# Proof 28: Plate’s moment-of-inerta

Plate rotating perpendicularly about its centre-of-mass

With a single point having the moment-of-inertia $I_\text{point}=r^2m$ ($m$ and $r$ are mass and distance from the axis-of-rotation),1 infinitely many “particles” making up a rotating object gives us the infinite sum, written as an integral:

$$I=\int r^2\,\mathrm dm =\int \underbrace{r_x^2+r_y^2}_{r^2}\,\mathrm dm$$

Rewritten with Pythagoras’ Theorem to its two components on the plate. (All points are equally high up, so there is no third component, $r_z=0$.)

Imagine cutting the plate into many tiny bits. Each bit carries a small piece $\mathrm dm$ of the total mass $m$ and covers a small area $\mathrm dA$ of the total area $A$. The mass is uniform (evenly spread), so $\mathrm dm$ covers the same amount of the total mass $m$ as $\mathrm d A$ covers of the total area $A$. Meaning, they are both the same fraction of the total:

$$\frac{\mathrm dm}{m}= \frac{\mathrm dA}{A}\quad\Leftrightarrow \quad \mathrm dm= m\frac{\mathrm dA}{A} = \frac{m}{A} \mathrm dA$$

Plug this in:

$$I=\int r_x^2+r_y^2\,\mathrm dm= \int (r_x^2+r_y^2)\, \underbrace{\frac{m}{A} \,\mathrm dA}_{\mathrm dm}= \int (r_x^2+r_y^2)\, \frac{m}{wl} \,\underbrace{\mathrm dr_x\,\mathrm dr_y}_{\mathrm dA} =\frac{m}{wl} \int r_x^2+r_y^2 \,\mathrm dr_x\,\mathrm dr_y$$

If cut into rectangular bits, the total area is width times length $A=lw$ and each small area-bit is a small width-bit times a small length-bit $\mathrm dA=\mathrm dr_x\mathrm dr_y$. This was used here. Now, we see that this is a 2D case: We must integrate over both the length and width – integrate to both $\mathrm dr_x$ and $\mathrm dr_y$. Length-wise we must integrate the whole length $l$, meaning $\frac 12 l$ on either side, meaning from $-\frac 12l$ to $\frac 12l$. Same width-wise from $-\frac 12w$ to $\frac 12w$:

\begin{align} I&= \frac{m}{wl} \int_{-\frac 12l}^{\frac 12l} \int_{-\frac 12w}^{\frac 12w} r_x^2+r_y^2 \,\mathrm dr_x\,\mathrm dr_y\\ &= \frac{m}{wl} \int_{-\frac 12l}^{\frac 12l} \left[\frac 13 r_x^3+r_y^2r_x\right]_{-\frac 12w}^{\frac 12w} \,\mathrm dr_y\\ &= \frac{m}{wl} \int_{-\frac 12l}^{\frac 12l} \left(\frac 13 \left(\frac 12w \right)^3+r_y^2\left(\frac 12w \right) \right)- \left(\frac 13 \left( -\frac 12w \right)^3+r_y^2\left( -\frac 12w \right) \right) \,\mathrm dr_y\\ &= \frac{m}{wl} \int_{-\frac 12l}^{\frac 12l} \left(\frac 13 \frac 18w^3+r_y^2\frac 12w \right)- \left(-\frac 13 \frac 18w^3-r_y^2\frac 12w \right) \,\mathrm dr_y\\ &= \frac{m}{wl} \int_{-\frac 12l}^{\frac 12l} \frac 1{24}w^3+ \frac 12 r_y^2w + \frac 1{24}w^3+ \frac 12 r_y^2w \,\mathrm dr_y\\ &= \frac{m}{wl} \int_{-\frac 12l}^{\frac 12l} 2\frac 1{24}w^3+ 2\frac 12 r_y^2w \,\mathrm dr_y\\ &= \frac{m}{wl} \int_{-\frac 12l}^{\frac 12l} \frac 1{12}w^3+ r_y^2w \,\mathrm dr_y\\ &= \frac{m}{wl}\left[\frac 1{12}w^3r_y+ \frac 13r_y^3w\right]_{-\frac 12l}^{\frac 12l} \\ &= \frac{m}{wl}\left(\left(\frac 1{12}w^3\left( \frac 12l \right)+ \frac 13 \left( \frac 12l \right)^3w\right)- \left(\frac 1{12}w^3\left( -\frac 12l \right)+ \frac 13 \left(- \frac 12l \right)^3w\right) \right) \\ &= \frac{m}{wl}\left(\left(\frac 1{12}w^3\frac 12l + \frac 13 \frac 18l^3w\right)- \left(-\frac 1{12}w^3\frac 12l – \frac 13 \frac 18l^3w\right) \right) \\ &= \frac{m}{wl}\left(\frac 1{24}w^3l + \frac 1{24}l^3w+\frac 1{24}w^3l +\frac 1{24}l^3w \right) \\ &= m\left(\frac{\frac 1{24}w^3\cancel l }{w\cancel l}+ \frac{\frac 1{24}l^3\cancel w }{\cancel wl} + \frac{ \frac 1{24}w^3\cancel l }{w\cancel l} + \frac{ \frac 1{24}l^3\cancel w }{\cancel wl} \right) \\ &= m\left(\frac 1{24}w^2+ \frac 1{24}l^2 + \frac 1{24}w^2 + \frac 1{24}l^2 \right) \\ &= m\left(2\frac 1{24}w^2+ 2\frac 1{24}l^2 \right) \\ &= m\left(\frac 1{12}w^2+ \frac 1{12}l^2 \right) \\ &= \frac 1{12} m\left(w^2+ l^2 \right) \qquad \blacksquare\\ \end{align}

Starting over with the integral formula for moment-of-inertia, only one dimension will be needed, since the straight distance $r$ from any point to the axes is a equal to $r_x$:

$$I=\int r\,\mathrm dm= \int r_x^2\,\mathrm dm$$

This time we cut the plate in many parallel stripes that cover the whole length but are very thin. As before, the ratio between the small mass $\mathrm dm$ of a stripe and the total mass $m$ is the same as the ratio between thickness $\mathrm dr_x$ of a stripe and total width $w$:

$$\frac{\mathrm dm}{m}= \frac{\mathrm dr_x}{w}\quad\Leftrightarrow \quad \mathrm dm= m\frac{\mathrm dr_x}{w} = \frac{m}{w} \mathrm dr_x$$

We plug this in and integrate from $0$ to the full width of $w$:

$$I= \int r_x^2\,\mathrm dm = \int r_x^2\, \frac{m}{w}\; \mathrm dr_x= \frac{m}{w} \int r_x^2\,\mathrm dr_x \quad \Leftrightarrow\\ I= \frac{m}{w} \int_0^w r_x^2\, \mathrm dr_x=\frac{m}{w} \left[\frac 13 r_x^3\right]_0^w= \frac{m}{w} \left( \left(\frac 13 w^3\right)- \cancel{\left(\frac 13 \cdot 0^3\right)} \right)= \frac{m}{w} \frac 13 w^3= \frac 13 mw^2 \qquad\blacksquare$$