This is a work in progress...please do not read until I am finished. Thanks
[The CM-Gyroscope_Nutation.pdf file provided below does contain the full and complete derivation of the quantitative formulae and qualitative behaviors (precessional and nutational) of gyroscopes.]
This page discusses the full derivation of the motion of a gyroscope. It is supplementary material for the March 15, 2011 Fun Physics Facts for the Family Demonstration of the gyroscope...
Doesn't it appear that gyroscopes violate the laws of physics?!? Unfortunately, a full derivation of the dynamical behaviors of the gyroscope requires some rather sophisticated mathematics. In this topic we approach the problem using a nonhomogeneous second-order differential equation that allows us to derive specific forms for the formulae describing all of the behaviors of the gyroscope...
Above is a frame from the Gyroscope-Nutation-1080p.mov video showing a gyroscope repeatedly reversing the direction of its precessional motion producing a motion predominantly in one direction but with a back-and-forth oscillation superimposed on the unidirectional precession: a kind of cycloidal precessional motion.
(This video is uploaded to YouTube: http://www.youtube.com/watch?v=5Sn2J1Vn4zU )
Realize that the mathematics required for a full treatment of the nutational motion of a symmetric top also requires some sophisticated trigonometry as well as the solution of nonhomogeneous second-order differential equations. Thus the following link is to a wiki page reviewing the requisite mathematics:Review of the Mathematics for NutationThe mathematical details for deriving the nutational and precessional motions of a heavy symmetric top with one fixed point are extensive, and often the physics gets lost in the mathematics. Below we just list and give a very brief accounting of the formulae that we derive in the PDF files covering the nutational behaviors of the fast spinning top, and, in particular, listed are the formulae giving the motion in the \alpha angle (precession), both time dependent and time averaged, and the motion in the \beta angle (nutation). If you are interested in the full mathematical derivation of precession and nutation, then you must read the PDF files.But first, here is a quick summary of the territory through which we proceed during our derivation. We derive the mathematics for the spinning symmetric top with one fixed point from the Lagrangian function. This mathematics yields an Elliptic Integral of the First Kind and their inverses, the Jacobi Elliptic Functions, as the exact solutions to the nutational motions in the \beta angle. These elliptic integrals cannot be written in terms of simpler expressions, however they can be calculated to any desired degree of accuracy through series expansions (see the "Appendix: Elliptic Integrals" topic for details). Unfortunately, for most of us, this ellitpic integral exact solution provides little in the way of physical insight into nutational motions. It is of interest, as we will discover in the Math for the Motivated section, ``The Mathematics of Pendulums'', that precisely the same Elliptic Integrals of the First Kind and the Jacobi Elliptic Functions also constitute the exact solutions for the motion of a simple pendulum. And, just as we will carry out in the pendulum case, here for the spinning top we can simplify the mathematics by performing a "linearization" in which only the first-order terms in the series expansion of the force function are kept and all higher-order terms are dropped. Thus this "linearization" is an approximation that is valid and produces useful results when the motion is restricted to a small neighborhood of the equilibrium point. Another way of thinking about this approximation is that it means that only terms out to order two are kept for the potential energy function [Keeping first-order terms in the force function is equivalent to keeping second-order terms in the potential function since the force is the gradient (derivative) of the potential.], with all higher-order terms in the potential dropped. This means that under the linearization approximation, the resulting motions are described by second-order linear differential equations whose solutions are described by sinusoidal functions. It is these solutions of the linearized problem that give us insight into the nature and behaviors for nutational motions of spinning tops and gyroscopes.
Game Plan
Before we get started, let's discuss the target of the next few sections ending with the "Precession and Nutation in the Spinning of a Symmetric Top" subsection and its appendices. The overall goal is to derive the functional form for the nutation of a spinning top and use it to explain the top's qualitative behaviors. We accomplish this task in stages, the first of which is to derive the behavior of the free rotation of a rigid body using Euler's Equations. And in order to accomplish this, we must first understand the details of rigid bodies and their Euler Angle descriptions. From this Euler's Equations treatment we will find the expression describing the free precession of a spinning body, and explain in a topological sense why precession does not occur around one of the principal axes of the body.
