A uniform rod `AB` of mass `M` and length `sqrt(2)R` is moving in a veritcal plane inside a hollow sphere of radius `R`. The sphere is rolling on a fixed horizontal surface without slipping with velocity of its centre of mass `2v`. When the end `B` is the lowest position, its speed is found to be `v` as shown in the figure. If the kinetic energy of the rod at this instant is `(4)/(K)Mv^(2)`. Find `K`.

Explanation

Let the velocity of the point `A` have components `2V` along horizontal, `v_(1)` along the vertical.
Since the rod `AB` is rigid, taking the velocities along the rod `AB`,
`2vcos45^(@)-v_(1)cos45^(@)=vcos45^(@)`
or `v_(1)=v`
The angular velocity of the rod `AB` equals (in the frame in which end `B` is at rest)
`omega=(sqrt(2)v)/(sqrt(2)R)=(v)/(R )`
The velocity of the `C.M.` of `AB`
Where `hati`,`hatj` are the unit vectors along the horizontal and vertical (upward) direction respectively.
The `K.E.` of the rod `AB`
`=(1)/(2)Mv_(CM)^(2)+(1)/(2)I_(CM)omega^(2)=(1)/(2)M((9v^(2))/(4)+(v^(2))/(4))+(1)/(2)(1)/(12)M(sqrt(@)R)^(2)((v)/(R ))^(2)=(4)/(3)Mv^(2)`