A particle moving in a circular path has an angular momentum of L. If the frequency of rotation is halved, then its angular momentum becomes

Options

A. `(L)/(2)`
B. L
C. `(L)/(3)`
D. `(L)/(4)`

Answer

A

Explanation

We knoe that `L=I omega=2pi mr^(2)f`
Now, `omega'=omega//2`
Hence, `L'=Iomega'=mr^(2)=(omega)/(2)=pimr^(2)f rArr (L)/(L')=(2m pi r^(2)f)/(pi mr^(2)f)rArr L'=(L)/(2)`.