A solid cylinder rolls down an inclined plane. Its mass is `2 kg and radius 0.1 m`. It the height of the inclined plane is `4m`, what is its rotational `K.E.` when it reaches foot of the plane ? Assume that the surfaces are smooth. Take `M.I.` of solid cylinder about its axis = `mr^(2)//2`.

Explanation

`K_(T) + K_(R) = K`
`(1)/(2)mv^(2) + (1)/(2)I omega^(2) = K`
But `(1)/(2)I omega^(2) = (1)/(2) ((1)/(2)mr^(2)) omega^(2) = (1)/(4)mv^(2)`
`:. (1)/(2)mv^(2) + (1)/(4)mv^(2) = K or (3)/(4)mv^(2) = K`
Rotational `KE, (1)/(2)I omega^(2) = (1)/(4)mv^(2) = (K)/(3)`
`=(mgh)/(3) = (2 xx 9.8 xx 4)/(3) = 26 .13 J`