We then extend this treatment to the case where a torque is being applied to the axis of a spinning symmetric body. Initially, we will assume that the \beta angle (the torque angle, the angle from vertical for a gravitational torque) is constant. This allows us to derive an expression for the torque-induced precession of a spinning top. Now the generic motion of a top entails nutation in addition to its precession, and therefore \beta is not constant in general, so this constant \beta expression is not valid for the general motion unless the top's movement is initiated in a very specific beginning state where \beta may then remain constant. At the same time, we realize that even though a nonconstant \beta is the generic motion, friction in the gimbal mount or the fixed point pivot often quickly dampens the \beta motion (nutation). After dampening of the nutation, then \beta indeed is constant, so we see that even though a constant \beta is not generic for the spinning top, it occurs often in practice because friction has damped out the generic nutational motion.
We will then use Lagrangian Mechanics to derive the exact form for the nutational motion, and we find that Elliptic Integrals of the First Kind and their inverses, the Jacobi Elliptic Functions, describe the exact motions of the angle \beta, the nutational motion. Now while this is indeed the exact motion, Elliptic Integrals are not all that useful for understanding the qualitative nature of the nutational motion. Thus we then "linearize" the problem, meaning that we approximate its potential energy function to second-order terms (and thus first-order in the force). In other words, a potential well, for small motions, is approximated closely by a quadratic well (Taylor Expansion of the potential function to second-order terms). This allows us to model the "linearized" problem with a nonhomogeneous linear second-order differential equation, the solutions of which provide us with both the quantitative and the qualitative explanations for the various nutational motions of the spinning top. In particular, we will derive the time-dependent nutational function, \beta(t), under the approximation of small motions in the \beta angle (the "linearized" approximation), and show that the time average of this function gives the torque-induced precessional rate that we derived earlier under the assumption of a constant \beta, thereby verifying that our time-dependent result gives the proper time-averaged result. Overall, this is a rather long and mathematically complex story, but because we split it into stages in the end we will gain a physical intuition into the precession and nutation for a spinning symmetric top.
- Rigid Body Rotation
Rigid Body Rotation <-- click to go to the Rigid Body Rotation page.
- Nutation
Until I have time to add the discussion of nutation to this wiki page, the mathematical details are provided in the following PDF file.Here is the PDF file giving a full derivation of Nutation:
Here is the PDF file giving the references for the Classical Mechanics Chapter:The following video (Gyroscope-Nutation-1080p.mov) explains and demonstrates the Classical Mechanics Explanation for the nutational motions of a gyroscope. This explanation predicts, both qualitatively and quantitatively, the nutational behaviors when a torque is applied to the gyroscope's axle.Click the following link to view the video on YouTube: http://www.youtube.com/watch?v=5Sn2J1Vn4zU(YouTube encodes the video at a number of different resolutions.)Description of the video's experiments: This video discusses the Classical Mechanics Explanation for nutation (gyroscope at a given energy moving in a potential energy well) and demonstrates the various nutational and precessional behaviors of a gyroscope.
To save time for now, the following text portions include the LaTeX markup language instead of the math formulae.
The strategy in this section is to employ the Lagrangian to calculate the dynamics for the heavy symmetric top. In order to do so, we need to find the kinetic energy $T=\frac{1}{2}\sum_{\kappa=\{\quat{i},\quat{j},\quat{k}\}}\lambda_{\kappa}\omega_{\kappa}^2$, specified in the body coordinate system. Now the Euler Angles $(\alpha, \beta, \gamma)$ specify the orientation of the top. To find the angular velocity $\bsym{\omega}$ starting from the space coordinate system $(\quat{\hat{x}}, \quat{\hat{y}}, \quat{\hat{z}})$ we apply each Euler angle rotation in turn to arrive at the final body coordinate system $(\quat{i}, \quat{j}, \quat{k})$. Now we could employ ${\mathbf R}_{\gamma\beta\alpha}$ to perform the required Euler angle rotations, but this leads to a relatively large amount of algebra which is unnecessary because of the special symmetry of the top. There are a couple ways to understand why this is unnecessary. First of all, the symmetric top has two degenerate eigenvalues, say $\lambda_{\quat{i}}=\lambda_{\quat{j}}$. This means that any two orthogonal vectors will span the same subspace and thus can serve as eigenvectors. So, instead of having to write the Euler angle transformations in terms of $(\quat{i}, \quat{j}, \quat{k})$, the final body coordinates, we know that any two orthogonal vectors will span the same subspace as $\quat{i}$ and $\quat{j}$, and, in particular, we could just as easily use $\quat{i''}$ and $\quat{j''}$ instead, eliminating this last coordinate transformation. This results in not needing to perform the required full coordinate transformations. And this is mathematically equivalent to just taking $\alpha=0=\gamma$. When we do this, the coordinate transformations are much simplified; in addition we also find a much simpler form for the Lagrangian function. When we arrive at this simpler Lagrangian, we will then give second interpretation for this simplification due to the symmetry of the top and its degenerate eigenvalues.
