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The maximum acceleration experienced by your accelerometer (assuming that it does not automatically zero out the acceleration due to gravity) will be the same as the maximum acceleration experienced by any other object in the helicopter, as long as the helicopter attitude is maintained (no rotation). You would simply take the instantaneous acceleration ...


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I doubt this is your answer: what is the frequency $f$ doing, in the problem of a magnet sliding down an incline? What you should ask yourself is: how is changing the thickness of the aluminum foil going to modify the time down the incline? Increasing the thickness of the aluminum foil increases/decreases/leaves unaltered the foil's resistance (you pick ...


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Basically, it appears that, according to the opinions expressed in the answers in this page, there is a difference between acceleration by a distant gravity field, and acceleration, say, against a wall as we crash with our car. Let me restate the opinions expressed here in this form: we lock a person inside a windowless spaceship, then subject the ...


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You are correct. The concern over accelerations is with respect to a force applied on the surface of your body. Even with something like a uniform fluid to apply nearly even pressure across the body, your interior will always have density differences. Any density differences will create internal forces when the outside of the body is given a net force. A ...


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In short, you're correct: it's not the fall (uniform acceleration) that kills you, but the sudden stop at the bottom (large contact acceleration). But just for fun I'll point out that it's not clear what "uniform acceleration" even means. To operationally define (i.e., measure) acceleration you need an accelerometer. An example is a rigid sphere with a ...


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You will answer your question if you understand inertia: mass tend to oppose resistance to movement. Example: In a lift going upwards e.g., your feet are lifted while your head "wants" to remain at the same place (in a galilean reference frame). In the frame of the lift, everything happens as if a force ($m_\text{head}\ddot x$, roughly) was exerted on your ...


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I have a feeling, that you may be asking whether gravity is a force or an acceleration? In the confines of Newtonian mechanics, it's much better to talk about gravity in terms of acceleration, because point-like, free falling test masses responding to a large, gravitating body do not experience any actual forces acting on them. It's only a slight of hand, ...


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If you experience such a uniform force, e.g. when an astronaut on a space walk near the ISS (just earth's gravity), you don't experience any forces at all. That's freefall. Even with 10G, you'd experience a rapid freefall, but that is still harmless. It's the hitting the ground which kills you - that's not a uniform force.


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To make this easy, I will assume that all speeds are small compared to the speed of light $c$. The motion is in one dimension, so the trajectory of the space vehicle will be the following: $$x(t) = x_0 + v_0 t + \frac12 a t^2$$ We can easily set $x_0 = 0$ if the put the origin of the coordinate system at the start point where the space vehicle started. If ...


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Your friend is correct. The acceleration of the projectile is determined by the thrust its rocket motor can produce. If the acceleration of the projectile is $2g$ then the thrust of its motor would be $2mg$. But launching the projectile from your space vehicle can't increase it's thrust. You can increase its initial velocity, but once the projectile has ...


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A simple pendulum would be a good experiment to detect both non-inertial frames - rotating and linearly accelerating: If you know weight of the pendulum in inertial frame, in rotating frame, its weight would decrease because of radially outward centrifugal force acting on it. (If earth stops rotating, our weight would increase!) Think: what would happen if ...


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Yes, I can think of two ways to do this and there may be more. In a rotating frame the acceleration is a function of distance from the pivot. If B has sufficiently precise instruments the variation of acceleration with position will be detectable. However if B is confined to a very small space, or doesn't have precise enough instruments the variation of ...


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The friend rotating and experiencing the centrifugal force may observe several effects that his linearly accelerating friend doesn't: the acceleration at different points of the box is slightly different i.e. the apparent gravitational field is non-uniform there is the extra Coriolis force acting on objects that are moving relatively to the rotating frame ...


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You probably just didn't throw around enough weight to overcome friction. While technically, if you were on a completely frictionless skateboard, by simply moving the barbell around you would roll a bit (the center of mass of you and the barbell would remain stationary), in the real world you have to overcome some friction to start rolling. Friction is ...


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You are correct, acceleration due to movement cannot be separated from acceleration due to gravity. Inclinometers and accelerometers both measure acceleration the difference is in how they do so and what they are used for. Typically an accelerometer measures acceleration using small proof masses on springs. One mass for each axis. The masses are constrained ...


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From the path you need to find the radius of curvature $\rho$ at each point. This would be kind of noisy unless you have really precise data. Your best bet into input all the x and y points into cubic spline in order to get what the derivatives $x'$ and $y'$ are (in units of length per frame). In addition, you need to get the kinematic accelerations $x''$ ...


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If you don't care about the direction of the horizontal acceleration, the answer is yes. When the car is stationary (user acceleration very small, below some limit you define for the RMS of the three axes) you measure the vector $\vec g$ for the total acceleration - this is "down". Now during motion you find the user acceleration perpendicular to this ...


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In your case, lets $\Delta t = 0.03s $ By the method alemi explained, $$a_{x}(t)=\frac{x(t-\Delta t)-2x(t)+x(t+\Delta t)}{(\Delta t)^{2}}$$ and $$a_{y}(t)=\frac{y(t-\Delta t)-2y(t)+y(t+\Delta t)}{(\Delta t)^{2}}$$


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Yes. Provided you are only interested in the direction of the acceleration, and not it's magnitude. And further assuming your time samples are equally spaced, you can take the second derivative of the path and this will be proportional to the acceleration. A decent method in practice would be to use a second order central finite difference scheme wherein ...


