A particle is moving in a circle of radius r under the action of a force F = α r2 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) :

Question

A particle is moving in a circle of radius r under the action of a force F = α r2 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) α r3 (B)  (1/2) α r3 (C)  (4/3) α r3 (D)  (5/6) α r3

Tardigrade Admin · Accepted Answer

F=α r2=(m v2/r) (radial or centripetal force) K E=(1/2) m v2=(1/2) α r3 ; P E=-∫ F text et d r=-∫ α r2 d r=+(α r3/3) Total mechanical energy =K E+P E=(5/6) α r3

Q. A particle is moving in a circle of radius r under the action of a force $F = α r^{2}$ which is directed towards centre of the circle. Total mechanical energy (kinetic energy + potential energy) of the particle is (take potential energy = 0 for $r = 0$ ) :

$α r^{3}$

$\frac{1}{2} α r^{3}$

$\frac{4}{3} α r^{3}$

$\frac{5}{6} α r^{3}$

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

Q. A particle is moving in a circle of radius r under the action of a force F=αr2 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) :

αr3

21​αr3

34​αr3

65​αr3