The fixed point of the heavy symmetric top sits at the origin of the space coordinates, $(x,y,z)=(0,0,0)$.
Let the top's orientation be specified by the three Euler angles, $\alpha$, $\beta$, and $\gamma$, as drawn in the above Figure. Let $m$ be the total mass of the top and $\bsym{r}_{\text{CM}}$ the vector from the origin to the top's center of mass, thus $r_{\text{CM}}$ is the distance from the origin to the center of mass of the top.
The gravitational potential energy $V$ of the top is thus computed bywhere $dm$ is the infinitesimal mass element for the top and the volume integration is over any volume containing the top. Moreover, $z_{\text{CM}}$ is the height of the top's center of mass above the origin. From the geometry diagrammed in the Figure, we see that $z_{\text{CM}}=r_{\text{CM}}\cos\beta$ and thus
Similarly, from the geometry in the Figure we see that the $\quat{\hat{z}$ coordinate is given by
For a rotating rigid body, the velocity of a point on the body at a distance $r$ from the axis is given by $\bsym{v}=\bsym{\omega}\times\bsym{r}$. If we let $\bsym{r}(t)$ be the position of a point on the top at time $t$, then at time $t+dt$ the point has rotated to
where $d\alpha = \dot{\alpha}dt$, $d\beta = \dot{\beta}dt$, and $d\gamma = \dot{\gamma}dt$. This calculation is accurate to within second-order terms, $(dt)^2$. Employing $\bsym{v}=\bsym{\omega}\times\bsym{r}$ to calculate $\bsym{r}(t+dt)=\bsym{v}dt=\bsym{\omega}\times\bsym{r}(t)dt$, the position vector results from the sum of
and
The total velocity of the point is thus
which means that the total angular velocity of the rigid body is defined by
This reminds us that in Newtonian Mechanics, velocities add vectorially, thus $\bsym{v}=\bsym{v}_{\alpha}+\bsym{v}_{\beta}+\bsym{v}_{\gamma}$ where $\bsym{v}_{\alpha}=\bsym{\omega}_{\alpha}\times\bsym{r}$, $\bsym{v}_{\beta}=\bsym{\omega}_{\beta}\times\bsym{r}$, and $\bsym{v}_{\gamma}=\bsym{\omega}_{\gamma}\times\bsym{r}$.
In order to compute the kinetic energy of the top, we need the angular velocity $\bsym{\omega}$ expressed in terms of the body coordinates, $({\omega}_{\quat{i}}, {\omega}_{\quat{j}}, {\omega}_{\quat{k}})$, however. Once again, this could be accomplished through use of ${\mathbf R}_{\gamma\beta\alpha}$ to rotate the space angular velocity vector to the body coordinates, but this full calculation is unnecessary when $\alpha=0=\gamma$. Looking at the Figure, we see that in this case the $\alpha$ rotation ${\mathbf R}_{\gamma\beta\alpha}(\alpha+d\alpha,\beta,\gamma){\mathbf R}_{\gamma\beta\alpha}^{-1}(\alpha,\beta,\gamma)$ is just a rotation through an angle of $d\alpha$ around the $\quat{\hat{z}}$ axis, therefore
the $\beta$ rotation ${\mathbf R}_{\gamma\beta\alpha}(\alpha,\beta+d\beta,\gamma){\mathbf R}_{\gamma\beta\alpha}^{-1}(\alpha,\beta,\gamma)$ is a rotation through an angle of $d\beta$ around the $\quat{j''}$ axis, thus
while the $\gamma$ rotation ${\mathbf R}_{\gamma\beta\alpha}(\alpha,\beta,\gamma+d\gamma){\mathbf R}_{\gamma\beta\alpha}^{-1}(\alpha,\beta,\gamma)$ is a rotation through an angle of $d\gamma$ around the $\quat{k}$ axis and hence
From Equation for $\quat{\hat{z}}$, we see that when $\alpha=0=\gamma$, then $\quat{i''}=\quat{i}$, $\quat{j''}=\quat{j}$, and we have
Hence we may rewrite the total angular velocity in body coordinates as
and thus
Now we may calculate the kinetic energy $T$, remembering that $\lambda_{\quat{j}}=\lambda_{\quat{i}}$, as
