Find the torque of a rotating sheet and the upper bound of the torque

1. Apr 26, 2014

sikrut

1. The problem statement, all variables and given/known data
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?

$x\ \rightarrow\ a\ ,\ y\ \rightarrow\ b\ , z\ =\ 0$
$A\ =\ ab$

2. Relevant equations

Inertia Tensor
$$\textbf{I} = \rho \begin{bmatrix} \int_V (y^2 + z^2) dV & -\int_V xy dV & -\int_V xz dV \\ -\int_V xy dV & \int_V (x^2+z^2) dV & -\int_V (yz) dV \\ -\int_V xz dV & -\int_V yz dV & \int_V (x^2 + z^2) dV \end{bmatrix}$$
$$\vec{L} = \textbf{I} \vec{w}$$
$$\vec{\tau} = \frac{d\vec{L}}{dt} = \vec{w} \times \vec{L}$$

3. 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!

Last edited: Apr 26, 2014
2. Apr 26, 2014

TSny

Your work looks good to me up to where you are trying to maximize the torque. Should the angle $\phi$ be treated as a constant when taking the derivative with respect to $a$?

[Edit: I think you can simplify your expression for the torque in terms of $a$ and $b$ if you express the trig functions in terms of $a$ and $b$.]

Last edited: Apr 26, 2014
3. Apr 26, 2014

sikrut

Dam. You are completely right.

So i basically would make $\phi = \phi(a)$, and take the respective derivatives.

What I'm wondering about this is that since for example, $sin(\phi(a)) = \frac{b}{\sqrt{a^2+b^2}}$, could i just say that $\frac{d}{da}sin(\phi(a)) = cos(\phi(a)) \frac{d\phi(a)}{da}$, or would I have to first plug in the prior, then in $b = A/a$ and then take the derivative?

4. Apr 26, 2014

TSny

I wouldn't recommend doing that. I think it would be messy.

Yes. Using $\sin\phi = \frac{b}{\sqrt{a^2+b^2}}$ (and similar expression for $\cos \phi$), I find that your expression for the torque can be compactly expressed in terms of just $a$ and $b$. Then you can let $b = A/a$ and take the derivative with respect to $a$.

5. Apr 26, 2014

sikrut

Looks like i'm still running into a problem. When I plug in everything, it simplifies to this.

then when I take the derivative.

I can't solve for a in this answer. Could my initial $\tau$ be incorrect then?

edit: I've solved and resolved this solution by hand a few times now and I keep coming up with $\frac{d\tau}{da} = \frac{2}{3}mw^2\frac{a^3A^2}{(a^4 + A^2)^2}$. I even redid the problem by setting my origin as the centroid of the sheet, so that $\textbf{I} = \begin{smallmatrix} I_{xx} & 0 & 0 \\ 0 & I_{yy} & 0 \\ 0 & 0 & I_{zz} \end{smallmatrix}$ and since $\vec{w} = \begin{smallmatrix} w_x \\ w_y \\ 0 \end{smallmatrix}$, then $\vec{L} = I_{xx}w_x\hat{x} + I_{yy}w_y\hat{y}$. From there it's pretty straight forward, but I end up with the same answer as the one above (which kind of proves the symmetry argument being able to use only the principle moments). Is something off in my setup you think?

Last edited: Apr 27, 2014
6. Apr 27, 2014

TSny

I don't see anything wrong. Your calculations look correct to me. Assuming your answer is correct, how would you answer the question? A graph of the torque as a function of $a$ might help.

7. Apr 27, 2014

sikrut

I guess I really wanted a more finite answer, but that's not always true in physics.

My derivative $\rightarrow 0$ when $a \rightarrow \inf$, which mean this sheet starts to look like an infinitely long and thin line along the axis of the diagonal and the angular velocity vector.
Solving $\tau$ when $a >> A$, then $\tau = \frac{m}{12}w^2A$ . That's a nice clean answer!

I really appreciate your help! This problem has been eating away at me for a couple days now.