A classical mechanics problem by Rocco Dalto

Let $B:(x,y,z)$ be any point not on the given line and choose the points $0:(x_{0},y_{0},z_{0})$ and $A:(x_0 + a,y_{0} + b,z_{0} + c)$ on the given line. $$

The distance from point $B$ to the given line is $d = \dfrac{|\vec{OB} \: X \: \vec{OA}|}{|\vec{OA}|}$
$$

$\vec{0A} = a \vec{i} + b \vec{j} + c \vec{k}$ $$

$\vec{0B} = (x - x_{0}) \vec{i} + (y - y_{0}) \vec{j} + (z - z_{0}) \vec{k}$ $$

$\vec{OB} \: X \: \vec{OA} = (c(y - y_{0}) - b(z - z_{0})) \vec{i} + (a(z - z_{0}) - c(x - x_{0})) \vec{j} + (b(x - x_{0}) - a(y - y_{0})) \vec{k}$ $$

$$

$\implies d^2 = \dfrac{(c(y - y_{0}) - b(z - z_{0}))^2 + (a(z - z_{0}) - c(x - x_{0}))^2 + (b(x - x_{0}) - a(y - y_{0}))^2}{a^2 + b^2 + c^2}$ $$

$\vec{r}(\theta,z) = cos(\theta)\vec{i} + sin(\theta)\vec{j} + z\vec{k}$ where $(0 <= \theta <= 2\pi)$ and $(0 <= z <= h) \implies$

$$

$\dfrac{\partial \vec{r}}{\partial \theta} = -sin(\theta)\vec{i} + cos(\theta)\vec{j} + 0\vec{k},$ and $\dfrac{\partial \vec{r}}{\partial z} = 0\vec{i} + 0\vec{j} + \vec{k} \implies$

$$

$\vec{N} = \dfrac{\partial \vec{r}}{\partial \theta} \: X \: \dfrac{\partial \vec{r}}{\partial z} = cos(\theta)\vec{i} + sin(\theta)\vec{j} + 0\vec{k} \implies |\vec{N}| =1 \implies$ $$

$d^2 = \dfrac{(c(sin\theta - y_{0}) - b(z - z_{0}))^2 + (a(z - z_{0}) - c(cos\theta - x_{0}))^2 + (b(cos\theta - x_{0}) - a(sin\theta - y_{0}))^2}{a^2 + b^2 + c^2}$ $$

Simplifying we obtain:
$$

Let $f(z) = 2(y_{0} c^2 + bc(z - z_{0}) +y_{0} a^2 - ab x_{0})$ $$

$g(z) = 2(x_{0} c^2 + ac (z - z_{0})+ + x_{0} b^2 - ab y_{0})$ $$

$m(z) = (a^2 + b^2) (z - z_{0})^2 + 2c(b y_{0} + a x_{0}) (z - z_{0})$

$K = x_{0}^2 (b^2 + c^2) + y_{0}^2 (a^2 + c^2) - 2ab x_{0} y_{0} + \dfrac{a^2 + b^2}{2} + c^2$

$d^2(\theta,z) = \dfrac{(\dfrac{b^2 - a^2}{2}) cos(2\theta) - f(z) sin\theta - g(z) cos\theta - 2 ab sin\theta cos\theta + m(z) + K}{a^2 + b^2 + c^2}$ $$ $$

$I = \int_{0}^{h} \int_{0}^{2\pi} d^2(\theta,z) \: d\theta \: dz =$ , $$

$\dfrac{1}{a^2 + b^2 + c^2} * \int_{0}^{h} ((\dfrac{b^2 - a^2}{4}) sin(2\theta) + f(z) cos\theta - g(z) sin\theta - ab (sin\theta)^2 + (m(z) + K) \theta)|_{0}^{2\pi} \: dz =$ $$

$\dfrac{2\pi}{a^2 + b^2 + c^2} \int_{0}^{h} m(z) + K \: dz =$ $$

$\dfrac{2\pi}{a^2 + b^2 + c^2} ((\dfrac{a^2 + b^2}{3}) (z - z_{0})^3 + c(b y_{0} + a x_{0}) (z - z_{0})^2 + K z)|_{0}^{h} =$ $$

$\dfrac{2\pi}{a^2 + b^2 + c^2} ((\dfrac{a^2 + b^2}{3}) (h - z_{0})^3 + c(b y_{0} + a x_{0}) (h - z_{0})^2 + K h + (\dfrac{a^2 + b^2}{3}) z_{0}^3 - c(b y_{0}+ a x_{0}) z_{0}^2 )$ , $$

where $K = x_{0}^2 (b^2 + c^2) + y_{0}^2 (a^2 + c^2) - 2ab x_{0} y_{0} + \dfrac{a^2 + b^2}{2} + c^2$ $$

Now letting $x_{0} = y_{0} = z_{0} = 0$ and $a = b = c$ we obtain: $$

$I = \dfrac{ 4\pi h}{9} ( 3 + h^2) = \dfrac{ \alpha \pi h}{\beta} ( \lambda + h^2)$ $$

$\implies \alpha + \beta + \lambda = \boxed{16}$ $$

Again, my intention was to have the person find a general formula for $I$ given lamina $S: x^2 + y^2 = 1, 0 \leq z \leq h$ of unit density, and axis $A:$ the line $x = x_{0} + at, y = y_{0} +bt, z = z_{0} + ct$ . $$

The inertia tensor of a cylindrical shell of radius $1$ and height $h$ about its centre of mass is $\mathcal{I} \; =\; \left( \begin{array}{ccc} \frac{1}{12}m(6+h^2) & 0 & 0 \\ 0 & \tfrac{1}{12}m(6+h^2) & 0 \\ 0 & 0 & m \end{array} \right)$ and so the moment of inertia of the cylinder about the axis through the centre of mass in the direction of the unit vector $\mathbf{v} \; = \; \left(\begin{array}{c} a \\ b \\ c \end{array} \right)$ is $I_0 \; = \; \mathbf{v}^T \mathcal{I} \mathbf{v} \; = \; \tfrac{1}{12}m(6 + h^2)(a^2 + b^2) + m c^2 \; = \; \tfrac16\pi h(6 + h^2)(a^2 + b^2) + 2\pi h c^2$ since the mass of the cylinder is $m = 2\pi h$ .

The distance from the origin $\mathbf{0}$ to the above axis through the centre of mass, with equation $\mathbf{r} \; = \; \lambda\mathbf{v} + \mathbf{a} \; = \; \lambda\mathbf{v} + \left(\begin{array}{c} 0 \\ 0 \\ \frac12h\end{array} \right) \hspace{1cm} \lambda \in \mathbb{R}$ is $\sqrt{|\mathbf{a}|^2 - (\mathbf{a} \cdot \mathbf{v})^2} \; = \; \tfrac12h\sqrt{1 - c^2}$ and hence the moment of inertia of the cylinder about the axis in the direction of the unit vector $\mathbf{v}$ that passes through the origin is $I \; = \; I_0 + \tfrac14mh^2(1 - c^2) \; = \; I_0 + \tfrac12\pi h^3(1 - c^2)$ With $a = b = c = \tfrac{1}{\sqrt{3}}$ we have $I \; = \; \tfrac19\pi h(6 + h^2) + \tfrac23\pi h + \tfrac13\pi h^3 \; = \; \tfrac43\pi h + \tfrac49\pi h^3 \; =\; \tfrac49\pi h(3 + h^2)$ making the answer $\boxed{16}$ .