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Suppose we have two particles [...] How should they instead accelerate to have the distance Lorentz-contract? A solution is arguably described in J. Franklin, "Rigid body motion in special relativity" (arxiv.org/abs/1105.3899). Using the notation from the question above: Given the trajectory, wrt. the (inertial) "laboratory frame", of "particle $2$" ...


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Using accelerometers as a basis for determining the position of an Earth-based object is a non-trivial exercise. Using a simple numerical quadrature technique such as trapezium almost certainly is not going to cut it. Here are some of the challenges: Accelerometers report acceleration in terms of the accelerometer case frame. It is invalid to compare, for ...


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I think you have made some mistake in taking the initial condition. $$v(t)=u(0)+\int_{0}^{t}a(t)dt$$ and $$s(t)=s(0)+\int_{0}^{t}v(t)dt$$. I think you have not used $u(0)$ and $s(0)$ properly.


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"Length contraction", is the measure of a umproper length of an object, in a inertial frame $O'$ which has a relative velocity to a inertial frame $O$, where the coordinates of the object are fixed in this frame $O$ More precisely, one measures a proper time $\Delta \tau'$ in $O'$ (so at $x'=0$), which corresponds to the synchronisation with the beginning ...


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To make the example (and the question) clear, the particles should be on different frames, where when the 2 frames are at rest and coincide, the particles have the given separation. Then when one frame moves relatively (or accelerates) to the other with a given velocity (what velocity exactly?) the Lorentz-contraction appears but only with respect to the ...


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A simple example, particle moving in one dimension, say $x$: A particle has a (negative) acceleration equal to $-2m/s^2$ (note it is constant acceleration, yet negative, i.e opposite to direction of motion or to direction of cummulant force applied to particle) Another particle has a (decreasing) acceleration equal to $1/(kt)^2 m/s^2$ (note it is positive ...


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While aliasing is a real concern for anything where sampling is involved1, it is really unlikely for this to cause a big zero-peak in frequency space, for an accelerometer on an animal (where it's unlikely that there's an oscillation precisely in sync with the sampling clock). Of course it's possible that a lot of the measured signal had nothing to do with ...


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Without knowing more about the specific situation, the first likely culprit is aliasing. This occurs when you are sampling at a frequency $\nu_s$ that is too low compared to the largest frequency that is sizeably present in your timeseries. More specifically, Nyquist's theorem guarantees you that if the highest frequency present is $\nu_M$, then a sampling ...


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The question is phrased in terms of dynamical concepts like force and mass, but there's a more fundamental kinematical answer that trumps these issues. If an object is moving with speed $u$, and you then apply a boost $v$, the object's new speed is not $u+v$ but rather $(u+v)/(1+uv/c^2)$. This is always less than $c$. Therefore it's not possible to ...


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In special relativity, you are better off thinking of things as four-vectors, rather than three-vectors. In that case, you generalize momentum to 4-momentum ${}^{1}$(and you take time derivatives with respect to the clock of the spaceship). Then, you have $\bf F = \frac{d{\bf p}}{dt}$ Since momentum is given by $p = \frac{mv}{\sqrt{1-v^{2}/c^{2}}}$, and ...


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When a particle moves along a prescribed path, with tangent vector $\hat{e}(t)$ and normal vector $\hat{n}(t)$ then the velocity and acceleration vectors are decomposed as such: $$ \vec{v} = v(t) \hat{e}(t) $$ $$ \vec{a} = \dot{v}(t) \hat{e}(t) + \frac{v(t)^2}{\rho(t)} \hat{n}(t) $$ which is interpreted as The magnitude of the velocity vector is the ...


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The problem with supercavitation is that it occurs at a pressure drop, not an increase. The thing that kills you on impact is the pressure in front of you - the water that cannot get out of the way fast enough. Skin drag on the body happens later - when more of the body is already submerged. Most likely you are dead by then... Let me try (with my rusty ...


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In the context of linear motion (as BMS correctly points out in a comment of a different answer), the magnitude of acceleration is a measure of how much speed you are gaining per second. The difference with the acceleration vector is that the vector form also encapsulates the direction in which this gain in speed is happening. So as an example an ...


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Does it need to "slow down" to the new speed of light? Or does a new photon get generated? One has to keep clear in this case the difference between a photon and an electromagnetic wave. An individual photon is an elementary particle, its wavefunction is given by the quantum mechanical form of Maxwell's equations and the square of this wave function ...


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You could imagine this. Say you have a railgun on the moon (to avoid atmosphere) that fires a projectile at the spacecraft. The projectile bounces elastically off the spacecraft, transferring twice its momentum. If you are happy to accelerate in exactly the correct direction, it could return to the moon to be fired again. The momentum is transferred to ...


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The issue here is that the initial momentum needs to come from somewhere. I suppose, if we could track the momentum of various things in space, we could "hitch a ride" from it. Like netting a comet or taking advantage of space debris striking the craft. That's a properly analogous situation to your basketball/skateboard system. However, you seem to be ...



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