Notice that the kinetic energy does not depend upon $\alpha$ or $\gamma$, as it shouldn't since these two coordinates are cyclic coordinates. Therefore we may pick the origins for the $\alpha$ and $\gamma$ coordinates (equivalent to choosing any orthogonal set of vectors for the eigenvectors spanning the two-dimensional subspace corresponding to the degenerate eigenvalues of the inertia tensor) so that $\alpha=0=\gamma$ without changing the value of $T$. Hence the formula that we have derived for $T$ under the restriction that $\alpha=0=\gamma$ is in fact valid for all $\alpha$ and $\gamma$. This is the second interpretation for understanding what the symmetry of the top, the degenerate eigenvalues $\lambda_{\quat{i}}=\lambda_{\quat{j}}$ of the inertia tensor, provides for the Lagrangian. Hence we may now write the Lagrangian function as
and all of the dynamics for a system may then be derived from the Lagrangian function through what are known as the Euler-Lagrange Equations:
where ${q}_j$ are the general coordinates and the $\dot{q}_j$ are the generalized velocities. In our case for the spinning rigid symmetric top, the generalized coordinates are the three Euler angles $\alpha$, $\beta$, and $\gamma$ and the generalized velocities are the corresponding angular velocities $\dot{\alpha}$, $\dot{\beta}$, and $\dot{\gamma}$. These equations stem directly from Newton's Laws and the d'Alembert Principle. For a detailed mathematical derivation, see "The d'Alembert Principle" Section (beginning on page 4249) and especially the immediate pages leading up to the Euler-Lagrange Equations.
For now, if we simply accept the Euler-Lagrange Equations as providing the dynamics for the spinning rigid rotor, then we first notice that the Lagrangian function, $\mathcal{L}$, does not depend upon either the $\alpha$ nor the $\gamma$ coordinates. This means that
Coordinates that obey this condition are known as ignorable, or cyclic, coordinates. From the Euler-Lagrange Equations, we see that
Now when the time derivative of a function is zero, then the function itself is constant, so we have
These constants are given special names, they are called generalized momenta, denoted by $p_{\alpha}$ and $p_{\gamma}$, respectively. Yes, the generalized momenta are closely related to the linear momentum of Newtonian Mechanics. And, if fact, we see from the last two equations that $p_{\alpha}$ and $p_{\gamma}$ are conserved. What are these generalized momenta? Well, let's calculate them to see just what they are, shall we? We compute the partial derivatives as
and
Now these are simple enough expressions, but exactly what are they?
We first calculate an expression for the angular momentum in body coordinates,
and using Equation for the angular velocity in body coordinates, we find
Wow! This is equal to $p_{\gamma}$, and thus $p_{\gamma}$ is none other than the angular momentum of the spinning top around its spin axis, ${L}_{\quat{k}}$. Since $p_{\gamma}$ is conserved, then the angular momentum around the spin axis must be conserved:
Since ${L}_{\quat{k}}$ is constant and $\lambda_{\quat{k}}$ is a constant, then
On the other hand, employing the Equations for $\boldsymbol{L}$ and $\quat{\hat{z}}$, we may also calculate the component of $\bsym{L}$ along the space axis $\quat{\hat{z}}$,
which we now recognize is equivalent to $p_{\alpha}$! Since $p_{\alpha}$ is conserved, then
In words, the component of the angular momentum along the space $\quat{\hat{z}}$-axis is also conserved.
The reason that ${L}_{\quat{k}}$ and ${L}_{\quat{\hat{z}}}$ are conserved stems from the fact that the gravity torque, $\bsym{\Gamma} = \bsym{r}_{\text{CM}}\times{m\bsym{g}}$, is perpendicular to both $\quat{k}$ $(\bsym{r}_{\text{CM}}\parallel\quat{k})$ and $\quat{\hat{z}}$ $(\bsym{g}\parallel\quat{\hat{z}})$ (it lies along the line of nodes) and thus there is no component of this torque along either the $\quat{k}$-axis or the $\quat{\hat{z}}$-axis to alter the angular momenta around these axes.
