A rod of mass `m` is kept on a cylinder and sphere each of radius `R`. The masses of the sphere and cylinder are `m_(1)` and `m_(2)` respectively. If the speed of the rod is `V`, find the `KE` of the system `("rod"+"cylider"+"sphere")`. Assume that the surfaces do not slide relative to each other.

Explanation

Kinetic energy is case of rolling `KE=(1+(k^(2))/(R^(2)))(mv^(2))/2`
where `k=`radius of gyration. For cylinder `k^(2)=1/2`
`K_("cylinder")=(1+1/2)(m_(1)v^('2))/2=3/4m_(1)v^('2)`
For rolling `K_("cylinder")=3/4 m_(1)v^('2)`

and `K_("sphere")=1/2(1+2/5)m_(v)^('2)=7/10m_(2)v^('2)`,then
`K_("rod")=K_("cylinder")+K_("sphere")=1/2mv^(2)+3/4m_(1)v^('2)+7/10m_(2)v^('2)`
where `v'=v/2` and `m_(1)=m_(2)=m=69/80mv^(2)`