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The problem is that your analysis is all done from the perspective of the frame that measures the box to be moving at 0.1c -- in this frame, it's true that the time for the light to get from the source to the wall is different from the time for the light to get from the wall back to the source. But if these same events are measured by someone inside the box ...


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From physics we know that the net force on a mass moving in a circular path at constant speed always points towards the center and has magnitude $$|F_{net}| = m*a_{c}=\frac{mv^2}{r}$$ For an object moving in a vertical circle, when the object reaches the side the net force must be pointing towards the center (west). This implies that the force of weight is ...


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The reasoning is the same as the two-particle system: $$ E_i = E_{1i} + \frac{p_0^2}{2m_1} + E_i' + \frac{p_0^2}{2(M-m_1)}, $$ so that $$ E_i - E_{1i} - E_i' = \frac{p_0^2}{2}\left(\frac{1}{m_1} + \frac{1}{M-m_1}\right) = \frac{p_0^2}{2}\frac{M}{m_1(M-m_1)}. $$ Therefore $$ \frac{p_0^2}{2m_1} = \frac{M-m_1}{M}(E_i - E_{1i} - E_i'). $$


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I'm not sure what you're referring to by 'straight' acceleration. If that angle is the angle at which a force, $F$, is being applied, then the horizontal force is: $$ F_x=F\cdot cos(\theta)\\ F_y=F\cdot sin(\theta) $$ These will get the forces along the X and Y axis respectively. Assuming width, $w$, length, $l$, height, $h$, mass, $m$ and $\text{'some ...


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Here is one take on how to understand the relation between force and potential energy, which I think is the closest modern version of how it would have been seen originally. Let's take as the conditions for a force to be conservative $$\nabla \times \mathbf{F} = 0$$ and $\mathbf{F}$ is a function of position only (this leaves out the magnetic force on a ...


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We can start at the relationship: $W=-\Delta U$, which is work done by a conservative force. The math A (conservative) force $F$ will do this work on an object when doing a displacement $\Delta x$, and $W=F \Delta x$. In the general case, the force might be different at different points as the object is moved (the force of gravity is not constant along ...


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As the first question has received sufficient exposition, I would like to make a point with regard to the second one. First thing to understand is that integrability and non-linearity of a system are two different concepts. It is true though that all linear systems in classical mechanics (i.e those that are described by systems of linear equations, be them ...


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I'll write here a list of my personal favorites plus some commonly used books. I wouldn't be surprised if your teacher chose either one of the books below as a textbook: i) Mechanics, the first volume of the Landau course on Theoretical Physics; ii) Goldstein's book "Classical Mechanics"; iii) Taylor's book "Classical Mechanics"; iv) Marion's book ...


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The essential idea of a Poincaré map is to boil down the way you represent a dynamical system. For this, the system has to have certain properties, namely to return to some region in it’s state space from time to time. This is fulfilled if the dynamics is periodic, but it also works with chaotic dynamics. To give a simple example, instead of analysing the ...


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(1) In general, what is meant by non-linear system in classical mechanics? A linear system is described by a set of differential equations that are a linear combination of the dependent variable and its derivatives. Some examples of linear systems in classical mechanics: A damped harmonic oscillator, $$m \frac{d^2 x(t)}{dt^2} + c \frac{d x(t)}{dt} + k ...


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As requested by the OP, I gather my points in an answer. Linear systems Linear systems are systems which are linear with respect to a physical quantity. Mathematically, their evolution can be written as a (possibly differential) equation. Examples: A linear spring is linear in the sense that is produces a force proportional to the displacement it ...


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A linear system is one whose dynamics obeys linear differential equations, in contrast with those that are non-linear whose dynamics obeys non-linear differential equations. So if the dyanmics of the variable $x(t)$ obeys a a differential equation $$f\left(x(t),\frac{d}{dt}x(t),\dots,\frac{d^n}{dt^n}x(t),t\right)=0,$$ if $x_1(t)$ and $x_2(t)$ are differente ...


