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sikrut
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Homework Statement
Introduction to Classical Mechanics by David Morin - problem 9.43, page 424
A uniform flat rectangular sheet of mass m and side lengths a and b rotates with angular speed w around a diagonal. What torque is required? Given a fixed area A, what should the rectangle look like if you want the required torque to be as large as possible? What is the upper bound on the torque?
##x\ \rightarrow\ a\ ,\ y\ \rightarrow\ b\ , z\ =\ 0 ##
##A\ =\ ab ##
Homework Equations
Inertia Tensor
$$\textbf{I} = \rho \begin{bmatrix} \int_V (y^2 + z^2) dV & -\int_V xy dV & -\int_V xz dV \\ -\int_V xy dV & \int_V (x^2+z^2) dV & -\int_V (yz) dV \\ -\int_V xz dV & -\int_V yz dV & \int_V (x^2 + z^2) dV \end{bmatrix} $$
$$ \vec{L} = \textbf{I} \vec{w} $$
$$ \vec{\tau} = \frac{d\vec{L}}{dt} = \vec{w} \times \vec{L} $$
The Attempt at a Solution
Because our z=0, we can immediately simplify our inertia tensor as follows:
$$ \textbf{I} = \rho \begin{bmatrix} I_{xx} & I_{xy} & 0 \\ I_{yx} & I_{yy} & 0 \\ 0 & 0 & I_{zz} \end{bmatrix} $$
Thus our angular momentum vector (L) takes the form:
$$\vec{L} =\ (I_{xx}w_1 + I_{xy}w_2)\hat{x} + (I_{yx}w_x + I_{yy}w_y)\hat{y} + (I_{zz}w_z)\hat{z}$$
We defined ##\phi## as the angle from the x-axis, such that:
$$\vec{w} =\ wcos(\phi)\hat{x} + wsin(\phi)\hat{y} + 0\hat{z} $$
Which is just equivalent to ##w\hat{r}## in spherical coordinates, where ##\theta =\ \pi/2##
After solving the integrals in the inertia tensor (where ##\rho = \frac{m}{ab}##, we get:
$$\textbf{I} =\ \begin{bmatrix} \frac{m}{3}b^2 & -\frac{m}{4}ab & 0 \\ -\frac{m}{4}ab & \frac{m}{3}a^2 & 0 \\ 0 & 0 & \frac{m}{3}(a^2 + b^2) \end{bmatrix} $$
Thus our L vector becomes:
$$\vec{L} =\ mw\left[(\frac{b^2}{3}cos(\phi) - \frac{ab}{4}sin(\phi))\hat{x} + (\frac{a^2}{3}sin(\phi) - \frac{ab}{4}cos(\phi))\hat{y}\right]$$
Then solving for torque:
$$\vec{\tau} = \vec{w} \times \vec{L} = mw^2\left[\frac{1}{3}sin(\phi)cos(\phi)(a^2 - b^2) + \frac{1}{4}ab(sin^2(\phi) - cos^2(\phi)\right]\hat{z}$$
After solving for torque, we need to differentiate it with respect to one of the sides (a or b). We define a said with respect to the other side and the area:
$$ A = ab\ \rightarrow b = A/a$$
$$\tau = mw^2\left[\frac{1}{3}sin{\phi}cos{\phi}(a^2 - A^2/a^2) + \frac{1}{4}A(sin^2(\phi) - cos^2(\phi))\right]$$
From wolfram:
Wolfram solution
$$\frac{d\tau}{da} = m*w^2\left(\frac{A^2}{3a^3} + \frac{a}{3}\right)sin(2\phi)$$
Setting the derivative to 0 and solving for a with Wolfram:
Wolfram solution
$$a = \pm \sqrt[4]{-1} \sqrt{A}$$I seem to get a complex solution, which I don't expect to be correct, but I can't quite point out were it all went wrong. Is my angular speed correct? I've been stuck on this for quite a while and any help would be much appreciated!
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