Question
A particle is moving in a circular path of radius a under the action of an attractive potential $$U = - \frac{k}{{2{r^2}}}.$$ Its total energy is:
A.
$$ - \frac{k}{{4{a^2}}}$$
B.
$$\frac{k}{{2{a^2}}}$$
C.
$$Zero$$
D.
$$ - \frac{3}{2}\frac{k}{{{a^2}}}$$
Answer :
$$Zero$$
Solution :
$$F = - \frac{{\partial u}}{{\partial r}}\hat r = \frac{K}{{{r^3}}}\hat r$$
Since particle is moving in circular path
$$\eqalign{
& F = \frac{{m{v^2}}}{r} = \frac{K}{{{r^3}}}\,\,\,\,\, \Rightarrow m{v^2} = \frac{K}{{{r^2}}} \cr
& \therefore K.E. = \frac{1}{2}m{v^2} = \frac{K}{{2{r^2}}} \cr} $$
Total energy $$=PE.+K.E.$$
$$ = - \frac{K}{{2{r^2}}} + \frac{K}{{2{r^2}}} = {\text{Zero}}\left[ {\because P.E. = - \frac{K}{{2{r^2}}}{\text{ given}}} \right]$$