# Moment of inerta of square frame [closed]

A rod of length $$l$$ and mass $$m$$ has $$ml^2/12$$ as moment of inertia about an axis through its center of mass.

(source: draw.to)

Say now we take four identical copies of the rod above and form a square frame, whose center of mass lies exactly at the geometric center of the square. How can we then use the moment of inertia of a single rod to calculate the moment of inertia of the entire square frame?

(source: draw.to)

I recently began learning about moment of inertias so I am not really sure how to go about doing this.

## closed as off-topic by rob♦Aug 16 at 17:30

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You can use the parallel axis theorem to work out the moment of inertia of a rod of length $l$ with it's centre of mass displaced from the axis of rotation by $\frac{l}{2}$ then multiply this value by four to get the moment of inertia of the whole square.

The parallel axis theorem is:

$$I = I_{cm} + md^2$$

Where $I$ is the moment of inertia when the object has been displaced, $I_{cm}$ is the moment of inertia of the object when the axis of rotation passes through the centre of mass and $d$ is the distance it is displaced.

For each rod in your square we have:

\begin{align} I_{cm} &= \frac{1}{12} ml^2 \\ d &= \frac{l}{2}\\ I &= ml^2 \left(\frac{1}{12} + \frac{1}{4} \right) = \frac{ml^2}{3} \end{align}

So multiplying by four gives:

$$I_{square} = \frac{4ml^2}{3}$$

• or1426's answer is mostly correct but $m$ is the mass of one rod so the mass of the square is $M=4m$. So the correct equation should be $$I=\frac{ML^2}{3}$$ – Someone Curious Jun 25 '17 at 15:49
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