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5

The problem here is that you've neglected the effect of firing the bullet on the plane itself. It turns out we can account for the bullet's extra energy by examining the energy lost by the plane. To resolve this apparent contradiction, we need to examine the problem in terms of conservation of momentum: In the second case, where the bullet is fired from a ...


5

Yes, $B$ does rotate when seen from a static frame of coordinates outside the disk: As to velocities and accelerations, see the article in Wikipedia. It says, $$\vec {v_s} = \vec {v_r} + \vec {\Omega} \times \vec r,$$ where $v_s$ is the velocity in the static frame and $v_r$ in the rotating. If you apply this formula for both points $A$ and $B$, their ...


5

If I understand your question correctly you are saying that: $$ v = r\omega $$ and therefore: $$\begin{align} v_A &= r\omega \\ v_B &= \tfrac{1}{2}r\omega \\ v_A &= 2v_B \end{align} $$ but how can $A$ and $B$ have different velocities when they are both attached to the disk so the separation between is fixed? The answer is that $A$ and ...


3

Things will be bigger/smaller/slower with speed as $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$. Thus, as $v \rightarrow c$, Mass (or energy) goes to infinite, Time goes to zero. But it is only a limit, which is unreachable (at least in the special relativity), because of (1).


2

No, for several reasons. First, the idea of time "slowing down" is a little bit of a misnomer. If you were traveling at relativistic speeds, you would not perceive the passage of time any differently than you do right now. It's only when you compare your clocks to an observer in another reference frame (let's me, sitting in my living room, at rest with ...


2

If you are looking at the Earth's rotation from the outside, the direction of rotation would suggest a z-axis that points upwards to make a right-handed coordinate frame. A possible explanation for the reversed z-axis could be, that when you are stationary on Earth, you see the sky move in the opposite direction; forming a right-handed coordinate frame ...


2

I don't know what papers you are referring to, but in the normal definition of the Earth-Centered Inertial reference frame, the Z-axis is defined by the Earth's spin, and therefore points up through the North Pole. Unless you are standing on the North or South Pole, the Z-axis won't be vertical either up or down. There are several other coordinate systems ...


1

If we assume that mechanical energy (K+U) is conserved in both the earth frame and the initially co-moving, constant velocity frame, then it's not the differences in velocities which are the same; it's the differences in the squares of the velocities which are the same. $$\frac{1}{2}v_{1e}^2=\frac{1}{2}v_{0e}^2+2gh$$ and ...


1

A rotating reference frame is an accelerated reference frame so $A$ and $B$ are at rest in an accelerated reference frame. Assume an inertial reference frame $S_0$ and another reference frame $S$, with a common origin and rotating with respect to $S_0$. Let the (constant) angular velocity vector of $S$ be $\mathbf \Omega$. Then, the time rate of change of ...


1

Einstein's equivalence principal states that an accelerated reference point is indecipherable from a reference frame in a gravitational field, so an accelerated reference frame will act in the same way that a gravitational field with the same acceleration would act. As for if all reference points are equally valid, the answer is generally "yes" with some ...


1

But then someone talked to me about Principle of Equivalence and not possibly being able to identify what is proper acceleration and what is coordinate acceleration with an accelerometer. Is it true ? That's not true. By definition, an ideal accelerometer measures proper acceleration. It appears you (and possibly the acquaintance who talked to you) are ...


1

It appears that the confusion both in your question and in the question linked stem from a confusion of reference frames, leading to an apparent difference in outcomes across multiple tests of the "same" process. The original question regards a cannon that fires a 2kg shell at 1km/s muzzle velocity, i.e. imparts 1MJ energy to the projectile. The question ...


1

The thing about Lagrange points is there is always two ways of viewing the system. Method 1: The Inertial Frame. Position yourself "above" the Solar System, looking down at the objects moving in ellipses. Here the force of gravity is the only force. Earth doesn't crash into the Sun because the constant inward acceleration merely bends its path into an ...


1

An object at L3 orbits the Sun through exactly the same mechanisms that the Earth does: it feels the Sun's gravitational attraction, and this is exactly the centripetal force it needs to perform a uniform circular motion about the Sun. The slight problem, as you point out, is that it also needs to contend with the Earth's gravity, which at L3 also pulls it ...



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