Mathwizurd.com is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.

# Point

MathJax TeX Test Page The moment of inertia of a point particle is $\boxed{mr^2}$. $r$ is the distance from the axis of rotation. For future reference, it is important to note that moments of inertia always add.

MathJax TeX Test Page Because moments of inertia always add, we can think of a road as a lot of individual points, so we create this integral. So, the moment of inertia of one small point is $$\mathrm{d}I = r^2 \mathrm{d}m$$ $$\mathrm{d}m = \rho (\text{density}) * \mathrm{d}x = \dfrac{M}{L}\mathrm{d}x$$ Now, we can express $dI$. $$\mathrm{d}I = r^2 \dfrac{M}{L}dx$$ $$\int_0^L \dfrac{M}{L}r^2\, \mathrm{d}x = \dfrac{M}{L}\dfrac{L^3}{3} = \boxed{\dfrac{ML^2}{3}}$$

MathJax TeX Test Page We keep coming back to one principle which is that moment of inertias add. The moment of inertia of a rotated rod about its center equals 2 * (rod cut in half about its end) That's just what we do! The length is cut in half, and because the density of the rod doesn't change, the mass is halved as well. $$2 * \dfrac{\left(\dfrac{M}{2}\right)\left(\dfrac{L}{2}\right)^2}{3} = \boxed{\dfrac{ML^2}{12}}$$

# Solid Disk

MathJax TeX Test Page

# Solid Cone

MathJax TeX Test Page The important thing to remember is that the total inertia is the sum of many moment of inertias of solid discs. $$dI = \dfrac{1}{2}dm*r^2$$ Why is this true? Because the total moment of inertia is the sum of moment of inertias of solid disc, and the formula is $\frac{1}{2}mr^2$. $$dm = \rho \text{ (density) } * \pi{}r^2 \text{ (area) } \mathrm{d}x \text{ (width) }$$ $$\rho = \dfrac{M}{\pi{}R^2h\frac{1}{3}} = \dfrac{3M}{\pi{}R^2h}$$ $$\mathrm{d}m = \dfrac{3M}{hR^2}*r^2\mathrm{d}x$$ $$\int_0^h \frac{1}{2}\left(\dfrac{3M}{hR^2}*r^2\right)* r^2\mathrm{d}x$$ $r$ is equal to $y$, which is equal to $\frac{R}{h}x$ $$\dfrac{3M}{2hR^2}\int_0^h y^4 \,\mathrm{d}x$$ Looking at the picture above, $y = \dfrac{R}{h}x$, so the integral is $$\dfrac{3M}{2hR^2}\int_0^h \frac{R^4}{h^4}x^2 \,\mathrm{d}x$$ $$\dfrac{3MR^2}{2h^5}\int_0^h x^4 \,\mathrm{d}x$$ $$\dfrac{3MR^2}{2h^5} * \frac{h^5}{5} = \boxed{\frac{3MR^2}{10}}$$
David Witten