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A square plate of mass M and edge L . the moment of inertia of the plate about the axis in the plane of plate passing through one of its vertex making an angle 15 degree from horizontal is.
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Calculation of Moment of Inertia of a Square Plate

To calculate the moment of inertia of a square plate about an axis in the plane of the plate passing through one of its vertices making an angle of 15 degrees from the horizontal, we can use the following steps:

Step 1: Determine the coordinates of the center of mass of the square plate.

The center of mass of a square plate is located at the intersection of its diagonals, which is at a distance of L/2 from each side of the square. Therefore, the coordinates of the center of mass are (L/2, L/2).

Step 2: Calculate the distance of the axis of rotation from the center of mass.

The distance of the axis of rotation from the center of mass can be calculated using the distance formula, which is given by:

d = sqrt[(x2 - x1)^2 + (y2 - y1)^2]

where d is the distance, (x1, y1) are the coordinates of the center of mass, and (x2, y2) are the coordinates of the axis of rotation.

In this case, the axis of rotation passes through one of the vertices of the square plate, which has coordinates (0, 0) in the xy-plane. Therefore, the distance of the axis of rotation from the center of mass is:

d = sqrt[(0 - L/2)^2 + (0 - L/2)^2] = L/2 * sqrt(2)

Step 3: Use the parallel axis theorem to calculate the moment of inertia.

The parallel axis theorem states that the moment of inertia of a rigid body about any axis parallel to its center of mass axis is equal to the moment of inertia about its center of mass axis 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 plate about an axis passing through its center of mass is:

Icm = (1/6) * M * L^2

Therefore, the moment of inertia of the square plate about an axis passing through one of its vertices and making an angle of 15 degrees with the horizontal is:

I = Icm + Md^2 = (1/6) * M * L^2 + M * (L/2 * sqrt(2))^2

I = (1/6) * M * L^2 + (1/2) * M * L^2 = (2/3) * M * L^2

Therefore, the moment of inertia of the square plate about the given axis is (2/3) * M * L^2.
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