The moment of inertia of a square frame of side length a and mass M about an axis passing through one of its
comers and perpendicular to its plane is βMa². The value of β is________(upto two decimal places)
Correct answer is between '0.82,0.84'. Can you explain this answer?

Question

The moment of inertia of a square frame of side length a and mass M about an axis passing through one of its
comers and perpendicular to its plane is βMa². The value of β is________(upto two decimal places)

Correct answer is between '0.82,0.84'. Can you explain this answer?

Answer

To calculate the moment of inertia of a square frame about an axis passing through one of its corners and perpendicular to its plane, we can use the parallel axis theorem.

The parallel axis theorem states that the moment of inertia of a body about any axis parallel to an axis passing through its center of mass is equal to the moment of inertia about the center of mass plus the product of the mass of the body and the square of the distance between the two axes.

In this case, the moment of inertia of the square frame about an axis passing through one of its corners can be calculated as follows:

1. First, we need to find the moment of inertia of the square frame about an axis passing through its center of mass. Since the frame is symmetric, its center of mass is located at the center of the square. The moment of inertia of a square about an axis passing through its center and perpendicular to its plane is given by:

I_cm = (1/6) * M * a^2

2. Next, we need to calculate the distance between the center of mass and the corner of the square. This distance is equal to half the length of the diagonal of the square, which can be calculated using the Pythagorean theorem:

d = (a/2) * sqrt(2)

3. Finally, we can use the parallel axis theorem to calculate the moment of inertia about the axis passing through one of the corners:

I_corner = I_cm + M * d^2

Substituting the values we found:

I_corner = (1/6) * M * a^2 + M * [(a/2) * sqrt(2)]^2

Simplifying:

I_corner = (1/6) * M * a^2 + M * (a^2/4) * 2

I_corner = (1/6) * M * a^2 + (1/2) * M * a^2

I_corner = (1/6 + 1/2) * M * a^2

I_corner = (2/3) * M * a^2

Therefore, the moment of inertia of a square frame of side length a and mass M about an axis passing through one of its corners and perpendicular to its plane is (2/3) * M * a^2.

Answer

The mass of the square frame is M
The sides ot square frame can be treated as rods ot mass 3/4 and length a.
Moment of inertia of one rod about an axis passing through its centre and perpendicular to its plane

Using parallel axis theorem, moment of inertia of the rod about an axis passing through centre of square frame and perpendicular to its plane.

By symmetry clue to all rock, total M.I.,

Now, using parallel axis theorem again, to find moment of inertia about the axis given in question,

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