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This is because in deep space everything is being attracted equally in all directions since the universe is homogeneous. Hence, all of these tiny gravitational forces acting on the objects in your ship cancel out. Therefore you are left with a net force equal to zero on these objects in your spacecraft. On reflection, forget deep space for the time being ...

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Your question is actually quite a complicated one as there are lots of different factors at play. However if you're asking why the objects inside the rocket don't attract each other then the main reason is that gravity is actually a very weak force. If a book is floating a metre away from me then the gravitational acceleration of the book due to my mass ...

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Your answer of gravity being weak is exactly correct. The forces involved can be calculated with the following equation: $$F = \frac{GMm}{d^2}$$ A fueled Saturn V was about $3.0 \times 10^6 kg$. Most of it doesn't reach space but we'll be conservative and say your spaceship is that massive. If you had a 1kg book that could somehow be only 1 meter away ...

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The rocket is in free fall along with the book. The nearest gravitating bodies are very far away, so whatever meager acceleration they cause will be almost exactly the same on the rocket and on the book. Suppose you instead went on a very close flyby of a neutron star. Now the book will fall rapidly, and away from the rocket's center of mass. The only way ...

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The angular momentum $L_{A/B}$ of a rigid body $A/B$ about its center of mass is $$L_{A/B} = I_{A/B} \omega_{A/B},$$ where $I_{A/B}$ is the inertia matrix of $A/B$ about its center of mass in the world frame and $\omega_{A/B}$ is the angular velocity of $A/B$. The angular momentum $L_{A/B}^0$ of a rigid body $A/B$ about the origin of the world frame is ...

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when low mass object hits high mass object it is reflected gaining opposite velocity almost the same as initial velocity. If I jump onto the wall why my body is not reflected? I know that collision is not fully elastic but it should be at least similar. Human body is not elastic: it cannot be deformed/ compressed in any way and then return to ...

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Changing shape could still be an elastic deformation (of a rigid body). So obviously there are also plastic deformations involved, when jumping against a wall.

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No. As @Jim said, the heat would weaken the rock, which would cause a tunnel collapse before any sublimation could occur. Also, remember that the air in the tunnel would generally be at the same temperature as the rock (unless a large cooling system was put in), so thermal equilibrium would be maintained without any sublimation.

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My answer is the same as Chris', but formulated in a different way (it's essentially the same as this wiki article): In polar coordinates, the position vector is $$\mathbf{r} = r\,(\cos\varphi,\sin\varphi) = r\,\mathbf{\hat{r}},$$ with $\mathbf{\hat{r}}$ the radial unit vector. The velocity is then $$\mathbf{v} = \dot{r}\,(\cos\varphi,\sin\varphi) + ... 1 Some conservation laws are related to conservation of angular momentum. There is a famous example (from Feynman if I recall correctly), where you assume an infinite flat space and conservation of angular momentum about any point, and then you get conservation of linear momentum for free. Intuitively, to get say the x component of linear momentum is ... 1 You are introducing some irrelevant variables, as F_{M_3}, T_1, T_2. Let us make the assumption that T_1=T_2=T (the pulley doesn't rotate and the string is massless). The whole has mass M=m_1+m_2+m_3, accelerates with a and the force on M is$$F=Ma.$$The tension T equals m_2 a and also m_3g, so$$a=\frac{m_3}{m_2}g.$$Hence ... 1 Hint. I think the error is in your first equation, adding the forces for mass M1: \vec{F}_{M_3}+\vec{F}=m_{1}\vec{a}_1\\. There is an additional force on M1 that you have omitted. Edit: The pulley exerts a force on M1. 2 The "associated scalar equation" is just the formula for the time evolution of the scalar magnitude of the displacement, r, rather than all its vector components. It really only makes sense to write such an equation if the right-hand side can be expressed in terms of r only, and not \mathbf{r}. Then you can use it to analyze the evolution of r in ... 1 Thermally supported, self-gravitating bodies (those to which the Virial theorem applies) qualify if you are willing to neglect the radiative energy loss. Depending on the time-scales that interest you this can be quite a good approximation. Stellar nebulae, brown dwarfs and so on. 0 I would say three degrees of freedom, and you can use, e.g., the Euler angles (http://en.wikipedia.org/wiki/Euler_angles ). 5 It can be shown via simple dimensional analysis. We know that [u]=m/s, so just multiply by 1 in terms of an area:$$ [u]=\frac{m}{s}\cdot\frac{m^2}{m^2}=\frac{m^3}{m^2\cdot s}=\color{red}{\frac{1}{m^2}}\cdot\color{blue}{\frac{m^3}{s}} $$The blue term is the volumetric flow rate while the red term is the area, thus we have a volumetric flow rate per unit ... 3 You could express flow rate as a velocity. But if you want to have a quick measure of how much material flows through (for example) a pipe, you need to know both the velocity and the area - a quick diagram shows you that velocity x area = volume that passes through the area per unit time. So if$$v \cdot A = Vol/time$$Then it follows that$$v = ...

