A pencil that's made of homogeneous wood is shaped like a regular hexagonal prism, with hexagon side $s$ and mass $M$. The pencil rolls down an inclined plane. Find the moment of inertia.

Assume that the sides are slightly concave so that only the vertices of the hexagons are touching the plane at any given moment.

### Solution

First we find the moment of inertia of a equilateral prism about its center. Suppose the triangle side is $s$. By dimensional analysis, we know that the moment inertia $I$ must have the form $I = kMs^2$ for some $k$. Also note that the equilateral triangle is a superposition of 4 smaller triangles.The moment of inertia from the triangle in the center is $k.M/4.(s/2)^2 = kMs^2 / 16$.

The moment of inertia from each of the outer smaller triangle can be found using parallel axis theorem. From geometry, we know the displacement of the axes is $s / \sqrt{3}$. So: $$I_t = \frac{kMs^2}{16} + \frac{M}{4} . (\frac{s}{2\sqrt{3}})^2 = \frac{Ms^2}{16} (k + \frac{1}{3})$$ Therefore our original moment of inertia should be equal to the superposition of four triangles: $$kMs^2 = \frac{kMs^2}{16} + 3\frac{Ms^2}{16} (k + \frac{1}{3})$$ $$k = \frac{k}{16} + \frac{3k+1}{16}$$ $$k = \frac{1}{12}$$ So the moment of inertia of an equilateral prism about its tip is $$\frac{Ms^2}{12} + M(\frac{s}{\sqrt{3}})^2 = \frac{5Ms^2}{12}$$ Thus the moment of inertia for six of these triangles about their tip (i.e. hexagon about the center) is $6 . 5/12 . M/6 s^2 = 5Ms^2/12$ So the moment of inertia of the hexagon about the edge is $$ \frac{5Ms^2}{12} + Ms^2 = \frac{17}{12}Ms^2$$

### Second Solution

Alternatively we can find the moment of inertia of a triangular prism about its tip by dividing it into a series of slabs, where each slab does not touch the rotational axis. Each slab is infinitesimally thin (having thickness of $dx$) where $x$ is the distance from the axis to the slab. The slab would have a dimension of $l \times dx \times s$ where $x = \sqrt{3}s / 2$ so $s = 2x / \sqrt{3}$.Suppose the density of the wood is $\rho$. The mass of the slab is then $m=\rho.l.s.dx$. The m.o.i of the slab around the center is $$\frac{m}{12}(dx^2 + s^2) \simeq \frac{ms^2}{12}$$ The m.o.i of the slab around the axis that's $x$ away is then $$\frac{ms^2}{12} + mx^2 = \rho.l.s.dx.(\frac{s^2}{12} +x^2) = \rho.l.\frac{20}{9\sqrt{3}} x^3 dx$$ The triangle is a superposition of these slabs, with $x$ ranging from $0$ to $\frac{\sqrt{3}}{2}S$ $$I = \int_0^\frac{\sqrt{3}S}{2} \rho.l.\frac{20}{9\sqrt{3}} x^3 dx = \rho.l.\frac{20}{9\sqrt{3}} . \frac{9/16 S^4}{4} = \frac{5}{16\sqrt{3}} \rho.l.S^4$$ But then $\rho = M / (l.S^2.\sqrt{3}/4) = 4M / (\sqrt{3}lS^2)$ so: $$I = \frac{5}{16\sqrt{3}} l.S^4 . \frac{4M}{\sqrt{3}lS^2} = \frac{5MS^2}{12}$$ The moment of inertia for the hexagon about the edge follows easily just like in the first solution.

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