The potential energy of a particle of mass ? at a distance ? from a fixed point ? is given by `V(r)=kr^(2)//2`, where ? is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius ? about the point ?. If ? is the speed of the particle and ? is the magnitude of its angular momentum about ?, which of the following statements is (are) true?

Options

A. `v= sqrt((k)/(2m))R`
B. `v= sqrt((k)/(m))R`
C. `L=R^(2) sqrt(mK)`
D. `L=sqrt((mk)/(2))R^(2)`

Answer

B, C

Explanation

`U=(kr^(2))/(3)`
`|F|=(dU)/(dr) rArr |F|=kr`
The above-calculated force is responsible for the circular motion of particle, hence we can write the following:
`(mv^(2))/(R)= kR rArr v= R sqrt((k)/(m))`
Hence option (b) is correct.
Angular momentum `L= mvR=m (R sqrt((k)/(M)))R`
`=R^(2) sqrt(mK)`