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Because of gravitational forces you liquid is at the bottom of both phases. A pressure gradient, due to those forces exist in both phases; the pressure increases linearly through bot fluid, with a discontinuous ramp at the interface liquid/gaz. So if you push down the piston, by increasing the volume you will modify the equilibrium point and so decrease ...

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Yes lagrangians and hamiltonians are indeed used by engineers. To be precise, used by some types of engineers like aeronautical engineers, aerodynamics etc.. For example: http://www.osti.gov/eprints/topicpages/documents/starturl/47/566.html As far as i know electrical engineers dont use the lagrangian nor hamiltonian forms of mechanics nor ...

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I'm a electrical engineer, and have never used either one in over 30 years of designing circuits. I vaguely remember that we went over them briefly in school, but since I haven't used them (knowingly) since, I can't tell you what the physical meaning of either is, which of course perpetuates the fact that I'm not going to use them.

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In civil engineering they use it for structures, and strength of materials in the elastic realm. It goes by the name of the enegy method. Google books might give an indication. Some authors are Beer and the mechanical engineer Stephen Timoshenko. This is for some what "static" indeterminant structures. So, there is no time element. But, I am sure it ...

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I don't know where you're getting those $m$s from, or what substitution you're making. The appropriate substitution to perform is $$p = \frac{\partial L}{\partial \dot{q}} = \omega \dot{q}.$$ If you do this, then the hamiltonian becomes $$H = p\dot{q} - L = \frac{p^2}{\omega} - \frac{p^2}{2\omega} + \frac{1}{2} \omega q^2 = \frac{1}{2\omega}\left(p^2 + ... 0 Calculating a Hamiltonian from a Lagrangian leads us to a new quantity, which is a function of coordinates and momenta. If you simply substitute velocity as a function of momentum into the Lagrangian, then the "Hamiltonian" will depend on velocity implicitly. This is not what we want. Let me show you an example to explain why Hamiltonian (the usual one) ... 1 Simplifying the problem slightly, I think it can be solved. Assumptions: Pushing force is applied at the bottom of the box - so there is no net torque about the horizontal axis Weight distribution in the box is even Force distribution (normal force) is even - imagine 1000's of tiny springs touching the ground Coefficient of friction is constant Now we ... 1 A lever can work in zero-gravity conditions as long as the fulcrum and lever arm are physically attached. They also work regardless of atmospheric presence. 1 Note that the notion of work W=\int_{\gamma} {\bf F}\cdot \mathrm{d}{\bf r} depends on the force {\bf F}. Here {\bf F} could e.g. be a gravitational force or a frictional force, etc., leading to a gravitational work or a frictional work, respectively. The phrase work done against a force is shorthand for work W_1 done by a force {\bf ... 0 Assume you have a force field (it can be caused by eletric, magnetic, whatever fields). So, if you have a object in position \vec{r}, it will feel a force \vec{F}(\vec{r}). So.. inside the vector field, now comes the question: What's the work I have to do to move a particle from here to there? Or, equivalently: What's the work I have todo against the ... 0 Both cases are the same. Let's assign some axes. Say z points up through the helicopter, y forward, and x to the right. And let's agree that the blades are spinning counterclockwise when seen from above. That is, the angular frequency \vec{\omega} and angular momentum \vec{L} of the blades are initially both in the +z-direction, and the ... 1 A shorter way to do this would be to use the concept of a pseudo-force. When you observe the motion from the inclined plane's frame, you should be applying a pseudo-force (virtual force in the direction opposite to acceleration) because it is an accelerating frame. Considering this, there are three forces on the block in this frame (which is seen ... 0 As @NeuroFuzzy pointed out, since the blade will rotate with that pitch for 180 degrees before you change the pitch again, the average position in which the upward forces act during those 180 degrees will be behind axis of rotation, not to the left of it, which only is where the force acts in the beginning of those 180 degrees. 1 Alright, I think I solved the problem: 1 Work is done by something or someone, expending energy by exerting a force on an object. I'm not quite sure what the context is of your quote, but I'd imagine that it could be a scenario much like the following one: Imagine a ball with some mass m being lifted by a man. The gravitational force from Earth is pulling the ball down, but the person ... 0 You don't need to express t in terms of z and p, explicit time dependence is permissible in the Hamiltonian. (It would not even be possible without restoring to \dot z.) In the other question it was not mentioned because it was not needed, the Lagrangian was time-symmetric and consequently so was the Hamiltonian. This is very often the case so many ... 