# Tag Info

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I would attack this problem this way: Locate the center of gravity of the upper hemisphere; specifically determine how far up the central radius the CofG lies Assume the upper sphere rotates ("wobbles") through an angle $\theta$; find the new point of contact between the upper and lower hemispheres in terms of $\theta$ From these new points of contacts, ...

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It will turn out that there is no gravitational field inside a hollow sphere. I'm not going to do the integral to prove this for you, but here are some tips. $dV = r^2 sin \theta dr d\theta d\phi$, not with an $R$. You are integrating over the entire volume of the sphere. You are approaching the integral in the wrong way by converting to (x,y,z) space, ...

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There are two forces acting on the mass (omitting any frictional forces). The first one is the centrifugal force. The second one is the gravitational force. At the moment the string slackens, the sum of the forces in the direction of the string will be zero. In a formula: m*v^2/a - m * g / cos(pi/3) = 0 From here you can work out the relation between v^2 ...

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The key is the statement that the string slackens. This means that the entirety of the centripetal force needed to make the bend of radius $a$ comes from the component of the weight which is along $a$. From kinematics, you know that the centripetal acceleration at the top point is $$a_c=\frac{v^2}{a},$$ and you know that the component of gravity which is ...

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I think this problem is trying to get you to use generalized coordinates. If you make your coordinates the center of mass and the seperation, you will get decoupled equations for the free particle and the harmonic oscillator respectively. Being agnostic about the coordinates is the secret superpower of the Lagrangean and Hamiltonian approaches.

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Firstly we divide the work up into internal and external components: $$\sum _i\int ^{r_2}_{r_1}\vec{F_i}^{(e)}\cdot d\vec s _i+\sum _{i,j}\int ^{r_2}_{r_1}\vec {F_{ij}}\cdot d\vec s _i$$ The factor of a half comes in since we are summing over both $i$ and $j$, and (I think) we can assume that $V_{ij}=V_{ji}$ (why?). They ...

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There's a sign error in your equations of motion. The Lagrangian of the system will be $$L=T-U= \frac{m}{2} \left( \dot{x_1}^2 + \dot{x_2}^2 \right)-\frac{k}{2} \left( L + x_1 - x_2 \right)^2$$ So the equation of motion for $x_1$ is: $$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}_1}-\frac{\partial L}{\partial x_1}=0 \\ m\ddot{x}_1+k( L + x_1 - x_2 )=0 \\ ... 3 I would say pressure is better defined by$$ \vec{F} = P \vec{A}. $$Yes, we are defining a quantity without having it all alone on the left-hand side. And yes, area is a vector. And as you guessed trying to divide one vector by another leads to trouble, so we won't do it. Let me explain where this comes from and what it is shorthand for. In continuum ... 0 There are different mathematical way to define pressure (all equivalent), but perhaps the most common one is using the component of the force normal to the surface. That is why in the definition you are using you are actually dividing scalars. For this way to define pressure, you see http://en.wikipedia.org/wiki/Pressure. You can also consider area as a ... 1 Yes, area is a vector, which is the normal to the surface. (\vec{A}=A\vec{n}) \vec{F} = -P\vec{A} In this case P is simply the proportionality constant to the vectors F and A, which also mean that F and A has to be in the same direction (F is the normal force and not shear forces). The negative sign accounts for the fact that the force and normal ... 1 REMARK. Perhaps I wrongly interpreted the question. I interpreted it as if were referred to the total volume of phase space. The answer is negative if the question regards general changes in time of topology of the total space of phases and if you do not impose any generic restriction on the topology of the spaces, like compactness (see the final ... 5 Here's something I believe is a simple proof. Unfortunately it uses a little bit of cohomology. Consider the canonical 2-form in extended phase space T^*M \times \mathbb{R}$$\omega = \sum_{i=1}^N dq_i \wedge dp_i - dH(\vec{q},\vec{p},t) \wedge dt ,$$where N = dim(M). A function f: M \to M is said to be a canonical transformation iff f^* \omega = ... 2 By the main theorem of connectedness in general topology, continuous maps preserve connectedness. Time evolution of Hamiltonian systems preserves connectedness because it is continuous. I think it is independent of from Liouville's theorem, it just requires the proving Hamiltonian time evolution is continuous. This is just a formal way of restating ... 0 The problem is that your analysis is all done from the perspective of the frame that measures the box to be moving at 0.1c -- in this frame, it's true that the time for the light to get from the source to the wall is different from the time for the light to get from the wall back to the source. But if these same events are measured by someone inside the box ... 0 From physics we know that the net force on a mass moving in a circular path at constant speed always points towards the center and has magnitude$$|F_{net}| = m*a_{c}=\frac{mv^2}{r}$$For an object moving in a vertical circle, when the object reaches the side the net force must be pointing towards the center (west). This implies that the force of weight is ... 1 The reasoning is the same as the two-particle system:$$ 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'). $$0 I'm not sure what you're referring to by 'straight' acceleration. If that angle is the angle at which a force, F, is being applied, then the horizontal force is:$$ F_x=F\cdot cos(\theta)\\ F_y=F\cdot sin(\theta) $$These will get the forces along the X and Y axis respectively. Assuming width, w, length, l, height, h, mass, m and \text{'some ... 0 Here is one take on how to understand the relation between force and potential energy, which I think is the closest modern version of how it would have been seen originally. Let's take as the conditions for a force to be conservative$$\nabla \times \mathbf{F} = 0$$and \mathbf{F} is a function of position only (this leaves out the magnetic force on a ... 1 We can start at the relationship: W=-\Delta U, which is work done by a conservative force. The math A (conservative) force F will do this work on an object when doing a displacement \Delta x, and W=F \Delta x. In the general case, the force might be different at different points as the object is moved (the force of gravity is not constant along ... 3 As the first question has received sufficient exposition, I would like to make a point with regard to the second one. First thing to understand is that integrability and non-linearity of a system are two different concepts. It is true though that all linear systems in classical mechanics (i.e those that are described by systems of linear equations, be them ... 7 I'll write here a list of my personal favorites plus some commonly used books. I wouldn't be surprised if your teacher chose either one of the books below as a textbook: i) Mechanics, the first volume of the Landau course on Theoretical Physics; ii) Goldstein's book "Classical Mechanics"; iii) Taylor's book "Classical Mechanics"; iv) Marion's book ... 3 The essential idea of a Poincaré map is to boil down the way you represent a dynamical system. For this, the system has to have certain properties, namely to return to some region in it’s state space from time to time. This is fulfilled if the dynamics is periodic, but it also works with chaotic dynamics. To give a simple example, instead of analysing the ... 4 (1) In general, what is meant by non-linear system in classical mechanics? A linear system is described by a set of differential equations that are a linear combination of the dependent variable and its derivatives. Some examples of linear systems in classical mechanics: A damped harmonic oscillator,$$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 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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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 ...

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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 $(2)$: ...

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$m_2$ 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 ...

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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 ...

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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.

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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 $$\int_A^B \frac{ds}{R^2}$$ It is a problem of ...

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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 ...

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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. ...

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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.

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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 ...

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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 ...

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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: ...

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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

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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 ...

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