# Tag Info

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I find no trouble in thinking of velocity and acceleration vectors as arrows. First some definitions: Velocity is a vector. Speed is it's magnitude. Acceleration is a vector. It's magnitude has no new name. We agree that acceleration is present if there is a change in velocity: $$\vec a= d \vec v /dt$$ That is, any change. So if magnitude (speed) and/...

3

Centripetal force is the name of the force that points towards the centre. This is in the radial direction. Tangential velocity is, as the name suggests, a velocity direction tangent to the circle. The radial and tangential directions are by definition always perpendicular - in the same way that the x and y axes are. You are of course right that if any ...

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You can't meaningfully make that substitution. Newton's second law (for constant mass) is actually: $$\sum F = ma$$ In other words, the acceleration is proportional to the total force.

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I think the OP has made a mistake in applying second law of Newton. The law (about a particle) says: $$\Sigma \vec F=m\vec a$$ As it is seen, this is a vector equation. This means that corresponding components of both side of the equation must be equal. Although it is not said in the law's body, but it is obvious that we must write the equation above with ...

2

I want a more mathematical way to see it. Acceleration is defined as the rate of change of velocity with time. Velocity, being vector, can change just by changing direction keeping the magnitude constant. In a circular motion if the angular frequency $\omega$ is constant, then the magnitude of the velocity i.e., speed is constant but the velocity changes ...

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It seems to me that the key to this trick is to build up enough elastic energy in the rope - which requires you to "build tension" by riding the edge hard, as explained in this video. If you do the trick too close to point A, there is limited lateral motion needed to build tension - but as you take off, the force you are looking for will disappear as the ...

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In short, you are to think of the direction of the torque as pointing along the axis of the rotation it would induce in a rigid body initially at rest. But if the conception of torque as a vector out of the page seems artificial, that's because it is. Torque is not fundamentally a quantity that is a vector but a directed plane or directed area. Such an ...

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There are probably lots of duplicates, so my apologies, but for clarity I will try a short answer, as the graphic from Wikipedia is particularly illustrative. The torque is perpendicular, ( orthogonal) to the other two vectors, so it could be the line where the hinges are located, depending on the direction of the other two forces. From Wikipedia ...

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This used to be covered in textbooks. A fairly recent article about it is "Why do forces add vectorially? A forgotten controversy in the foundations of classical mechanics" by Marc Lange in the American Journal of Physics 79(4) 380-388 (2011); http://dx.doi.org/10.1119/1.3534836 And there are two common answers. In dynamics you can used Newton's Second ...

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I don't think that "satisfying the laws of vector addition" is necessary for something to be a vector, depending on what you mean by that. Take velocities in special relativity. They are vectors; the vector sum of velocities is well defined. But it's rarely useful. More commonly, when you have two velocities and need to combine them somehow, the combination ...

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1) Usually special relativity is taught at the end of the semester, after the class got through rotations, which they, on average, don't understand; torques, which are pseudo vectors and for this reason blow their minds up... By the time they get to relativity they are done! If you teach the same kind of course, it is unavoidable, that they will get confused....

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Your substitution is almost correct. You can have multiple forces each supplying their own acceleration vector, but it doesn't make a lot of sense to include the mass in the sum because it would mean you have to distribute the mass among all the constituent forces just to add them again. $$\sum _{i}{\vec F_i}=\vec F_{total}$$ $$\vec F_{total}=m\vec a_{total}=... 1 Proper time is the reference time all observers can agree on. Makes for calculating things much much easier. Since tau is attached to the particle the space differentials are zero Imagine we used some other reference time (which you are more than welcome to do). Then the spacetime line element would have dx, dy, and dz not equal to zero. Furthermore ... 1 4-vectors have invariant length as defined by$$\vec{v}\cdot \vec{v} = g_{ab}v^a v^b. Coordinate velocity $\frac{\mathrm{d}\vec{x}}{\mathrm{d}t}$ does not have this property. Proper velocity $\frac{\mathrm{d}\vec{x}}{\mathrm{d}\tau}$ does. Coordinate velocity is defined. It's not a 4-vector, so it's not that useful.

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