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This is a collection of resources related to complexity. Many physical systems can be represented with graphs: Quantum graphs Mechanical systems Complexity of a graph or a weighted graph is a notion established by algebraic geometry. http://arxiv.org/abs/0705.2284 (The weighted complexity and the determinant functions of graphs) Audrey Terras. Zeta ...


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I understand that you're not talking about entropy (which is not the same as complexity or disorder), but $\text{complexity}$ is a rather subjective term. I would assume that the relative complexity of a system (which is broad in and of itself, by the way) would have to do with statistical analysis of its parts, the energy required to achieve a certain ...


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Direct integration give you the time it takes to go between $v_1$ and $v_2$. $$ t = \int \limits_{v_1}^{v_2} \frac{1}{ \dot{v}(t)} \, {\rm d}v $$ So if $\dot{v}(t) = a_0 - \beta v^2$ then $$ t = \frac{1}{2 \sqrt{a_0 \beta}} \ln \left( \frac{a_0+\sqrt{a_0 \beta}(v_2-v_1)-\beta v_1 v_2 }{a_0-\sqrt{a_0 \beta}(v_2-v_1)-\beta v_1 v_2} \right) $$


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I do not know if it is "plausible" (I do not think so), however a trivial model can be constructed for the one-dimensional case with continuous forces depending on velocities, for $c>0$ constant: $$F_{12}(v_1,v_2) = c\sqrt{|v_1-v_2|} \quad\mbox{and}\quad F_{21}(v_1,v_2) = -c\sqrt{|v_1-v_2|}$$ The system of these two particles does not admit a unique ...


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The first Newton's laws of motion states that if the sum of all the external forces applied to the body cancel each other, the latter will be in continuous and uniform motion. Consequently, if you can think about all the external forces applied to the body you are considering, you just have to sum them (vectorially) and see if the result is the null vector ...


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The concept of an isolated system is an approximation. Like all approximations, it applies to some systems better than others. A billiard ball feels only a weak frictional force in the form of rolling resistance, whereas a wall typically has foundations buried in the ground which can provide a strong resistive force. Thus, the former system is closer to the ...


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It all depends on what you want to study. The billiard balls are generally viewed as an isolated system for the purposes of explaining elastic collisions, but you could as well introduce friction with the pool table, and the consider the system balls+table as the isolated one. This just means you have to consider the friction. In the case of the car hitting ...


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$\textbf{T}=-T\hat r$ : the direction of the centrifugal force is radially outward ($\hat r$ direction), but the string force is counteracting this force, so it is pointing radially inward ($- \hat r$) direction. The magnitude of the string force is $T$, so we have for our force vector $\textbf{T}=-T\hat r$ $\dot \theta=v/R$ is the definition of radial ...


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If you didn't have the string, the mass would have continued along a straight line. The string pulls the mass towards the center at every moment. Therefore, the force due to the string, also called tension, is along the radial direction. Further, since it is directed inwards, it is along $-\hat{r}$. There is another way to understand this point. If you ...


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I suspect the result depends on how fast you increase the pulling force. If you pull abruptly, I would expect the rope will tear at the weakened point that is the farthest from you, as the tension wave from the wall will first arrive at that point, whereas, if the pulling force is increased very slowly, I would guess the point of tear can be determined by ...


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I would say that not in the middle, either one of the other points. This is my argument: The string when pulled from both extremes (the same in both cases you ask) it will vibrate in its eigen-frequencies, shown in the picture below taken from Wikipedia. This oscillation will add strain to the points of rope. But as can be seen, while the point in the ...


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On the off chance there was any doubt about the numerics, I too wrote a code to follow these orbits. I use RK4 to take the first half timestep (a full timestep here is always $0.001$), and then perform the remaining integration via leapfrog. Below is the evolution of an asymmetric orbit. The potential is given by $v_0 = 0.9$, $q = 0.7$, $L = 0.05$. I set ...


