| A particle is moving in a circle of radius r under the action of a force \[F=\alpha {{r}^{2}}\]which is directed towards centre of the circle. [JEE MAIN 11-04-2015] |
| Total mechanical energy (kinetic energy + potential energy) of the particle is (take potential energy=0 for r=0): |
A) \[\alpha {{r}^{3}}\]
B) \[\frac{1}{2}\alpha {{r}^{3}}\]
C) \[\frac{4}{3}\alpha {{r}^{3}}\]
D) \[\frac{5}{6}\alpha {{r}^{3}}\]
Correct Answer: D
Solution :
| [d]\[\frac{m{{V}^{2}}}{r}=\alpha {{r}^{2}}\] |
| \[\therefore \]\[K.E.=\frac{\alpha {{r}^{3}}}{2}\] |
| \[\Delta P.E=\int\limits_{0}^{r}{\alpha {{r}^{2}}.dr}\] \[P.E=\frac{\alpha {{r}^{3}}}{3}\] |
| \[T.E=\frac{\alpha {{r}^{3}}}{2}+\frac{\alpha {{r}^{3}}}{3}\] \[T.E=\frac{5}{6}\alpha {{r}^{3}}\] |
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