Symmetric square

Classical Mechanics Level 3

A uniform square plate is centered at $O$ . It has a moment of inertia $I_{\text{AB}}$ about the axis $\text{AB}$ going through two opposite vertices of the square. $\text{CD}$ is another axis lying in the plane of the square and making an angle of $30^\circ$ with the axis $\text{AB}$ .

What is the moment of inertia of the square about the axis $\text{CD}$ ?

Less than

I_{\text{AB}}

Equal to

I_{\text{AB}}

Greater than

I_{\text{AB}}

1 solution

Sarthak Agrawal
Mar 20, 2018

Let $I_\text{O}$ denote the moment of inertia of the given plate about the axis passing through O and perpendicular to the plane of paper.

Now, consider another axis perpendicular to axis AB in the plane of paper. Call it $A'B'$ . It can be concluded, by the argument of symmetry, that $I_\text{A'B'}$ must be equal to $I_\text{AB}$ .

$I_\text{A'B'} = I_\text{AB}$

Now, by perpendicular axis theorem we assert that -

$I_\text{A'B'}+I_\text{AB}=I_\text{O}$

$\Rightarrow 2I_\text{AB}=I_\text{O}$

Similarly, It can be shown for $I_\text{CD}$ and hence they are equal.

Symmetric square

1 solution

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