A particle is moving in a circular path of radius a under the action of an attractive potential U = - (k/2r2). Its total energy is -

Question

A particle is moving in a circular path of radius a under the action of an attractive potential U = - (k/2r2). Its total energy is - (A)  - (k/4a2) (B)  (k/2a2) (C) zero (D)  - (3/2) (k/a2)

Tardigrade Admin · Accepted Answer

F=(-d U/d r) [U=-(k/2 r2)] (m v2/r)=(k/r3) [This force provides necessary centripetal force] ⇒ m v2=(k/r2) ⇒ K . E=(k/2 r2) ⇒ P . E=-(k/2 r2) Total energy = Zero

Q. 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 -

$- \frac{k}{4 a ^{2}}$

$\frac{k}{2 a ^{2}}$

zero

$- \frac{3}{2} \frac{k}{a ^{2}}$

$F = \frac{- d U}{d r} [U = - \frac{k}{2 r ^{2}}]$
$\frac{m v ^{2}}{r} = \frac{k}{r ^{3}}$ [This force provides necessary centripetal force]
$\Rightarrow m v^{2} = \frac{k}{r ^{2}}$
$\Rightarrow K . E = \frac{k}{2 r ^{2}}$
$\Rightarrow P . E = - \frac{k}{2 r ^{2}}$
Total energy $=$ Zero

Q. A particle is moving in a circular path of radius a under the action of an attractive potential U=−2r2k​. Its total energy is -

−4a2k​

2a2k​