# Fracture because of high-speed rotation

I was watching a rerun of an early MythBusters episode, where they look at whether CDs in high-speed drives can explode / fail simply because of being rotated too fast. The following are some calculations I did:

Let the thickness of the CD be $t$, made of material with density $\rho$. Consider a small element of the CD at a distance $r$ from the center. The element subtends an angle $d\theta$ at the center, and is of length (along the radius) $dr$. The centripetal force needed to pull this in at angular velocity $\omega$ is $m r \omega^{2}=\rho \left(r d\theta dr t\right) r \omega^{2}$.

The CD has a central spindle hole of radius $r=1.5\ cm$, and is itself of radius $R=12\ cm$. The force then exerted on an elemental area subtending angle $d\theta$ at the inner edge of the CD is (integrating the previous quantity from $r=r$ to $R$) $\frac{\rho \left(R^3 - r^3\right) d\theta t \omega^2}{3}$, which means the inside edge has pressure approximately $\frac{\rho R^3 \omega^2}{3r}$ (dividing the previous quantity by the area $t r d\theta$, and assuming $R^3$ dominates $r^3$).

Fracture now happens when $\frac{\rho R^3 \omega^2}{3r} > T$, the fracture strength of polycarbonate plastic (which being brittle, is approximately its ultimate tensile strength). Substituting $\rho = 1.22\ g\ cm^{-3}$, and $T = 75\ MPa$ (all approximate values obtained from Wikipedia), I got $\omega > 17000\ rpm$.

My question is, is the calculation correct? Or is there something I'm missing or misinterpreting? The value is the same order of magnitude as what MythBusters reported -- fracture at $\approx 25000\ rpm$. This is heartening, but it would be nicer to know that the number is off only because of incorrect inputs (strength of plastic, density etc.).

As nice as MythBusters sometimes is, it would be nicer still if they did a little more hard science.

Correction: As akhmeteli points out below, the CD dimensions are incorrect. The central spindle hole has radius $7.5\ mm$, and the CD itself has radius $60\ mm$. There's also an arithmetic mistake in the above calculation. The updated answer is therefore $\approx 24000\ rpm$. That would ordinarily make me very happy, but as both user758556 and ja72 argue, this calculation would be useful as nothing more than a sanity check. What's really happening is a little too complicated for these arguments to properly explain.

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You probably need to do some kind of advanced finite element analysis if you wanted to get closer. Your calculation is a good back-of-the-envelope type calculation but fracture mechanics and stress-strain curves near the breaking regime can often be quite nonlinear. Perhaps this might give you a flavor for how complex this problem might be: arxiv.org/abs/physics/0211004 Other things you aren't accounting for are the aluminum surface coating on the CD, a protective varnish layer on the play surface, or variations in the mechanical properties of polycarbonate. – user758556 Jun 26 '12 at 5:41
I would guess that you could account for a lot of your "error" with reasonable estimations of the impact of these since you have the order of magnitude right. In general, this kind of rough estimation probably won't get you more than order of magnitude, at least as far as I know. One final thing that I noticed when watching the video was the instability of rotation - the vertical deformation of the CD is quite evident from the hi-speed footage and I think without this instability higher angular velocities could be achieved. Hopefully someone far smarter than I can give you a better answer! – user758556 Jun 26 '12 at 5:44

Opening my mechanical engineering book on the rotating rings section and I get

$$\sigma_t(r) = \rho \omega^2 \frac{3+\nu}{8} \left( r_i^2 + r_o^2 + \frac{r_o^2 r_i^2}{r^2} - \frac{1+3\nu}{3+\nu} r^2 \right)$$ $$\sigma_r(r) = \rho \omega^2 \frac{3+\nu}{8} \left( r_i^2 + r_o^2 - \frac{r_o^2 r_i^2}{r^2} - r^2 \right)$$

Inner Edge Von-Mises Stress $\sigma^2_{vm} = \sigma_r^2 + \sigma_t^2 - \sigma_r \sigma_t$, $r=r_i$, $\sigma_r(r_i)=0$

$$\frac{\sigma_{vm}}{\rho \omega^2 r_o^2} = \frac{(1-\nu) r_i^2}{4 r_o^2} + \frac{\nu+3}{4}$$

Outer Edge Von-Mises Stress $\sigma^2_{vm} = \sigma_r^2 + \sigma_t^2 - \sigma_r \sigma_t$, $r=r_o$, $\sigma_r(r_o)=0$

$$\frac{\sigma_{vm}}{\rho \omega^2 r_o^2} = \frac{(3+\nu) r_i^2}{4 r_o^2} + \frac{1-\nu}{4}$$

with the inner edge usually having the higher stress. Here $\nu=0.4$ is Poisson's ratio.

The result I get from MathCAD is 34,200 rpm.

The difference being, a) The overall density is higher because the CD contains metals also, and b) Being brittle there are bound to be micro-cracks already in place before spinning starts. Assuming the $75\,\rm MPa$ figure is correct the density has to be $\rho=2.3\,\rm gm/cm^3$ for it crack at 25,000 rpm.

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With $r_0 = 60\ mm$, I get $\omega=42500\ rpm$ from your final formula. Even though its farther away from experiment, I'm inclined to accept your calculations as suffering only from input error, as opposed to what I did. – Anonymous Jun 26 '12 at 22:20