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The swing of a pendulum is often approximated by Simple Harmonic Motion (SHM) for small swings, i.e. if $\theta$ is small. SHM takes place over a straight line. Even though the pendulum really moves through a distance $2L\theta$, it's similar to $2L\sin\theta$ as $\theta$ is almost the same as $\sin\theta$ (in radians) if the angle is small. The $2L\sin\... 1 I think you are misreading the page/slide that you show. I don't think it is saying that linear displacement is the length of the arc. It says linear displacement is$\Delta x$(not$\Delta s$). Although$x$is not shown in the diagram, I imagine it is referring to a one dimensional linear motion$x=f(t)$where displacement between times$t_1$and$t_2$is$...

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First I think it is useful to give a brief definition of displacement and distance. Distance depends on the path which an object moves whereas displacement is the shortest distance between two points, usually initial and final positions. Displacement and Linear distance imply the same. You have mentioned in your question that you have used equations of ...

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Just wanted to add a picture to better visualize the equation of constraint. Hope it helps someone out there. I've called the angle related to the rotation of the sphere $\gamma = \pi$ and I've drawn the situation for when $\gamma=\pi$.

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Your thought experiment needs to start with a more basic questions: "What is a straight line?" and "What is a right angle?" and "How to measure distance?" If we are doing real physics experiment, and we say use a beam of laser as a "straight line" and an atomic clock to measure the time of transit between points thus ...

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As an alternative to the other answers, you could just use the principle of relativity rather than any explicit form of the Lorentz transformation (which is deduced from it): this says that the laws of physics can be formulated in a way that is the same in all inertial frames. This implies that there is no experiment that would tell you your state of (...

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I have made a diagram to clarify the situation black is for the $S$ coordinate system red is for the $S'$ coordinate system with motion $v'$ relative to $S$ The black triangle is what $S$ sees. The red triangle with smaller reduced base $b'$ is what $S'$ sees. $S$ and $S'$ will both see right triangle and both will obey the Pythagorean Theorem. a^2+b^2=c^... 6 In all inertial frames the metric is ds^2=-c^2 dt^2 + dx^2 + dy^2 + dz^2. For any moment in time we can set dt=0 and recover the Pythagorean theorem. So the Pythagorean theorem holds for space in all inertial frames in special relativity. So, if one leg of a right triangle is length contracted, then you can use the Pythagorean theorem to determine the ... 15 If you draw a right triangle on a sheet of rubber, and you uniformly stretch the rubber in a direction aligned with one of the non-hypotenuse sides...you still have a right triangle. However, the angle has changed. Do you suppose there could be any parallels with your thought experiment? 3 The unit normal vector is part of the definition of a surface. To completely define a surface you must give (choice) its unit normal vector. Consequently you must be careful using it for calculations. For example the flux of a vector function for one choice would be the opposite of that of the other choice etc. In Figure-01 above we see a surface in \:\... 0 Assuming that the radius of the spool of cable decreases linearly with rotation angle as the drum is unwound. Then\text{ total length of cable unwound} =\text{average of initial and final radii} \times \text{ angle turned through} Note that this does not depend on the cable thickness. This is because a larger thickness results in a larger decrease in ...

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You can determine the center of mass of your example bottle using one string method experiment plus one logical assertion. The logical assertion is that the bottle is rotationally symmetric, meaning that its center of gravity must lie on the symmetry center line. For the string method experiment, make sure that the string is not attached at the bottle's ...

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He must be talking about how the ideas of geometry need not correspond to real world objects. Geometry was initially conceived as the study of real world shapes that we see around us. Stuff like "The sum of angles of a triangle is 180 degrees", or "parallel lines never meet each other" were seen as self-evident truths. Euclid's axioms ...

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Einstein is pointing out the difference between Mathematics and Physics. In Mathematics axioms are assumed and then theorems are derived from this axioms. For example Euclidean geometry is derived from Euclid's five axioms. Assuming different axioms results in different geometries like hyperbolic geometry and elliptic geometry. The important point here is ...

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Suppose for a moment that there's no Earth under you, but the Moon remains where it is. You are at a very large altitude above the Moon: about $4\times10^8\,\mathrm m$. What does the border of the Moon look like? It's a circle. If you go "down" (towards the Moon), this circle will grow, eventually exceeding your field of view, but still remaining a ...

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Circles appear curved when viewed from somewhere not in the plane of the circle.

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Your concern about shifting vectors is well-founded. The problem originates in the sloppy introduction to vectors usually present in introductory physics courses. The video you refer to is not much different. Let me start stressing where are the problems. From the mathematical point of view, the idea that a parallel shifted vector is the same vector would ...

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He wasn't talking in that sense. It just means that a vector is describe by an arrow in space that has a specific direction and a length. Now it doesn't matter if this arrow is moved around in space without changing its direction and length, It will still remain the same vector. For example, Consider a vector which point from $(1,1,1)$ to $(2,2,2)$. So the ...

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