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There is something you should be careful of regarding Liouville's theorem. If there are momentum-dependent forces, then Liouville's theorem changes because phase density is no longer incompressible. Suppose we define $f_{s}$ = $f_{s}(\mathbf{x},\mathbf{p},t)$ $\equiv$ the particle distribution function of species $s$, which is non-negative, contains a ...


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The center of mass moves as though the string is connected there, but the disk is going to start rotating - this affects the motion. You need to write down a couple of equations: one that describes the motion of the center of mass, and another the describes the change in distance between the disk and the mass at the end of the string. For this you must note ...


0

Let $v_1$ and $v_2$ be the particle's velocities in the center of mass coordinate system after the collision. by conservation of momentum and energy we have \begin{gather}\tag{1}m_1v_1+m_2v_2=0 \\ \tag{2}m_1v_1^2+m_2v_2^2=m_1(v-v_c)^2+m_2v_c^2 \end{gather} Isolating $v_1$ in $(1)$ and substituting in $(2)$: ...


0

$m_2$ will leave with the same magnitude of momentum but opposite direction. Now the assertion is made that in an elastic collision, $m_1$ and $m_2$ have the same speeds leaving the collision as entering it. In other words, the speed of $m_1$ is $v-v_c$ and the speed of $m_2$ is $v_c$ after the collision. In order to simplify things and to ...


0

The actual shock wave is quite short lived (I think it's visible for less than a second near 0:14 as a white sheet around the smoke/dust cloud) and doesn't propagate very far in this case. When the shock dissipates what's left is a pressure wave, the "bang" or sonic boom, and that propagates at the speed of sound. So I guess your video uses a reasonable ...


2

If I'm not mistaken, this is the picture you have in mind (in this diagram, the hole is just a little bit bigger than it needs to be): In the limit, the size of the hole that just works has a diameter $$d = 2 r \sin(60˚)$$ as should be obvious from looking at the picture.


0

You just have to make sure you hit the center of mass to give it a velocity you need, irrespective of the position of the pivot, whether it is at the end of the rod or 0.1m away from it. If you hit the center of mass of the rod and make it move with a velocity that is just enough for it to reach the vertical position above the pivot, that velocity will be ...


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This problem was first formulated by Leonhard Euler in 1744 WPlink: "That among all curves of the same length which not only pass through the points A and B, but are also tangent to given straight lines at these points, that curve be determined in which minimizes the value of \begin{equation} \int_A^B \frac{ds}{R^2} \end{equation} It is a problem of ...


2

Any force not applied through the center of mass of an object will impart both linear motion and torque. You can consider the force applied at the center of mass for the computation of the linear motion, and you can then consider the torque as being the moment applied by the force at the center of mass. So for both linear and rotational motion, you end up ...


0

You can sum them, without taking their respective positions into account. For example: Lets say you have a 2D body, with two forces applied to them, forming a couple. When you reference point is exactly between the points where the forces are applied, you experience $ \tau = Fa $ with $F$ as one force and $a$ as the distance between the applied forces. ...


1

The accepted answer is (subtly) wrong. While the projectile will indeed initially have an upward velocity, the shape it traces out is that of an ellipse, not a parabola -- remember, the earth is not flat.


0

You're missing perspective in your question. With respect to the Earth, the box will travel in a parabolic arc, just as if you've thrown the box into the air with the same initial velocity. While the acceleration is certainly downwards after you drop it, its velocity is still upwards (and forward) until gravity changes that (and gravity is quite weak, so it ...


0

All objects that are in free fall (which is to say, no other forces are acting on said objects) will have a height above ground that can be predicted as such, until said objects reach the ground: h=-(g/2)(t^2) + vt + c where: g represents the acceleration due to gravity. On Earth, at sea level, this value is approximately 9.80665 m/s^2. t represents the ...


