A thin metal hoop of radius `0.25 m` and mass `2 kg` stars from rest and rolls down an inclined plane. If its linear velocity on reaching the foot of the plane is `2 m//s`, what is its rotational `KE` at that instant ?

Explanation

Here, `r = 0.25m, m = 2 kg, v = 2 m//s`
Rotational `KE = (1)/(2)I omega^(2) = (1)/(2) xx mr^(2) ((v)/(r ))^(2)`
`= (1)/(2)mv^(2) = (1)/(2)mv^(2) =(1)/(2) xx 2(2)^(2) = 4J`