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

A thin metal disc of radius of 0.25 m and mass 2 kg starts from rest a

1.	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
Answer» 2 `2 sqrt2` `2 sqrt3` `3 sqrt2` Solution :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