Four identical particles of mass 0.298 kg each are placed at the vertices of a 2
ID: 2265041 • Letter: F
Question
Four identical particles of mass 0.298 kg each are placed at the vertices of a 2.43 m x 2.43 m square and held there by four massless rods, which form the sides of the square. What is the rotational inertia of this rigid body about an axis that (a) passes through the midpoints of opposite sides and lies in the plane of the square, (b) passes through the midpoint of one of the sides and is perpendicular to the plane of the square, and (c) lies in the plane of the square and passes through two diagonally opposite particles?
Explanation / Answer
let d = 2.43 m
m = 0.298 kg
a)
I = 4*m*(d/2)^2 = 1.76 kg.m^2
b)
I = 2*m*(d/2)^2 + 2*m*(d^2+ (d/2)^2 ) = 5.279 kg.m^2
c)
I = 2*m*2*d^2 = 7.04 kg.m^2
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