Question icon
11 grade physics others

Four spheres of diameter 2a and mass M are placed with their centers on the four corners of a square of side b. Then the moment of inertia of the system about an axis along one of the sides of the square is:
(A) M a² + 2M b²
(B) M a²
(C) M a² + 4M b²
(D) (8/5) M a² + 2M b²

Profile image of Aniket Singh
1 Year agoGrade
Answers icon

1 Answer

Profile image of Askiitians Tutor Team
1 Year ago

To calculate the moment of inertia of the system about an axis passing through one of the sides of the square, we need to consider the individual moments of inertia of each sphere about that axis and then sum them up.

The moment of inertia of a sphere about an axis passing through its center and perpendicular to the diameter can be calculated using the formula:

I_sphere = (2/5) * m * r^2

where:

I_sphere is the moment of inertia of the sphere.
m is the mass of the sphere.
r is the radius of the sphere.
In this case, the diameter of each sphere is 2a, so the radius is a. The mass of each sphere is M.

I_sphere = (2/5) * M * (a^2)

Now, we have four such spheres, and they are placed at the corners of a square of side b. We want to calculate the moment of inertia of the system about an axis passing through one of the sides of the square.

The axis of rotation is parallel to the side of the square and passes through the center of each sphere. We can treat each sphere's moment of inertia as if it's rotating about its own axis parallel to the side of the square.

Now, consider the geometry:

Two spheres are on one side of the square, and two are on the other side. So, the distance of each pair of spheres from the axis of rotation (the side of the square) is b/2.
The total moment of inertia for the two spheres on one side is:

I_side1 = 2 * I_sphere (two spheres with respect to their own axis)

The total moment of inertia for the two spheres on the other side is also:

I_side2 = 2 * I_sphere (two spheres with respect to their own axis)

Now, to find the total moment of inertia of the system, we sum up the contributions from both sides:

Total moment of inertia = I_side1 + I_side2

Total moment of inertia = 2 * I_sphere + 2 * I_sphere

Total moment of inertia = 4 * I_sphere

Substitute the expression for I_sphere we found earlier:

Total moment of inertia = 4 * [(2/5) * M * (a^2)]

Total moment of inertia = (8/5) * M * (a^2)

So, the moment of inertia of the system about an axis passing through one of the sides of the square is:

M * (8/5) * (a^2)

This matches option (D):

(D) (8/5) * M * (a^2) + 2 * M * (b^2)

So, the correct answer is (D).