A hollow cylinder with inner radius `R`. Outer radius `2R` mass `M` is rolling with speed of it's axis `v`. What is its kinetic energy ?
.

Answer

`(13)/(16) Mv^(2)`

Explanation

`KE = (1)/(2) Mv^(2) + (1)/(2) I omega^(2)`
`KE = (1)/(2) Mv^(2) +(1)/(2) I((v)/(2 R))^(2)` ….(1)
Let we have solid cylinder of radius `2R` and mass `M` from which cylinder of radius `R` of mass `M' = (M')/(4)` is taken out from common axis
Mas of hollow cylinder `M = M' - M''`
`M =M' -(M')/(m) =(2M')/(4) rArr M=(3M')/(4)`
`rArr M' = (4M)/(3) & M'' = (M)/(3)`
`I_(2R) = I_(R) +I rArr I = I_(2R) -I_(R)`
`I = (1)/(2) M'(2R)^(2) -(1)/(2) M''R^(2)`
`I = [(1)/(2) xx (4M)/(3) xx 4R^(2) -(1)/(2) xx(M)/(3) R^(2)]`
`I = (5)/(2) MR^(2)` ....(2)
`KE = (13)/(16) Mv^(2)`
Alternatively : Let the length of cylinder is `l`
`I = int_(R)^(2R) dmr^(2) rArr I = int_(R)^(2R) (M)/([pi(2R)^(2) -piR^(2)]l) xx 2pi rdrl xx r^(2)`
`I = (2M)/(3R^(2)) int_(R)^(2R) r^(3) dr rArr I = (5)/(2) MR^(2)`.