New answers tagged

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By "acceleration is absolute" we mean that in reference frames which move with different accelerations the laws of physics (for instance, Maxwell's equations) will look different. In other words, you can determine your acceleration by conducting experiments locally, without looking at the outside world. If you are in a car, and the car is slowing ...


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There is no way to know if we are moving at constant speed in a straight line in space or at rest in space because there is no such thing as "absolute" motion. An object in space (or anywhere, for that matter) can be moving at constant speed in a straight line with respect to one inertial (non accelerating) observer and be at rest with respect to ...


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If I throw a ball vertically inside a moving train, there will be horizontal movement if the train accelerates/decelerates (ie is not an IRF) and no horizontal movement if it does not (ie is an IRF). Correct, in the reference frame of the thrower in the train. Surely, the ball should feel the acceleration of the train just as much as the person on the ...


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Surely, the ball should feel the acceleration of the train just as much as the person on the train, in which case the horizontal location of the ball relative to the thrower should be the remain the same? Yes, in the reference frame of the train both the ball and the thrower will be subject to the same fictitious force arising from the use of the non-...


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When you are on an accelerating train, you feel the acceleration because you are in direct contact with the train itself. As the train accelerates forward, the train's floor pushes on your shoes, making you accelerate along with the train. Both you and the train are rigid bodies in direct contact, with a high coefficient of friction between them, so any ...


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If you are talking about the inter atomic interactions, it is what makes rigid bodies (solids) different from fluids (liquids and gases) and that's why center of mass becomes very important. That is the reason why rigid bodies rotate when you apply force to someplace other than center of mass, because it tries to maintain it's rigidity.


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Whenever there is a normal force acting on the person in the elevator, whether the elevator is going up or going down, it is the force acting on the person due to the elevator. Hope it helps.


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When you apply a force to an object it can gain rotational motion or translation motion or both of them. The point you have to consider is the equilibrium of all the forces on the object. You don't have to think of it in atomic state (because all those forces are internal and they cancel to each other). If you don't have any opposite forces to the force that ...


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A. No, the shortest possible distance would go through the Earth. Humans don't inhabit two-dimensional positively-curved space. We just happen to live on a ball of rock, so our map of the world looks like that. And even "from the perspective of the passenger", the trip is not a straight line. The velocity is not uniform because they would be able ...


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why can't object translate linearly, when I apply force away from center of mass? Consider an object of mass $M$, which is pushed by the force $\vec F_\text{push}$. By Newton's second law, this object will accelerate with acceleration $$\vec a=\frac{\vec F_\text{push}}M.\tag1$$ Let's now look inside this object. Namely, suppose it's made of two balls each ...


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We are used to think of translation as a natural movement. This is the concept of inertia. Force is necessary to modify the velocity, (or more precisely the linear momentum), not just to put in movement something that is at rest. The idea is similar for rotation. It is a natural movement for a solid body, and torque is necessary to modify the state of ...


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Lets say you have coordinate system that rotate about the z-axis with the angle $~\Omega\,\tau~$ where $\Omega~$ is constant, thus the position vector relative to inertial system is $$\vec R=\left[ \begin {array}{ccc} \cos \left( \Omega\,\tau \right) &-\sin \left( \Omega\,\tau \right) &0\\ \sin \left( \Omega \,\tau \right) &\cos \left( \Omega\,\...


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A reference frame is simply a co-ordinate system - a self-consistent rule for assigning a unique set of four co-ordinates to every event in the universe. If Newton’s first law applies when co-ordinates of events are expressed in that reference frame (without needing to introduce artificial forces such as centrifugal force) then we call it an inertial ...


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Imagine one atom that is in a state of equilibrium with some other number of atoms. Presumably for this question in a solid state of matter. I think it boils down to: it is impossible for this atom to exert a net force on any other atom in the system that is not colinear with the line between the two atoms. EG, if you've got two atoms $a1$ and $a2$ in a ...


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It might help to answer your question by analogy. In short, the Euler-Lagrange equations can be likened to the general equation $\nabla f|_M = 0$ used to find (constrained) extrema, while Newton's laws are analogous to a precise form of $\nabla f|_M = 0$, in a specially chosen coordinate system (e.g. the coordinate system where $\nabla f = 0$ has the ...


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Yes when it is accelerating Upwards against gravity but If the elevator is moving upward at a constant speed then it’s no different than standing on the ground.


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"Homogeneity" in time means being invariant under time translations. Special relativity is invariant under the Poincaré group, and so in particular time translations.


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so does that mean that the person falling down is following the path of geodesic Yes as accleration is deviation from geodesic and falling person isn't acclerating and are the people on the building deviating from path of geodesics since they are acclerating? When we say, the falling person isn't accelerating, it means something specific. After all, he is ...


