Four Sphere of diameter 2a and mass M are placed with their centres on 4 corner of a square of side b . then the moment of inertia of the system about an axis along along one side of a square is

 

Ans 8/5Ma^2  +  2Mb^2

 

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for a sphere placed along the axis the

MI = 2/5 Ma² 

 

for 2 spheres

MI = 4/5 Ma²

 

 

MI of a sphere along the opposite side axis:

 = 2/5Ma²

 

the MI of this sphere about our required axis by parallel axis theorem:

 = 2/5 Ma² + Mb²

 

for 2 spheres MI abt reqd axis = 2(2/5 Ma² + Mb² )

 

so MI_net = 4/5 Ma² + 2(4/5 Ma² + Mb²)

             = 8/5 Ma² + 2Mb²

 

 

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may b this will help u!

 

 

one more thing ......dont even think of using perpendicular axis theorem as our bodies are all non-laminar......

 

2/5 Mr^2 is the moment of inertia of a sphere of radius 'r' about an axis passing through its centre.....i have taken THIS axis along the respective sides....and then applied parallel axis theorem

 

cheers!