Question

Four spheres each of diameter 2a and mass m are placed with their centres on the four corners of a square of the side b. Calculate the moment of inertia of the system about any side of the square.

Answer 1

diameter of sphere , d = 2a
so, radius of sphere , r = a
moment of inertia of sphere about an axis passing through its centre , I = 2/5ma² [ as sphere is solid ]

Moment of inertia of sphere which is located b distance from observations points , I' = mb² [ here we assume size of sphere is negligible compare of b ]

Now, see the figure,
$I_1=mb^2$
$I_2=\frac{2}{5}ma^2$
$I_3=\frac{2}{5}ma^2$
$I_4=mb^2$

so, moment of inertia of system of sphere,
I = $I_1+I_2+I_3+I_4$

= mb² + 2/5ma² + 2/5 ma² + mb²

= 4/5 ma² + 2mb²

Hence, answer is (4/5 ma² + 2mb²)

Answer 2

Explanation:

diameter of sphere , d = 2a

so, radius of sphere , r = a

moment of inertia of sphere about an axis passing through its centre , I = 2/5ma² [ as sphere is solid ]

Moment of inertia of sphere which is located b distance from observations points , I' = mb² [ here we assume size of sphere is negligible compare of b ]