0

Imagine a rod $AB$ pushed suddenly at $B$ by a force $F_1$, as in the top diagram below End $A$ doesn't move immediately (due to the finite speed of sound in the rod), but end $B$ moves a small distance in the direction of $F_1$, as shown in the second diagram. This causes the rod to stretch slightly, the length of the rod in the second diagram is longer ...


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Apparent forces are those that show up as a result of an accelerating reference frame, i.e., they are not forces over the system, but an effect of being in a reference system over which a force IS being applied. Therefore, if you accelerate in an accelerating reference frame, you would start already with some "virtual" acceleration.


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In a rotating frame, the Hamiltonian picks up an additional term proportional to $\vec\omega\cdot \vec L$. Thus typically \begin{align} \hat H=\frac{p^2}{2m}+V(r)-\vec\omega(t) \cdot \vec L\, , \end{align} and the extra term completely based on classical mechanics. This type of additional non-inertial term comes up quite a bit in the study of nuclear and ...


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The equation that is used in your linked answer $$ t^\prime = \gamma \left(t - \frac{v}{c^2}x\right) $$ comes from the Lorentz-transformations for the time variable. Deriviations of the Lorentz-transformations can be found on this Wikipedia page, especially section 7.


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The "time dilation" formula you are thinking of ($t'=\gamma t$ ?) accounts for the time interval between two events (your flashes of light) which occur at the same location in the rest frame of the events. If the events occur at different locations, you must use the full Lorentz transformation of the event timing. That equation is the Lorentz ...


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If you envision the rotating blender to be in outer space will be the same as removing the downward gravity on it. For a blender to operate properly, gravity is needed to pull all the stuff inside towards the blending blades. Only stuff at the bottom will be blended. Not too well though because it ends up at the wall of the container (out of reach of the ...


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For a spherical mass like the sun, the Schwarzschild metric applies: See this Wikipedia article. Here $\tau$ stands for the proper time, i.e. the time as measured by a clock following the Earth. This time is independent of coordinates so no matter from where you look, the metric will always be the same. Which means that also the curvature is independent ...


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One of the first things a person must get rid of when coming to GR from special relativity is the existence of a global reference frame for a particular observer. In general relativity you cannot ask a question "what is the state of a distant object relative to me?", because the question is ill-defined. You can talk about what you see and what you ...


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You need to take the Schwarzschild metric (in the Sun's frame): $$ g_{tt} = -\left(1-\frac{r_s} r\right)$$ $$ g_{rr} = \left(1-\frac{r_s} r\right)^{-1}$$ $$ g_{\theta\theta} = 1$$ $$ g_{\phi\phi} =\sin^2{\theta}$$ and transform it into the distant observer's frame, and see if the geodesic equation: $$ \frac{d^2x^{\mu}}{ds^2}+\Gamma^{\mu}_{\alpha\beta}\frac{...


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I assume that by "first postulate" you are referring to The laws of physics are invariant (that is, identical) in all inertial frames of reference (that is, frames of reference with no acceleration). This remains true in general relativity, but needs some modification. The difference between special relativity and general relativity is that, in ...


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Yes, you are correct. A sees B’s clocks run slow and B sees A’s clocks run fast. This is different than the usual case where each sees the other’s clocks running slow. In that usual case the situation is symmetric because both A and B are inertial. But in this case the two observers are not symmetric because A is inertial and B is non-inertial. In B's ...


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Envision a body of particles that are all held together by some force. Say a spherical body of tiny (as compared to the whole body) plastic balls that are glued together. On the surface, I apply a force parallel to the surface of this body. You can imagine differently formed bodies, and apply more forces in more directions, but this combination is the ...


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You can do this, although it may actually make it harder to develop such a ship rather than easier. The key equation is the equation for centripetal acceleration: $a_c=r\omega^2$, where $r$ is the radius of the rotation and $\omega$ is the angular velocity (how fast it is spinning). There's a few variants of that (some use velocity rather than angular ...


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I can't skip the mathematics bit of the puzzle. Rotation and angular momentum are fundamental properties of a "rigid body." This is the superficial answer, but its an important piece of the puzzle. It lets us reframe your question as to why a rigid body is such a good model of real life objects and their behaviors. This is fun because its so ...


0

Can you have in inside cylinder spinning at a faster rate than $x_1$? Yes, you can. Any sort of gear train connecting the two cylinders can enforce higher or lower (or reverse) speeds as seen from an outside observer. Will this affect the artificial gravity? I doubt it. Do the math model for the $g$ forces due to spinning you you find $g = R_1 x_1^2$ and in ...


0

The solution requires recognizing that: In the general case, force is not exactly equal to mass times acceleration. In the rest frame of the loop, the magnetic Lorentz force acting on the loop is zero, while the electric field is relativistically transformed. Due to the time dilation of the relatively moving loop as viewed in the rest frame of the external ...