In summary so far, the Euler-Lagrange Equations easily derive the conservation of ${L}_{\quat{\hat{z}}}$ and ${L}_{\quat{k}}$. Physically, these quantities are conserved since the gravity torque is perpendicular to both $\quat{\hat{z}}$ and $\quat{k}$ and thus has no component to alter these values.
For the third and last generalized coordinate, $\beta$, we have to return to the full Euler-Lagrange Equation,
where we calculate the derivatives by
and thus
By the same token we find
In general, this derivative is not identically zero, $\partial{\mathcal L}/\partial{\beta}\ne0$, and consequently this time derivative is nonzero in general which means that the generalized momentum $p_{\beta}$, given by the above Equation, is not in general constant either. In other words, $p_{\beta}$ is not conserved, in general, for the spinning symmetric top.
Substituting these derivatives into the $\beta$ Euler-Lagrange Equation, we have
But since our earlier informal argument found a steady solution for the precession rate, $\dot{\alpha}_{-}$, let's begin with trying to find this steady solution in this rigorous treatment. So we first assume that there exists a steady solution; this means that $\beta=\text{constant}$. If $\beta$ is constant, then $\ddot{\beta}=0$, which we substitute into the above equation and take the negative of the resulting expression to yield,
Assuming that the top is leaning at least a little bit, otherwise there would be no gravity torque on it, then $\beta\ne0$ and we can divide this last expression by $\sin\beta$,
I want to point out that this division eliminates the $\sin\beta$ factor from the torque term $r_{\text{CM}}mg\sin\beta$, and thus it is $r_{\text{CM}}mg$ and not $r_{\text{CM}}mg\sin\beta$ that ends up in the torque-induced precession rate expression, $\dot{\alpha}_{-}$, that we derive below.
And finally, we substitute $\omega_{\quat{k}} = \dot{\gamma}+\dot{\alpha}\cos\beta$ to find
This Equation is simply a quadratic equation in the unknown $\dot{\alpha}$, and it is easily solved by the quadratic root solution,
Now you may be thinking that these roots do not look like the solution we found earlier for the precession rate, the Equation for $\dot{alpha}$. But, in fact, the above minus root, $\dot{\alpha}_{-}$, is precisely this precession rate. We can see this though use of the approximations in the ``Quadratic Mystery Solutions'' Example (page 3999). Let's see how this is done, shall we?
To use the Quadratic Mystery Solutions, we have to assume $\omega_{\quat{k}}$ is very large, so large that $\lambda_{\quat{k}}^2\omega_{\quat{k}}^2 > 4\lambda_{\quat{i}}r_{\text{CM}}mg\cos\beta$. If this is true, then we can employ the mystery solutions. Now this is almost always true for spinning tops, otherwise the top would not operate as you would expect.
To get started, let's review Newton's Real Exponent Binomial Theorem,
valid for $\vert{y}\vert<1$ and $\alpha\in\Rrea$. If we let $y=-y$ and $\alpha=1/2$, then we find
and if we let $y=(4\lambda_{\quat{i}}\cos\beta{r_{\text{CM}}mg})/(\lambda_{\quat{k}}^2\omega_{\quat{k}}^2)$ and keep only the first two terms, we have
which we will use in the quadratic roots. In particular, notice that the quadratic roots can be rearranged to
into which we substitute Newton's Approximation for the radical to obtain
For the plus sign root, we find
and if $\lambda_{\quat{k}}\omega_{\quat{k}}\gg{r_{\text{CM}}mg}$ then the last fraction on the right-hand side may be neglected and we are left with the approximation,
On the other hand, for the negative sign we find
and therefore
Notice that the approximation for $\dot{\alpha}_{+}$ is proportional to $\omega_{\quat{k}}$, it describes a motion when $r_{\text{CM}}mg=0$ equivalent to free precession that we derived in the Rigid Body Rotation topic (and the CM-Gyroscope_SymmetricTop.pdf file). This is the precession of the rotation axis when no torques are applied. It stems from a slight misalignment of the angular momentum and the spin axis. On the other hand, the approximation for $\dot{\alpha}_{-}$ is proportional to $r_{\text{CM}}mg$, that is, it is proportional to the torque applied by the force of gravity at any given $\beta$ angle, and inversely proportional to $\omega_{\quat{k}}$. This is the average precessional motion of the spinning top caused by the torque from gravity. It is equivalent to the average precession rate that we derived earlier under the assumptions of a weak torque, fast spin rate, and thus a constant $\beta$ angle. [Our earlier derivation of this Equation stemmed directly from Euler's Equations for the rotation of a rigid body under the influence of a torque. Now, our derivation of the same precession rate, $\dot{\alpha}_{-}$, was through use of Lagrangian Mechanics. So, we find the same expression for the torque-induced precession rate from two alternative approaches, as we would expect, assuming that we performed the calculations correctly.] Notice that the greater the gravity torque (at any angle), the faster the precessional motion, while at the same time the faster the spin velocity, the slower the precessional motion. We thus see that the two approximate solutions to the quadratic equation describe two types of precessional motion, the free precession that takes place in the absence of torques and the precessional motion produced by a torque supplied by the force of gravity.
