Four small spheres, each of which you can regard as a point of mass 0.200 kg, are arranged in a square 0.400 m on a side and connected by light rods. Find the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane.

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Question:

Four small spheres, each of which you can regard as a point of mass 0.200 kg, are arranged in a square 0.400 m on a side and connected by light rods. Find the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane.

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Answer:

The moment of inertia of the system about an axis through the center of the square, perpendicular to its plane is ?f=0.

Given data:

The mass of each sphere is, ?f=m%20%3D%200.

Length of side of square is, ?f=L%20%3D%200.

The expression for the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane is,

?f=I%20%3D%204%20mR%5E%7B2%7D

Here,

R is the distance between center of the square and the sphere. And its value is,

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?f=R%20%3D%5Cdfrac%7B1%7D%7B2%7D%5Csqrt%7BL%5E%7B2%7D%2BL%5E%7B2%7D%7D%5C%5CR%20%3D%5Cdfrac%7B1%7D%7B2%7D%5Csqrt%7B0.400%5E%7B2%7D%2B0.400%5E%7B2%7D%7D%5C%5CR%20%3D%200

Then, moment of inertia is,

?f=I%20%3D%204%20mR%5E%7B2%7D%5C%5CI%20%3D%204%20%5Ctimes%200.200%20%5Ctimes%200.282%5E%7B2%7D%5C%5CI%20%3D%200

Thus, the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane is ?f=0.

Learn more about moment of inertia here:

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