2

A personal anecdote might help you. I went out with some friends on a small boat. When we returned, the boat was only about a foot from the quayside, so I decided to jump ashore. I was forgetting Newton's third law. When I went to leap over the gap, instead of my leg just propelling me forward, it propelled the boat away from the quayside too. I ended up in ...


2

Okay , so this a very common phenomena. There reason behind this , since there is no external force the spaceship and people as a system , the net position of center of mass remains same. To understand why does ship moves backward , you should analyze the problem by making a free body diagram of the man and the ship and try to find how / why does the people ...


0

Yes, if the space twin goes fast enough and goes far away enough before returning , then his twin brother on earth would be dead when he comes back


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So you've got a set of particles with trajectories $\vec{r}_i(t)$ and velocities $\vec{v}_i(t)$. Internal forces are the ones acting between the particles - lets denote $\vec{F}_{ij}$ the force acting on particle $i$ by particle $j$ . Crucially, they satisfy the 3rd law $\vec{F}_{ij} = - \vec{F}_{ji}$ as a result their sum over $i$ and $j$ should be zero: $$\...


1

I did not do it via Lagangians, but I have solved this problem as a homework question using "$F=ma$" methods. For what it is worth, here is my solution. It gives us a hint as to what the problem is with the Lagrangian approach. A ball of mass $m$, radius $a$ and moment of inertia $I$ rolls without slipping on a flat turntable that lies in the $...


1

Newton Euler Equations \begin{align*} & \underbrace{\left[ \begin {array}{cccc} m&0&0&0\\ 0&m&0&0 \\ 0&0&{\it I_s}&0\\ 0&0&0&{\it I_s} \end {array} \right]}_{\boldsymbol A} \begin{bmatrix} \ddot{x} \\ \ddot{y} \\ \dot{\omega_x} \\ \dot{\omega_y} \\ \end{bmatrix}= \underbrace{\begin{bmatrix} ...


-1

The derivation of the Lagrange equations from Newtonian mechanics relies on zero work done by the constraints (I couldn't find an English reference, but here is a Russian ones: Zhuravlev, Basics of theoretical mechanics, §24). In your case the constraints actually do work, since the table is rotating, and the power generated by the table is $W=\bar{v}_{table}...


0

VacuuM gave an excellent answer but I will say some more words to clarify some ideas. It is a postulate of physics that inertial observers should measures the same thing that is, if an inertial observer $A$ set up a experiment and an observer $B$ set up an identical experiment than their result should be the same. In special relativity we have global ...


1

Consider disc of uniform mass density for simplicity. Factors on which rotation will depend on frictionless surface: Torque about Centre of Mass [COM] Direction of Force There can be $3$ possible cases to apply the Force: Tangentially, Normally & Along a chord of disc Observe the situations shown in the image carefully. Case I: Tangentially ...


0

One of my teacher told me that without friction, the object will not rotate but another teacher told me that it will have both motion. Maybe the first teacher was talking about the friction between the rim and whatever is applying the tangential force to its rim. If there is no friction on its rim , then the tangential force will just "slide" ...


1

If the axis of the force does not pass through the center of gravity, then it will exert a torque and cause a rotational movement.


0

Note that in both cases the force acts for the same time. Per the work energy theorem: the net work done on an object equals its change in kinetic energy, or for a constant force acting through a distance $d$ $$W_{net}=F_{net}d=\frac{1}{2}mv_{f}^{2}-\frac{1}{2}mv_{i}^{2}$$ So if you want to compare the work done on the two objects, you must base it on an ...


0

Yes, the work will be larger if the object is already moving in the direction of the force. The mechanical power (work per time) is given as: $$P = \frac{\Delta W}{\Delta t} = \frac{Fd}{\Delta t} = F v$$ where $v = \frac{d}{\Delta t}$ is the velocity. So, yes, it costs extra energy to apply the same force to an already moving object than if it was at rest! ...


2

Here is one approach: The Lagrangian for a point particle in an accelerated reference frame is$^1$ $$ L ~=~\frac{1}{2}m\vec{v}^2+m\vec{v}\cdot (\vec{\Omega} \times \vec{r})-V(\vec{r}),$$ where $$ V(\vec{r})~=~m\vec{A}\cdot \vec{r} -\frac{m}{2} (\vec{\Omega} \times \vec{r})^2,$$ cf. my Phys.SE answer here. So the Hamiltonian becomes$^1$ $$H~=~ \frac{1}{2m}(\...


0

A better way to build intuition about angular momentum is not to think directly in terms of rotation but in terms of surface area. Indeed, it is quite simple to see that ${\bf L} \Delta t = {\bf r}\times m {\bf v} \Delta t$ is directly proportional to the area spanned by the position vector ${\bf r}$ in a short time $\Delta t$. Here, short time means a $\...


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