# Tag Info

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For c) You should think about what it means that an orbit is closed. It means that after some time $t$ the particle will return to it's original position. For this to happen $m_1\tau_{osc}$ and $m_2\tau_{orb}$ have to be equal for some integer values of $m_1$ and $m_2$. This can only happen in your example if $\sqrt(n+2)$ is rational.

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Why don't you straight away solve this question, assuming our system to be a variable mass system. $m\frac{dv}{dt}= F_{external}+ V_{relative}\frac{dm}{dt}$ Let the length of the left half of falling chain be $x$, and the length of the right half be $y$. Consider the left half as your system. If the initial length of chain was $L$, then that can be equated ...

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If $\Delta S = r cos\theta$ then $dl=ds=rd\theta$ and $F_g=mg$ $$W_{g}=\int_0^\pi mgrcos\theta d\theta$$ If you're taking the angle from the center of the circle (which you are, since you said that $\Delta S = r cos\theta$, then the initial position of the ball is $-R$, since displacement is a vector quantity (and the final ...

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An important thing to recall about alpha particles is that at energies up for a few tens of MeV they range out in very short distances (less than a milimeter in many cases). Multiple scattering can be expected to generate non-trivial scattering angles only over larger ranges than the penetration depth of all alpha-decay alphas and a great many alphas that ...

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Alpha particles are only significantly scattered by nuclei. Electrons are so much lighter than an alpha particle that it is hard for the alpha particle to transfer much momentum to them. But nuclei are small. The radius of a nucleus is of the order of $10^{-5}$ times the radius of an atom, so the cross-sectional area of the nucleus is of order $10^{-10}$ ...

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I'm posting here an extraction from a paper written by T. Padmanabhan (http://arxiv.org/abs/hep-th/0608120) : Consider a dynamical variable $q(t)$ in point mechanics described by a lagrangian $L_q(q,\dot q)$. Varying the action obtained from integrating this Lagrangian in the interval $(t_1,t_2)$ and keeping $q$ fixed at the endpoints, gives the Euler ...

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Let us say you have two frames of reference; frame $F$ and frame $F'$ such that $F'$ is moving at velocity $v$ in the positive $x$ direction of $F$. Given a space time event that occurs at $(ct,x,y,z)$ in frame $F$ the Lorentz transform helps us to find the space-time coordinates $(ct',x',y',z')$ of that event in frame $F'$. If, however, you know the event ...

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The following is rewritten after @Qmechanic's comment. While his observation was correct, I think the main point below holds on its own. The case @Cham considers is that of a Lagrangian $L' = L - \frac{d}{dt}q^ip_i$ modified by a total derivative for the purpose of implementing a change in the boundary conditions. Originally the $p_i$-s are assumed to be ...

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Not that I know of. However if you're fine with considering only small oscillations, then you can replace $\sin \theta$ by $\theta$ and $\cos \theta$ by $1-\frac{\theta^2}{2}$. This might make things simpler although the solution you get will be acceptable for small angles.

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Yes. The tear is initiated at stress concentrations around the holes, where stress is highest. After initiation, the tear continues to propagate along the line of highest stress. Stress is a function of force and geometry ($\sigma_{n} = \frac {F}{A_{n}}$). In a piece of paper without holes, the stress is uniform, and the paper tears when stress exceeds ...

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I guess what the exercise means is the following: The pendulum is a massless, stiff rod with a mass $m$ at one end. The other end of the rod is constrained to be at $\vec r(t) = a \begin{pmatrix} \cos(\omega t) \\ 0 \\ \sin(\omega t) \end{pmatrix}.$ If the pendulum can only move in the $x$-$z$-plane, this leaves only one degree of freedom (the angle ...

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I don't agree with the previous answer. Firstly, the OP's question isn't about the lagrangian formulation, it's about the Einstein equation : $$\tag{1} G_{\mu \nu} + \Lambda \, g_{\mu \nu} = -\; \kappa \, T_{\mu \nu}.$$ Secondly, there are stress-tensors that can't be derived from an action : fluids tensors (especially with ...

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The beam will be accelerating downwards after one guy lets go, and it will be rotating. I would ask myself - how fast is it accelerating? How fast is it rotating? There will be a torque $\Gamma$ due to the remaining force $F$ of the one man $\Gamma=F\ell/2$ resulting in angular acceleration $\dot \omega = \Gamma/I$; and a vertical acceleration of the ...

