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## Hot answers tagged reference-frames

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There are a few ways to justify it. First, you could look at the motion of the object as it rotates. In 2D, it turns out that all such motion can be decomposed into motion of the CM, and rotation about the CM. Therefore, rotating around the CM itself is the only way to guarantee the CM doesn't move, and hence is the least energetically costly. Another way ...

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Fictitious forces do not exist in inertial frames. Fictitious forces result from force-fitting Newtonian mechanics to non-inertial frames.

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I'll give you a couple of ways to think about this. First, geometrically, the circle you are thinking about drawing should contain the entire circular path of the car. If we're assuming that the car is remaining at a constant "elevation" on the banked surface, then the center of that circle has to be at the same elevation: otherwise you'd be drawing a cone ...

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Here is one way of looking at it via a velocity-dependent potential.$^1$ The Coriolis potential is $$\tag{1} U_{\rm cor} ~=~ -m({\bf v} \times {\bf \Omega})\cdot{\bf r} ~=~-{\bf v}\cdot ({\bf \Omega}\times{\bf r} ),$$ cf. Ref. 1. The factor $2$ comes from two different terms in the corresponding force formula $$\tag{2} {\bf F}~=~\frac{\mathrm d}{\mathrm ... 3 This is actually a famous theorem known as the Einstein Equivalence Postulate (sometimes Equivalence Principle). It's true that since Earth is spinning, acceleration in a spacecraft isn't quite the same situation we experience daily, but in general, yes, gravity is indistinguishable from uniform acceleration. Specifically, if you are in a box with no windows ... 2 You don't have to, but it makes the equations easier to deal with because you don't have to account for the moment of acceleration terms. See the 2nd part this this answer about deriving Newton's laws on an abitrary point not the center of mass. So finally the equations of motion of a rigid body, as described by a frame A not on the center of gravity C ... 2 Let the body rotate about the z-axis, then by the definition of angular momentum$$\vec{L}=\vec{\omega} I_z.$$where \omega is the angular velocity about the z-axis. So we could take the parallel axis theorem and multiply it by \omega:$$\vec{\omega}I_{z}=\vec{\omega}I_{cm}+\vec{\omega}ma^2$$Now ponder the terms in it. If I understand the ... 2 A rotating reference frame is not an inertial reference frame: In the rotating frame, objects accelerate even though there are no forces acting on them. In your example, you can in fact determine easily whether you are rotating or the universe is rotating around you. In the first case there is artificial gravity on the ship, and in the second case there is ... 2 It is possible as soon as one is sure to be very distant from every body (gravitational source) in the universe. This is because all inertial forces behave as gravitational forces. If one is confined to stay in a closed room and observes the motion of bodies therein, he/she cannot decide whether the observed accelerated motion is due to a gravitational field ... 2 Here is the question: what do we mean by "kinetic energy"? I assert that we mean "the amount of energy or 'work' I would need to make this object come to rest in my frame of reference". Consider a ball with mass m in free space$${\huge\circ}$$In it's inertial frame of reference it's velocity is zero so its kinetic energy is also zero. But all inertial ... 2 There isn't a "centripetal force" vector. As the car goes around the banked curve, the normal force on the car increases relative to what it would be on an un-banked straight road. The vertical component of the normal force supports the weight of the car, and the horizontal component of the normal force provides the centripetal force necessary to cause the ... 2 The first figure shows the parallel and normal components of a vector \mathbf{r} relatively to a direction \mathbf{n}. Based of this, the second figure shows the centripetal acceleration. In case of plane circular motion \omega R = v. 1 How is this resolved without implying a preferred reference frame? That you asked this implies you are thinking that reference frames have universal extent. While that is true in special relativity, it is not the case in general relativity. Reference frames are local in general relativity. That said, there is a frame in which cosmologists prefer to ... 1 That the moment of inertia about an axis passing through the CM is minimized, with respect to any other parallel axes, is a consequence of the quadratic (squared) dependence of the moment of inertia on distance. In other words, the {r^2} term in {I=mr^2} makes it so that masses at farther distances are preferentially weighted in their contribution to the ... 1 The \Omega \times terms signify change due to the moving orientation of the item. So the velocity vector when attached to a moving body will cause a \Omega \times v term in the acceleration vector. 1 The 'swinging in a circle' creates a centripetal force that accelerates all water away from the center of the circle, with the force increasing with the distance from the center. The result will be that the water will orient so it is maximally distant from the center of the rotation. Hindered by the bucket from simply flying away, the resulting surface will ... 1 I think sections 4.1.2 and 4.1.3 of this lecture on dynamics explains it by looking at each component separately. Since \frac{{\rm d}}{{\rm d}t} \sin \theta = \dot{\theta} \cos\theta and \frac{{\rm d}}{{\rm d}t} \cos \theta = -\dot{\theta} \sin\theta the components of {\rm d}u are perpendicular to u. Our first step is to choose cartesian axes, ... 1 But why must d\hat{u} be orthogonal to \vec{\Omega} too (i.e. be tangential to a circle orthogonal to \vec{\Omega})? To get such a precession there must be a clockwise torque in the plane of the screen acting on the system which means that the torque vector must be pointing into the screen. That torque produces a change in the angular momentum ... 1 Gravitational fields aren't homogeneous. Here on the Earth, a clock on the floor runs more slowly than a clock on the table, and we have clocks precise enough to measure such small differences due to the gravitational gradient. But doesn't a clock in an accelerating spaceship run at the same rate no matter where in the ship you put it? See page ... 1 Take a free particle moving on a plane in polar coordinates$$ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} r \cos \theta \\ r \sin \theta \end{pmatrix}$$The velocity is found from the chain rule, with clear separation for radial and tangential components:$$\begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} = \begin{vmatrix} \cos \theta & ...

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Correct; in general the speed of light is constant only as measured by local inertial observers. As an extreme example, consider a photon emitted from a galaxy far, far away, in our direction. Although it moves away from the galaxy in the direction of the Milky Way, the expansion of space makes it increase its distance from us. Eventually, however, it will ...

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Actually, it's NOT true that in SR the speed of light in vacuum is the same for all observers, regardless of the motion of the light source. This is true only for inertial observers. The same applies for GR, in which the generalization is a "freely falling frame" (a local inertial frame without effects of gravity). A good reference: Speed of Light

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Let two orthonormal systems $Oxyz$, $O'x'y'z'$ with a general motion (translational plus rotational) between each other and a point particle $\rm P$, see Figure. Symbol Conventions : 1.The vectors for position $\mathbf{R}$, velocity $\mathbf{U}$ and acceleration $\mathbf{A}$ of a particle with respect to $Oxyz$ expressed by coordinates of this same ...

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The most straightforward observation to show that the Earth moves is stellar parallax. If you take photographs of a groups of stars over a period of six months (half an orbit), some of the stars will seem to shift in position compared to the others. These stars are much closer to Earth and so seem to move more. This is similar to how, when you are riding in ...

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