If you're watching the 10th flavour of The Big Bang Theory, you lot must know that the latest episode started amongst Euler's disk, a supersized spinning coin. Here's a really helpful 2016 video close Euler's disk:
The disk is unremarkably sold equally a large in addition to heavy cylindrical steel amongst chrome on it along amongst a mirror that has a shallow hole in addition to then that the "big coin" stays close the center. You should definitely purchase the bestselling $35 Toysmith Euler's disk – which has 251 reviews (it almost looks similar the heroes of The Big Bang Theory were using this exact shiny $35 product, or was it this one for $40?) – in addition to also the #1 bestselling fragrange on Amazon.com, the Ivanka Trump spray. The #2 bestselling affair inwards beauty is the Ivanka Trump Roller Ball, whatsoever it is. Not bad for a adult woman who isn't fifty-fifty a existent climate skeptic in addition to who teaches her pocket-sized fry Chinese instead of Czech.
The video I embedded at the hand off explains the telephone commutation mathematics of Euler's disk. Well, the 2nd part of the video is divided to 2 half-screens. The intellectually challenged people are to a greater extent than probable to sentinel the right manus side where the disk is spinning in addition to completes the theatre piece the smarter viewers tend to await to the left where he describes how it industrial plant mathematically.
The spinning takes such a long fourth dimension that Rajesh almost suffocated because he promised non to breathe piece it was spinning, long plenty for Penny to lose her involvement long earlier the spinning ends, in addition to long plenty for the writer of the video at the hand off to explicate how it works. ;-) He drew a vector from the pump of the disk to the circumference, a radial vector \(\vec r\). When you lot honour the spinning of Euler's disk, you lot run across that the management of \(\vec r(t)\) is changing amongst time.
It's basically because the actual physical circumference of the disk whose length is \(2\pi R\) is beingness "drawn" on a shorter circle of circumference \(2\pi r\) which is smaller. You may really accurately assume that the pump of majority of the disk sits on a vertical axis at all times – well, it would last truthful if you lot managed to fully stabilize the horizontal motion of the disk along the mirror.
Why is the pump of majority non oscillating inwards the horizontal direction? Well, it's because alone vertical forces – gravity in addition to the pressure level from the mirror – are acting on the disk. There is no horizontal forcefulness – which agency that the velocity of the pump of majority inwards the horizontal management remains constant: it is only non periodically oscillating. This disceptation isn't quite truthful because the mirror underneath is a chip curved in addition to then the forcefulness from the mirror isn't quite vertical in addition to your Euler's disk is like shooting fish in a barrel moving along the mirror inwards diverse directions, too. But nosotros fail this motion – if you lot exercise it well, the motion may last ready to zero.
OK, the lower circumference of your Euler's disk is adiabatically moving along a circle of circumference \(2\pi r\) where \(r\) is a chip smaller than \(R\) – fifty-fifty though it approaches \(R\) at the terminate of the spinning performance. How much smaller \(r\) is relatively to \(R\)? Well, if the angle betwixt the horizontal surface of the mirror in addition to the apartment base of operations of the Euler's money is labeled \(\theta\), you lot may run across a triangle connecting the intersection of the spinning axis amongst the mirror, the pump of majority of Euler's disk, in addition to the alone indicate where the disk touches the mirror.
One of the angles of this triangle is ninety degrees, some other one, close the "touching point", is \(\theta\), in addition to then you lot run across past times the basic trigonometry that \(r=R_+ \cos\theta\). Well, if you lot were really careful, you lot must bespeak keep noticed that I had to usage \(R_+\) rather than \(R\) where \(R_+\) is the distance of the center of mass of Euler's disk from the lower circumference – in addition to because the disk is slightly thick, it's a chip longer than the radius \(R\) of the base of operations of the disk (the distance measured along the lower surface of the disk). OK, let's fail this divergence betwixt \(R\) in addition to \(R_+\) for a moment.
We run across that 1 "period" of the spinning tries to wind the circumference of the disk on the smaller circumference \(2\pi r\) drawn on the mirror but it doesn't quite wind the whole circumference of the disk. The divergence is\[
2\pi R - 2\pi r = 2\pi R (1-\cos\theta) \approx \pi R\theta^2
\] where I Taylor-expanded the cosine, cancelled the leading term, in addition to alone kept the subleading term proportional to \(\theta^2\). The deficit nosotros precisely advert causes the rotation of the radial arrow on the disk, the "wobbling". You may run across inwards what management it's wobbling. And the ratio of the wobbling in addition to the spin is therefore\[
\frac{2\pi (R-r)}{2\pi r} \approx \frac{1}{\cos\theta} - 1 \approx \frac{\theta^2}{2}
\] where the in conclusion approximation is OK for a sparse disk in addition to \(\theta\to 0\). In this limit, you lot run across that the wobbling becomes much slower than the spinning because the seem to a higher house goes to zero. Now, a funny realization is that the charge per unit of measurement of wobbling is constant basically because the angular momentum is conserved (at the end, the whole angular momentum is connected amongst the rotating management of the arrow he drew; the vertical cistron of the angular momentum is conserved because the physical occupation is rationally symmetric nether rotations some the vertical axis, cheers to Emmy Noether for the idea), in addition to that's why the charge per unit of measurement of spinning has to conk to infinity for \(\theta\to 0\).
If you lot furnish to 1 of the omissions, \(R\) vs \(R_+\), I retrieve that the upshot volition last that the ratio to a higher house – amongst \(2\pi R_+\) equally the commencement term inwards the divergence inwards the numerator – won't conk to nada for \(\theta\to 0\). Instead, it volition conk to a finite positive constant that is an increasing piece of work of the thickness of the disk. Because of the Pythagorean theorem (which really makes the equation below exact),\[
R_+ = R + \frac{t^2}{4R}
\] where \(t\) is the thickness of the disk. Note that \(R_+/R\) is alone greater than 1 past times \((t/2R)^2\) which is really pocket-sized for a pocket-sized plenty \(t\lt R\). But if you lot created a thinner Euler's disk, I retrieve that the genuinely concluding phase of the spinning would last fifty-fifty faster. Do you lot agree?
Equivalently, I could bespeak keep used \(R\) – the radius of the disk's base of operations – everywhere but I would bespeak keep to usage a dissimilar angle \(\theta_+\) which is the angle betwixt the horizontal mirror in addition to the line of piece of work connecting the "touch point" amongst the center of mass. This angle \(\theta_+\) doesn't quite conk to nada at the terminate of the spinning – in addition to this is why the ratio (and the spinning speed) doesn't quite conk to infinity. Again, the corrections conk similar \(t^2/R^2\).
Do you lot grip that inwards the final stages, the motion in addition to audio are almost to the chirping of a dark hole merger detected past times LIGO? ;-) I hope that the gravitational moving ridge detectors volition honour much longer in addition to clearer sounds sometime inwards the future.
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