Inertia of a square (1 Viewer)

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1. The problem statement, all variables and given/known data

Four identical particles of mass 0.351 kg each are placed at the vertices of a 3.50 m x 3.50 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?



2. Relevant equations
I = mr^2
I = Icom + Mh^2

3. The attempt at a solution
a) I think you would do I = mr^2 + mr^2
so, I = 2(.351)(3.5/2)^2 + 2(.351)(3.5/2)^2

b) I = .351(3.5/2)^2 + .351(3.5/2)^2 + 2(.351)(3.5)^2

c) Since the diagonal of the square is 4.95m
I = .351(4.95/2)^2 +.351(4.95/2)^2
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
 
Last edited:

tiny-tim

Science Advisor
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hi mmiller9913! :smile:

(try using the X2 icon just above the Reply box :wink:)
b) I = .351(3.5/2)^2 + .351(3.5/2)^2 + 2(.351)(3.5)^2
(a) and (c) are fine :smile:

try (b) again​
 

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