Four identical rods of mass `M = 6 kg` each are welded at their ends to form a square and then welded to a massive ring having mass `m = 4 kg` and radius `R = 1m`. If the system is allowed to roll down the incline of inclination `theta = 30^@`, determine the minimum value of the coefficient of static friction that will prevent slipping.

The moment of inertia of the system about the centre of ring will be

Options

A. `(g)/(2)`
B. `(g)/(4)`
C. `(7g)/(24)`
D. `(g)/(8)`

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))`