A unifrom solid disk of radius `R` and mass `M` is free to rotate on a frictionless pivot through a point on its rim. If the disk is released from rest in the position shown in figure. The speed of the lowest point on the disk in the dashed position is.
.

Options

A. `(2/3) (gR)^(1//2)`
B. `(gR)^(1//2)`
C. `(2gR)^(1//2)`
D. `2(gR//3)^(1//2)`

Answer

D

Explanation

Moment of inertia about the axis through the pivot on the rim of the disc is `(mR^(2)//2+mR^(2))=3mR^(2)//2`. Using the principle that change in potential energy (mgR) is equal to the gain of kinetic energy `(Iomega^(2)//2)` gives the answer. Also use `v=R omega`

`mgR=1/2 Iomega^(2)=1/2xx3/2mR^(2)omega^(2)`
`w=sqrt((4g)/3)`
`v_(c)=omegaR=sqrt((4gR)/3)`