I'm recreating a calculation from this paper. The authors calculated the angular momentum (L) of a spinning disc with mass = 114.7 grams (0.1147 kg) and radius = 4.75 cm (0.0475 m). The disc spins at 11,000 rpm (~1151 rads/s). They give L as "0.0474 J-s".

However, when I calculate L, I get something totally different. Here's my steps.

L = I * Omega, where I is moment of inertia(MOI) and Omega is angular velocity.

Using the equation for MOI of a disc using kilograms and meters I get I = 1/2 * (0.1147) * (0.0475^2)

To calculate Angular Velocity using radians/sec, I get L = 0.00012939 * 1151.917 = 0.14905 kg-m^2 / second

Which is nowhere near the authors' 0.0474 J-s. I suspect I am making an error in my use of units, but my understanding is that J-s (Joule-seconds) are equivalent to kg-m^2/sec (kilogram-meters squared per second). Am I confused about units, or did I make a calculation error?

EDIT: Solved! It appears the paper itself was wrong. Thanks to JEB for pointing this out.

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    $\begingroup$ You are not "way off", you are off by a factor of $\pi$. Actually, I think they are off by a factor of $\pi$. $\endgroup$ – JEB May 31 '18 at 18:08
  • $\begingroup$ Interesting... so just so I'm clear, you're saying they forgot to include pi in their calculation (In which case, I'm assuming they forgot to multiply by pi when converting rpms to radians/sec.)? $\endgroup$ – PhiloT May 31 '18 at 18:23
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    $\begingroup$ well, that's $2\pi$--so who knows--your numbers look good. $\endgroup$ – JEB May 31 '18 at 18:46
  • $\begingroup$ Great -- thanks very much for your help. This resolves hours of head scratching for me. It never occurred to that paper could be wrong. If you want to put your comment as an answer, I will go ahead and select it. $\endgroup$ – PhiloT May 31 '18 at 18:49

For folks who haven't followed the link, the paper is about using a modified hard disc drive to build a handheld rapid-response gyroscope, and the mass and radius in the question are for a stack of hard disc platters.

It's worth remembering that $I=\frac12mr^2$ is the moment of inertia for a uniform disk. When my students encounter these sorts of problems in the laboratory, it's most often because the assumption of disk uniformity is flawed, and their object has more of its mass near the spindle or near the edge. So let's give the authors the benefit of the doubt and see what moment of inertia they are implicitly using.

The moment of inertia implied by the angular momentum and angular frequency is

\begin{align} I_\text{reported} = \frac L\omega &= \frac{0.0474\rm\,J\,s}{2\pi \frac{\rm rad}{\rm turn} \times 11\rm k\frac{\rm turn}{\rm min}} \color{lightgrey}{\times \frac{60\rm \,s}{1\rm\,min}} = 41.1\rm\,\mu J\, s^2 \end{align}

The scale factor for the moment of inertia is

$$ I_\text{ring} = mr^2 = \rm 0.1147\,kg \cdot (0.0475\,m)^2 = 259\,\mu J\,s^2 $$

and the ratio between these two is

$$ \frac{I_\text{reported}}{I_\text{ring}} = 0.159 = \frac{1}{2.002\pi} $$

For an ideal disc you'd expect this ratio to be exactly $\frac12$. This doesn't look like the authors have done anything clever like taken the angular momentum of the motor and spindle into consideration; it looks like they thinko'd a factor of $\pi$ when they computed the angular frequency.


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