MHB Moment of Inertia of a plate with respect to one diagonal

[Fantini]

The problem is to show that the moment of inertia of a rectangular plate of mass $m$ and sides $2a$ and $2b$ about the diagonal is $\displaystyle \frac{2}{3} \frac{m a^2 b^2}{a^2+b^2}.$ I did it using the traditional definition of moment of inertia, that is, $$I = \int r^2 \, dm.$$ However, the book's solution uses a sleazy trick. He says $$I_x = \frac{1}{3} M b^2 \, \operatorname{and} \, I_y = \frac{1}{3} M a^2,$$ and claims that it can be shown that $$I_{BD} = \frac{1}{3} Mb^2 \cos^2(\theta) + \frac{1}{3} M a^2 \sin^2(\theta).$$

From the geometry, the result follows.

The question is: how can you show $$I_{BD} = \frac{1}{3} Mb^2 \cos^2(\theta) + \frac{1}{3} M a^2 \sin^2(\theta)?$$

 

[chisigma]

The problem is to show that the moment of inertia of a rectangular plate of mass $m$ and sides $2a$ and $2b$ about the diagonal is $\displaystyle \frac{2}{3} \frac{m a^2 b^2}{a^2+b^2}.$ I did it using the traditional definition of moment of inertia, that is, $$I = \int r^2 \, dm.$$ However, the book's solution uses a sleazy trick. He says $$I_x = \frac{1}{3} M b^2 \, \operatorname{and} \, I_y = \frac{1}{3} M a^2,$$ and claims that it can be shown that $$I_{BD} = \frac{1}{3} Mb^2 \cos^2(\theta) + \frac{1}{3} M a^2 \sin^2(\theta).$$
From the geometry, the result follows.

The question is: how can you show $$I_{BD} = \frac{1}{3} Mb^2 \cos^2(\theta) + \frac{1}{3} M a^2 \sin^2(\theta)?$$

Is...

$\displaystyle \cos \theta = \frac{a}{\sqrt{a^{2} + b^{2}}}, \sin \theta = \frac{b}{\sqrt{a^{2} + b^{2}}}\ (1)$

... and using the (1) You arrive at the identity...

$\displaystyle I = \frac{m}{3}\ (b^{2}\ \cos^{2} \theta + a^{2}\ \sin^{2} \theta) = \frac{2\ m}{3}\ \frac{a^{2}\ b^{2}}{a^{2} + b^{2}}\ (2)$

Kind regards

$\chi$ $\sigma$

 

[Fantini]

Thanks for replying, chisigma. However, I don't think you've understood my question. I know how to derive the result, assuming the relation the solution used.

My problem is how to prove that the relation holds. :)

 

[I like Serena]

Hey Fantini! ;)

The principal tensor of inertia is:
$$I_p = \begin{bmatrix}
I_x &0&0\\
0 &I_y&0 \\
0 & 0 & I_z
\end{bmatrix}$$

At an angle the tensor of inertia becomes:
$$R^T I_p R $$
where $R$ is the rotation matrix around the z-axis by an angle $\theta$.
(See Principal axes of the Moment of Inertia on wiki).

From this we can read off the new moment of inertia, which is what you have. (Mmm)

 

[Fantini]

Hey ILS. :) This is some help. Would you happen to have a more elementary derivation? Even if it's valid for this special case. Using tensors of inertia seems like overkill to me.

 

[I like Serena]

Hey ILS. :) This is some help. Would you happen to have a more elementary derivation? Even if it's valid for this special case. Using tensors of inertia seems like overkill to me.

Sure. (Sweating)

As you already know, we have:
$$I_x = \iint y^2 dm\tag 1$$
$$I_y = \iint x^2 dm\tag 2$$

What we want is:
$$I_{BD} = \iint d^2 dm$$
where $d$ is the distance of the point (x,y) to the line BD that makes an angle $\theta$ with the x-axis.

The distance $d$ is given by:
$$d = -\sin\theta \cdot x + \cos\theta \cdot y$$
Therefore:
$$I_{BD} = \iint (-\sin\theta \cdot x + \cos\theta \cdot y)^2 dm = \cos^2\theta\iint y^2 \,dm + \sin^2\theta\iint x^2 \,dm -2\sin\theta\cos\theta\iint xy\,dm \tag 3$$

The last term vanishes since:
$$\iint xy\,dm = \rho\iint xy\,dxdy = \rho\int \left[\frac 12 xy^2 \right]_{-b}^{b}dx = \rho\int \frac 12 x\Big(b^2 - (-b)^2\Big) = 0 \tag 4$$
More generally this is true for anybody that is symmetric in one of the axes.Combining $(1), (2), (3)$ and $(4)$ gives us:
$$I_{BD} = \cos^2\theta\cdot I_x + \sin^2\theta\cdot I_y$$Tada! (Party)(This is basically one of the derivations given on the wiki page - just limited to 2 dimensions. (Nerd))

 

[Fantini]

I see. It's basically the same derivation I did by applying the definition of moment of inertia, except that I used the center at the lower left corner of the plate. Still worked. Thanks, ILS! :)