A thin metal disc of radius of 0.25 m and mass 2 kg starts from rest and rolls down on an inclined plane. If its rotational kinetic energy is 4 J at the foot of inclined plane, then the linear velocity in m/s) at the same point is

Options

A. 2
B. `2 sqrt2`
C. `2 sqrt3`
D. `3 sqrt2`

Answer

B

Explanation

Rotational kinetic energy = `(1)/(2) I omega^(2)`
Here , ` I = (MR^(2))/(2) , therefore (1)/(2) (MR^(2))/(2) omega^(2) = 4 J`
or `(1)/(2) x (2)/(2) (R omega)^(2) = 4 J`
Since v = `R omega , therefore v^(2) = 8 or v = 2 sqrt2` m/s