Question

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. 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. $\frac {5}{6} \alpha r^3$

Answer 1

Answer: (D)

$F=\alpha r^{2}=\frac{m v^{2}}{r}$ (radial or centripetal force)
$K E=\frac{1}{2} m v^{2}=\frac{1}{2} \alpha r^{3} ;$
$P E=-\int F_{\text {et }} d r=-\int \alpha r^{2} d r=+\frac{\alpha r^{3}}{3}$
Total mechanical energy $=K E+P E=\frac{5}{6} \alpha r^{3}$