Four identical uniform rods of mass `M=6kg` each are welded at their ends to form a square and then welded to a uniform ring having mass m=4kg & radius R=1m the system is allowed to roll down on the rough and fixed incline of inclination `theta=30^(@)` (assume no sliding anywhere)
Q. The moment of inertia of system about the axis of ring will be-

Options

A. `20kgm^(2)`
B. `40kgm^(2)`
C. `10kgm^(2)`
D. `60kgm^(2)`

Explanation

`I=[(M(Rsqrt(2))^(2))/(12)+M((R)/(sqrt(2)))^(2)]xx4+mR^(2)`
`(4M+m)gsintheta-F=(4M+m)a`
`F.R.=I((a)/(R))`
solving `a=(7g)/(24)`
`F=20alemu(4M+m)gcos30`
`muge(5)/(12sqrt(3))`
`thereforemu_(min)=(5)/(12sqrt(3))`