Question

Four particles each of mass 1kg are at four corners of side 1m.Tge mi of the system about a normal axis through center of the square

Answer 1

The moment of inertia of the system about a normal axis through center of the square is I = 4 (mR^2).

Explanation:

Moment of inertia I  = ΣMR^2

R = √ 2 / 2 = 1 /√ 2

I = M1R1^2 + meR2^2 + m3R3^2 + m4R4^2

I = 4 (mR^2)

Thus the moment of inertia of the system about a normal axis through center of the square is I = 4 (mR^2).

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A stone is dropped from a height 300 m and at the same time another stone projected vertically upward with a velocity of 100 m/s . Find when and where the stone meet ?

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Answer 2

Answer:

2kg/m square is correct answer