You visited us 1 times! Enjoying our articles? Unlock Full Access!

Angular Momentum

A particle is...

Question

A particle is moving in a circular path of radius $^{'} a^{'}$ under the action of an attractive potential energy, $U = - \frac{k}{2 r^{2}}$ . Its total energy is:

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

z e r o

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $z e r o$
PE= $\frac{- k}{2 r^{2}}$
F= $\frac{- d U}{d r} = \frac{- k}{r^{3}}$
$m v \frac{d v}{d r} = \frac{- k}{r^{3}}$
on integration
$\frac{m v^{2}}{2} = \frac{k}{2 r^{2}} = K E$
TE=KE+PE=0

Suggest Corrections

Similar questions

a

U = - \frac{k}{2 r^{2}}

a

U = - \frac{k}{2 r^{2}}

a

U = - \frac{k}{2 r^{2}}

r

a

U = - \frac{k}{2 r^{2}}

r

a

U = \frac{- K}{2 r^{2}}

r

Join BYJU'S Learning Program