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. `5/7`
B. `5/(12sqrt(3))`
C. `(5sqrt(3))/7`
D. `7/(5sqrt(3))`

Answer

B

Explanation

`I={(M(Rsqrt2)^(2))/12+M(R/sqrt(2))^(2)}4+mR^(2)`
`=8/3MR^(2)+mR^(2)=20kgm^(2)`
`(4M+m)gsintheta-f-(4M+m)a`

`fR=Ia/R`
Solving eqn i and ii
`a=(7g)/24`
`f=20a=muxx28gcos30^@`
`mu=5/(12sqrt(3))`