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The orbits in the video you saw, for all intents and purposes, are about the same as the orbits for objects like Halley's comet -- thank's to https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion">Kepler's second law, when the object is close in, then it moves much more quickly than when it's far out. So, why does this show a black hole? Well, ...

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You need the total distance traveled by a point on the outer edge of the wheel to be equal to a predetermined distance. This can be found using the standard kinematic equation below: V^2 - (V0)^2 = 2*a*s where V is your final velocity, V0 is your initial velocity, a is your acceleration (negative for you wheel to be slowing down), and s is the distance ...

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In mathematics, complex numbers and rotation are intimately related. After rotation $\theta$ in the complex plane, the number $1$ becomes $e^{i\theta}=\cos\theta + i\sin\theta$. Although $e^{i(2n\pi)} = 1 = \cos(0) + i\sin(0)$, you actually moved by $\theta$ and not by $0$. If you rotated something by $\pi$, you could say the angle is now $-\pi$ - but you ...

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When an object is spinning, there are indeed stresses developing that keep the object from ripping apart. But every material has a finite strength (the yield stress), and when you exceed that it will deform plastically (it will not go back to its old form after you remove the stress), and may even break (when you reach the ultimate tensile strength). If you ...

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Let the cone lie on the $\hat{X}\wedge \hat{Y}$ plane (z=0) and let the $z$ axis pierce this plane at the cone's apex. If the cone's half angle is $\alpha$, then its axis of symmetry as a function of time is defined by the vector $$A(t)=\cos\alpha \left(\cos(\omega_0\,t) \hat{X} + \sin(\omega_0\,t) \hat{Y}\right)+\sin\alpha \hat{Z}$$ where $\omega_0 = ... 1 The answer is that at any moment a rigid body is rotating about an arbitrary axis, with an arbitrary angle. There are two components defining the direction of rotation axis, and one for the magnitude of rotation. Often we transform these three quantities into three sequential rotations called Euler Angles. If you look up rotation matrix you will find all ... 2 Made visualisations so you can see how it looks like. Rotating X: Rotating X and Y: Rotating X,Y and Z: 1 There is not a simple answer for this since it is both yes and no. From an internal frame of reference: Looking at the object, nothing is happening. Looking at the outside world, it is moving in a predictable pattern. From an external frame of reference: The rotation on one axis is easy to see. The rotation on 2 axes looks like a rotation on a moving axis. ... 0 The "math works it out" indeed. I try to write it as accessible as I can. A 0-domensional Euclidean space is just a point. The 1-dimensional is a line. The 2-dimensional is a plane. The 3-dimensional is the space as we know it. This can be continued to 4-5-6 whatever dimensions. On a plane you can draw lines and point, but not planes. In space you can ... 1 Yes, the shape of the Earth is consistent with general relativity. Thirring's equations are not an indication that general relativity predicts results that are inconsistent with the observed shape of the Earth, in large part because Thirring's equations applied general relativity incorrectly. For clarity, this question and answer are purely about ... 5 (The following answer was written to address the original version of this question, which was simply "Does General Relativity theory correctly explain the ellipsoidal shape of the earth?") Yes. General relativity predicts that the equator will bulge out just enough such that the reduction in gravitational time dilatation at the equator relative to the ... 3 Instead of working with angles (which are very easy to mess up in 3D), I would just go directly to a transformation matrix. I had to solve a very similar problem to this when tracking charged particles through magnetic fields and all the angles and rotations kept messing me up (probably due to subtle sign errors). So, I went with pure linear algebra. You ... 1 I believe you have a math error in your matrix multiply. Substituting the single prime vector into the second rotation and multiplying the matrices doesn't give that third matrix. 0$\sigma_{\vec{\imath}}$is the rotation matrix about the vector$\vec{\imath}$: $$\sigma_{\vec{\imath}} = \imath_x\sigma_{x}+\imath_y\sigma_{y}+\imath_z\sigma_{z}$$ After expanding it out completely and after some shuffling around, the solution is as follows:$\$e^{-i\theta/2 \sigma_{\vec{\imath}}^A} \otimes e^{-i\theta/2 \sigma_{\vec{\imath}}^B} ...

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