Question

The moment of inertia of a sphere (mass M and radius R) about it's diameter is I. Four such spheres are
arranged as shown in the figure. The moment of inertia of the system about axis XX' will be
(A) 3 I
(B) 5 I
(C) 7 I
(D) 9 I
ANSWER GIVEN IS D. PLEASE EXPLAIN IT...... ​

Answer 1

Given that the moment of inertia of each sphere is I, i.e., I = 2MR²/5, since it's a sphere.

So the moment of inertia of the second and fourth spheres each is same and that is equal to I, because the axis XX' is passing through the centers of these spheres so their moments of inertia is said to be about its diameter.

So their sum is 2I.

Moment of inertia of the first sphere is going to be found out by the theorem of parallel axes.

Its moment of inertia about the diameter is I, and the perpendicular distance between the center of the first sphere and the XX' axis is equal to R, the radius of the sphere. So its moment of inertia will be,

I + MR²

= 2MR²/5 + MR²

= 7MR²/5

= 7 × 2MR²/(5 × 2)

= (7/2) × (2MR²/5)

= 7I/2

Similarly, the moment of inertia of the third sphere is also 7I/2.

So the net moment of inertia will be 2I + 7I/2 + 7I/2 = 9I.

Hence (d) is the answer.