Question

A particle of mass m revolves round a horizontal circle of radius r under the influence of a centripetal force equal to $-\frac{K}{r^2}$ , where K stands for a constant. What is the total energy possessed by this particle?

• A. $+\frac{K}{r}$
• B. $-\frac{K}{2r}$
• C. $-\frac{2K}{r}$
• D. $+\frac{K}{r^2}$

Answer 1

The correct option is B $-\frac{K}{2r}$

$P.E=dU=-F.dr$ $=-(-\frac{k}{r^2})dr$

$\int dU=\int \frac{k}{r^2}dr$ $=\frac{-k}{r}$

$KE=\frac{1}{2}m{v}^2$

But given, $\frac{m{v}^2}{r}=|\frac{k}{r^2}|$

$v^2=|\frac{k}{mr}|$

$KE=\frac{1}{2}m.\frac{k}{mr}=\frac{1k}{2r}$

So, $TE=PE+KE=-\frac{k}{r}+\frac{k}{2r}$ $=-\frac{k}{2r}$