0 Usually people call configuration space, \mathcal{M}, to the space of all posible coordinates needed to determine your system (although,I would include the velocities too). In non-relativistic theories The coordinates are three... and we need to provide three coordinates per (point) particle, i.e., for n particles one needs 3n-coordinates describing ... 4 I will assume you are talking about the center of mass. If there's no external forces, the center of mass would conserve it's momentum. So, it would stay in constant speed, whatever what that speed is, with respect to whatever inertial frame of reference. This happens because Newton's third law. In the summation of all forces, the internal forces will ... 1 OP is essentially asking about terminology. As usual, be prepared that different authors call different notions differently. Well, here is a suggestion: Call the configuration space before (after) the constraints are implemented for the extended (physical) configuration space, respectively. More generally, if an author is talking about a configuration ... 3 I think that your description that the points of the configuration manifold are possible states of the system is as close to a precise definition as one will find. So for n particles in three dimensions, the configuration manifold is just (\mathbb{R}^3)^n. As for how this relates to constraints, consider the simplest example: two particles attached with ... 1 Not the exact title, but stripping some non-essential bits of the search to crash course lagrangian dynamics pdf, I found a 12-page document titled Crash Course in Discrete Lagrangian Dynamics (found here). This is likely not what you are looking for, but it could be useful nonetheless. 1 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 ... 0 "The non-radiating electron is a topic of its own and appears to be far from solved.". That's an interesting statement, since there is ample evidence, that there are no such things as "non-radiating electrons". The only thing that "exists" in nature at that scale is a quantum field, one solution of which are electrons. Under certain circumstances (when ... 1 In the idealized case of "the wheels of a moving car" and assuming this is an introductory mechanics course, the author of the question is probably looking for you to notice that the whole of the mechanical energy can be written as a rotational term (\frac12 I_{tangent} \omega^2) about the point of contact with the road. Now, by the parallel axis theorem ... 12 I've recently started trying to swim the butterfly. Unlike other swimming strokes, there doesn't seem to be any way to "go easy" and do a relaxing length of the pool: if I want to get my face out of the water to breathe, I essentially have to use the water to do a push-up. Now if a scrawny guy like me can lift his chin a few inches out of the water using ... 21 If you look closely at the crocodiles' tails you'll see that they wave their tails from side to side to provide propulsion for the jump. Compare this to a fish swimming: The side to side motion of the fish's tail propels it forward, and the crocodiles are using exactly the same sort of side to side motion to propel themselves upwards. 6 The upwards force comes from the rather violent tail movement. When the rest of the body is out of the water, the tail still acts sort of like a hydrofoil pushing the crocodile upwards, only not with a linear but oscillating motion, and obviously it's rather instable but enough to get the whole animal up in the air for a short while. 6 Let Q denote the set of all possible configurations of the system (the configuration manifold). Consider a point q_0\in Q. For the sake of conceptual clarity, and to make contact with physics notation, let's work in some local coordinate patch around q_0. Suppose that q_0 represents the position of the system under consideration at time t_0. ... 0 As I understand, it must be a displacement in generalized coordinates. If they are orthogonal space coordinates they are not virtual. 6 What Goldstein means by \nabla_iV_i is$$ \nabla_iV_i=\left(\frac{\partial}{\partial x_{1,i}}\hat{x}_{1,i}+\frac{\partial}{\partial x_{2,i}}\hat{x}_{2,i}+\frac{\partial}{\partial x_{3,i}}\hat{x}_{3,i}\right)V_i $$which is indeed a vector. Here, \mathbf r_i=(x_{1,i},\,x_{2,i},\,x_{3,i}) is the position of the ith particle (with respect to the origin), ... 5 This is typical to see in situations where the potential is a function of the coordinates of more than one particle:$$ V=V(\mathbf r_1,\ldots,\mathbf r_N)=V(x_1,y_1,z_1,\ldots,x_N,y_N,z_N). $$The force produced by such a potential on the ith particle is the gradient of this function with respect to that particle's coordinates, while holding all the other ... 1 In short: virtual displacement is "pretend you are moving, but don't really move". In other words - you move by such a small amount that you don't change the state of the system - but it gives you insight (through work done etc) in what would happen if you did move. In other words - if the system is really moving, you can look at an interval dt to see how ... 1 First of all, two tiny mistakes: your \theta equation is missing an \ell, and the sign of g \sin \theta is flipped. That is, it should be$$ 2 \dot{\ell} \dot{\theta} + \ell \ddot{\theta} = -g \sin \theta  Now, I have not seen Bessel equation emerge for pendulum dynamics for arbitrary length variability. It only arises, as far as I have seen, for ...

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