4

Your question has great practical significance: it is the very essense of loft bombing/LABS. Since nuclear weapons can damage/destroy an attacking plane, it was deemed necessary to devise a method to release the bomb and to increase the separation between the air burst and the aircraft. I recommend watching this training video: ...


6

A simpler example is Kepler's laws of planetary motion. In a spherically-symmetric gravitational field, the planets follow elliptical orbits. The orbits certainly do not have the full spherical symmetry of the potential. They may be extremely lopsided. :-D


22

It will travel along a parabola (ignoring drag from the air here), initially with upward velocity, as you describe in your first scenario. You're correct that the only force acting on the box is its weight, but this means it will have downward acceleration immediately, not necessarily downward velocity. Eventually the downward acceleration will lead to ...


0

Maybe in a perfect world with no thermal or atmospheric disturbance (yes simulation) this could happen. Zeno holds true regarding infinite time, but for all practical purposes the bob would appear to get there in a reasonably short time. In the real world control systems engineers accomplish this feat all the time using feedback to overcome the small ...


0

Why rubber is incompressible material? While an incompressible material must have a Poisson's ratio of 1/2, that rubber has a Poisson's ratio close to 1/2 does not mean rubber is incompressible. In fact, there is no such thing as a truly incompressible material. That rubber has a Poisson's ratio of 1/2 merely means that rubber is in some sense a bit ...


2

MartinG's answer gives the math, but try to imagine what happens when you squeeze a material. Poisson's ratio tells you how much of the contraction in one direction elongates the other directions. So if you squish a square horizontally by a distance $\delta$, the top and bottom surfaces will both move away from the center of mass by a distance $\nu\delta$. ...


2

Incompressibility implies Poisson's ratio $\nu = 1/2$, and vice versa. A cylinder of length $L$ and radius $r$ has volume $$ V = \pi r^2 L $$ For an incompressible material (constant $V$) differentiation gives $$ dV = 2\pi rL dr + \pi r^2 dL = 0 \\\nu \equiv -\frac{L}{r}\frac{dr}{dL} = \frac{1}{2} $$ Thus a material with $\nu = 0.3$ cannot be ...


1

A quick diagram: Key dimension here is the distance $d$. The weight of the robot $W = M \cdot g$ is carried equally by both legs, so we have a force $F$ along the lines $AC$ such that $$F = \frac{W}{2 \cos\alpha}$$ This force results in a torque at point $B$ because $B$ is not on the line $AC$ - it is displaced by distance $d$, given by $$d = Y ...


0

It ias a matter of selecting coordinates. In the inclined plane case the x axis is parallel to the slope. In the banked track case the x acis is horizontal.


3

If you initially give to the bob a velocity $\sqrt{4rg}$, it will actually take an infinite time for the bob to reach the top! A little lesser velocity will cause the bob to stop earlier and come back toward the initial point, while a little greater one will take the bob over the top (the motion will continue, with increasing velocity, to the other side). ...


0

Since it is a rigid rod, you are probably right. If the rigid rod is replaced with a string, then your teacher would be right, as the velocity at the top must be non zero in order for the string to remain tight and not collapse before reaching the top. In real life scenarios, however, it is nearly impossible to maintain an unstable equilibrium.


3

I have Marion-Thornton 4th ed. around here somewhere. It is an older book and presents some material differently than we are used to in more modern books (for instance they even use the old imaginary time method when discussing some things in special relativity, which I personally dislike). However I agree with DanielSank, different pedagogy does not equal ...


0

The Lorentz force $$ \vec F = q \vec E + \frac{q}{c} \vec v \times \vec B $$ Doesn't obey Newton third law and is one of the fundamental forces of nature (unlike drag for example). The magnetic part of the force satisfy that two charged particles exert a magnetic force with equal magnitute to each other, but the direction is not along the line that join the ...



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