarrow $$ $$m{v^2} = {k \over {{r^2}}}$$
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
Potential energy given as $$U = - {k \over {2{r^2}}}$$
We need to find Kinetic Energy.
As Force acting on the particle (F) = $$ - {{dU} \over {dr}}$$
$$ \Rightarrow F = - {d \over {dr}}\left( {{{ - k} \over {2{r^2}}}} \right)$$
$$= {k \over 2} \times \left( { - 2} \right) \times {r^{ - 3}}$$
$$ = - {k \over {{r^3}}}$$
Because of this force particle is having circular motion so it will provide possible centripetal force.
$$\left| F \right| = {{m{v^2}} \over r}$$
$$ \Rightarrow {{m{v^2}} \over r} = {k \over {{r^3}}}$$
$$ \Rightarrow $$ $$m{v^2} = {k \over {{r^2}}}$$
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
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