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Angular momentum, or a measure of rotation, in very large astrophysical or cosmological bodies and energy can become relativistic, and must be treated using General Relativity. In General Relativity (GR) angular momentum is not too different from momentum and energy. Different quantities, the last two are contained in the stress energy momentum tensor and ...


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This question leads to some subtleties. There are at least two distinct notions of "revolution" that could be meaningful in physics. Namely, "to revolve" can mean: To have angular momentum; To transform by a particular kind of Euclidean isometry (a rotation) (or, to be broader and more technical, a representation of that Euclidean isometry). As far as we ...


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$h=R(1-\cos\theta)=R(2\sin^2\frac{\theta}{2})$ $v=\sqrt{2gh}=\sqrt{4gR}\sin\frac{\theta}{2}$ $v$ is a maximum at $\theta=\pi$. The tangential acceleration is $\frac{dv}{dt}=\sqrt{4gR}\cos\frac{\theta}{2}.\frac12\frac{d\theta}{dt}=\sqrt{4gR}\cos\frac{\theta}{2}.\frac12.\frac{1}{R}\sqrt{4gR}\sin\frac{\theta}{2}=g\sin\theta$ as expected. This is a maximum at $\...


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This is a simple variation on the so called "Twin Paradox", which is not a paradox in the logical sense (i.e. not a logical contradiction). Each cycle of the oscillator's motion is like the journey of the spacefaring twin. One possible cycle on a spacetime diagram is drawn below (source: Wikipedia "Twin Paradox" Article with my own additions). The ...


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Lorentz transformations are transformations between inertial frames of reference, i.e. frames of reference that move with constant velocity relative to each other. An oscillation frame as you described it is not an inertial frame of reference, so Lorentz transformations do not apply here.


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The position vector of a particle in polar coordinates is given by $$\mathbf r ~=~ r\mathbf e_\mathrm r$$ Velocity $\bf v$ is \begin{align}\mathbf v&=\frac{\mathrm dr}{\mathrm dt}~\mathbf e_\mathrm r+ r\frac{\mathrm d\mathbf e_\mathrm r}{\mathrm dt}\\ &= \frac{\mathrm dr}{\mathrm dt}~\mathbf e_\mathrm r+ r\frac{\mathrm d\theta}{\mathrm dt}~\mathbf ...


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Point C is sure instant centre of rotation for both gears, otherwise they would get teeth broken if any relative slide to each other. Analogy is a wheel on a road having instant centre of rotation at the bottom point thus velocity of top point is twice more than of the car. As far as I understand, confusion point is that first you calculate VO in respect ...



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