Moment of Inertia Using Parallel Axis Theorem

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The moment of inertia of a cube about an axis through its center and the center of two opposing faces is given as (1/6)mQ^2. To find the moment of inertia when the axis is along one face of the cube, the parallel axis theorem is applied. The distance d from the center to the face is correctly identified as Q/2. The calculation results in Pface being (5/12)mQ^2. The final expression for the moment of inertia along the face of the cube is confirmed as accurate.
woaini
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Homework Statement



VlBTZHQ.png


Q is the edge length of the cube. The moment of inertia of the cube about an axis passing through its center and the center of two opposing faces is (1/6)mQ^2 (PCM in the diagram). Use the parallel axis theorem to find the moment of inertia if the axis is along one face of the cube (Pface).

PCM=(1/6)mQ^2
Pface=?
d=?

Homework Equations



Pface = PCM+md^2

The Attempt at a Solution



d=(1/2)Q^2

Pface=\frac{mQ^2}{6}+\frac{mQ^2}{4}

Pface=\frac{5mQ^2}{12}
 
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woaini said:

Homework Statement



VlBTZHQ.png


Q is the edge length of the cube. The moment of inertia of the cube about an axis passing through its center and the center of two opposing faces is (1/6)mQ^2 (PCM in the diagram). Use the parallel axis theorem to find the moment of inertia if the axis is along one face of the cube (Pface).

PCM=(1/6)mQ^2
Pface=?
d=?

Homework Equations



Pface = PCM+md^2

The Attempt at a Solution



d=(1/2)Q^2

Pface=\frac{mQ^2}{6}+\frac{mQ^2}{4}

Pface=\frac{5mQ^2}{12}

Correct...except d = Q/2
 
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