Note that the precessional rates calculated above when $\beta$ is a constant are steady. While these steady solutions do exist, this is a somewhat contrived situation. The general solutions to the Euler-Lagrange Equations entail nutational motions where $\beta$ oscillates back-and-forth and thus the precessional rates also fluctuating around their steady values calculated above. From a practical viewpoint, however, the oscillations in the $\beta$ angle typically dampen out quickly from friction in the gimbal mount of the gyroscope. Thus the nutational motion is a transient effect and the long term behavior of the gyroscope or spinning top is a steady precessional motion.
In summary, as we have already discussed, the Earth experiences a torque-induced precessional motion caused by the oblate spheroid shape of the Earth and its polar axis tilt of $23.4\degree$ that produces a torque on the rotational axis by the Sun's and Moon's gravitational forces.
The minus approximation, $\dot{\alpha}_{-}$, describes this motion, known as the Precession of the Equinoxes. Because the torque-induced precession is inversely proportional to the Earth's spin rate, it takes $25772\unit[\/]{years}$ to complete one revolution. At the same time, the other approximation, $\dot{\alpha}_{+}$, also describes a precessional motion for the Earth known as the Chandler Wobble, discovered in 1891 by the American amateur astronomer Seth Chandler. This precessional motion is proportional to the Earth's spin rate and thus is much faster. Because of the Earth's equitorial bulge, moment of interia about the polar axis is roughly one part in $300$ larger than the moment about the other two equitorial axes, $\lambda_{\quat{k}}=(1+1/300)\lambda_{\quat{i}}$. This means that the precession in the space frame, ${\lambda_{\quat{k}}}\omega_{\quat{k}}/{\lambda_{\quat{i}}\cos\beta}$, has a period of roughly a day. In the body frame, however, the precession rate, as given by $\kappa=\omega(I_1-I_2)/I_2$ (see the CM-Gyroscope_SymmetricTop.pdf file), is $(\lambda_{\quat{k}}-\lambda_{\quat{i}})\omega_{\quat{k}}/\lambda_{\quat{i}}$ which has a period of about $300$ days. In fact, it takes roughly $433\unit[\/]{days}$ to complete one revolution. At first this was a source of confusion, but then it was realized that because the motion of the oceans, the annual melt of snow, and the non-rigidity of the Earth itself, the free precession rate is modified from its expected $300$ day to its actual $433$ day period. [There are actually other polhode motions of the Earth's axis caused by the actions of tides and the superposition of the actions of the Moon's and Sun's gravity.]
Okay, we have now rigorously computed the steady precessional motion of a spinning top, and found the free precession and torque-induced precession rates. Next we extend this analysis to include the case when $\beta$ is not constant, and for this case we will find the motion known as nutation. Now the details of the mathematics become somewhat overwhelming, but not to worry because after the mathematical detail we will return to stress the physics of the spinning symmetric top.
We get started by recalling the Equations for the conserved ${L}_{\quat{k}}$ and ${L}_{\quat{\hat{z}}}$,
(...to be continued...)
- iPhone 5's Cycloramic.app: proposed mechanism
The recent Cycloramic app for the iPhone 5 will spin the iPhone standing on its bottom edge while taking a panoramic video. Explanations of the mechanism, the actual physics, behind this spinning motion so far have been limited to just stating that the spinning is due to the vibration without any explanation as to how that vibration produces rotary motion. We collected a number of observations from various videos posted on the Internet, and we propose a physical mechanism that explains all of these observations.
Gyroscope Precession: Cycloramic.app <-- click on this link to view the Cycloramic.app page.
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