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you have to notice that this motion is accelerated, so if you define velocity as $d/t$ you will get the wrong result. For uniformly accelerated motion (which is your case, the acceleration is contant: g), you have to use the following relationship: $y(t)=y_0+v_0t+\frac{1}{2}at^2$ When you release the ball, $y_0=H$, $v_0=0$ and $a=-g$ so you get $... 1 where spring's energy goes It does not go. From the conservation of energy,$\frac12Kx_\text{max}^2=\frac12mv_\text{max}^2$, and$m = 0$you get$v_\text{max}=\infty$. In conclusion, your spring will oscillate with$\text{Amplitude} = x_\text{max}\;,v_\text{max}=\infty \; \&\; \omega = \infty\; .$For more information see: ... 3 There are no ideal springs. Therefore the paradox with the infinite acceleration is not a physical one, but an artefact of the mathematical modelling. Conservation of energy does obviously not hold, when there are objects of mass zero in a system (because the kinetic energy will always be 0). So your setting simply does not fulfil the requirements for energy ... 0 Normally it is said that the spring has no mass, but in problems where the spring is attached to bodies with larger mass compared to the spring mass. So you pull the string from the side where you have a body of mass$m$(tha other side is attached to the wall). Then when you release this body with potential$\frac{1}{2} K x^2$it will be converted in ... 4 The Hamiltonian$H(\theta,p_\theta)$needs to be formulated in terms of the coordinate$\theta$and its canonically conjugate momentum$p_\theta = \frac{\partial L}{\partial \dot{\theta}} = R^2 \dot\theta. The correct expression for the Hamiltonian is \begin{align} H(\theta,p_\theta) & = p_\theta \dot{\theta}(\theta,p_\theta) - ... 1 I) Hamiltonian interpretation. Given a HamiltonianH(z;t)with canonical coordinates $$\tag{1} (z^1,\ldots,z^{2n}) ~=~ (q^1, \ldots, q^n;p_1,\ldots, p_n),$$ the Hamiltonian Lagrangian reads $$\tag{2} L_H(z,\dot{z};t) ~=~p_k \dot{q}^k - H(z;t).$$ Then OP's modified Hamiltonian Lagrangian becomes \tag{3}\tilde{L}_H(z,\dot{z};t) ~:=~L_H(z,\dot{z};t)- ... 0 Lets define the following action : $$\tag{1} S' = \int_{t_1}^{t_2} L' \, dt,$$ where $$\tag{2} L' = L - \lambda \, \frac{d}{dt} (q^k \, p_k).$$ An arbitrary variation \delta q^k gives the following : \begin{align} \delta S' &= \int_{t_1}^{t_2} \Big( \frac{\partial L}{\partial q^i} \; ... 0 First of all, you mention that when the instrument disconnects, it is no longer rotating. Why would that be? However, the bigger issue is that if you want to conserve angular momentum, you have to measure it in the same way each time. For angular momentum, that means you have to retain the same axis each time. The instrument may no longer be rotating ... 0 There has been an apprehension about whether the cyclist can take a turn without steering. In my opinion he can. There are two principles he can use. These are By shifting the line of normal reaction sideways to the line of force acting upon the center of gravity. By trying to rotate the cycle sideways. For simplicity consider the case of a standing man ... 0 I do not want to put too much focus on the gear systems. In general, P_{in} = P_{out} (assuming no power loss). Using, P = F v, one gets F_{in} v_{in} = F_{out} v_{out}. That is, one can "scale up" the output force by moving through a greater distance per unit time (i.e. since F_{out} = F_{in} (v_{in}/v_{out}), increase (v_{in}/v_{out})) What ... 0 Don't think of it as components of KE: rather think about it as that the total KE of the body is the summation of it's angular KE and Linear KE(which happens to be in the radial direction in this case)...Hope that helps.. 0 Everything in classical mechanics, momentum, angular momentum, torque, velocity etc. is measured about a point. Period. You can be sort of a Newtonian Nazi and complain that it is wrong to talk about torque about an axis and you'll be correct but here it means a completely different thing but in common language, we often make do with such words. So, coming ... 0 Before you read the rest of my answer, you must know that strictly speaking, we always calculate torque about a POINT. But, there is a way to find torque about an axis, provided that axis is the axis of rotation. But, when you find torque about a point, you don't even need any rotational motion. Since you have trouble understanding what torque about a point ... 0 Torque never acts over an axis it acts only at the point of contact whereas moment of inertia acts along an axis 0 It's like any balancing problem. You constantly move your point of support to cause yourself to fall one way or the other. If you don't like the way you are falling, you move your point of support to stop that fall, and then start falling the other way. Your point of support is never stationary. If it is, you fall over. On a bike, you move your point of ... -1 No, water moving out of a bottle rocket is not nearly fast enough to generate the energy sapping shock waves and mach which the converging-diverging nozzle seeks to eliminate. 0 A constant force pushes a mass m along a distance x. What is the final velocity? The acceleration is a=\frac{F}{m} and the kinematic relationship between speed and distance is \left. x = \frac{v^2}{2 a} \right\} v = \sqrt{2 a x} = \sqrt{ 2 \frac{F}{m} x}$$The final momentum is$$ p = m v = \sqrt{2 F x m} $$So momentum is proportional to ... -1 From definition, force is a momentum change rate in time \vec{F} = \frac{\text{d}\vec{p}}{\text{d}t}. So if the force working on both is equal then their momentum change would be the same so answer C, but... in my opinion the correct answer is D. There is no information given about friction (if we assume that blocks are moved on surface), whether is or is ... 1 In classical physics there exists no h, and everything was going happily along until the data/measurements showed that classical electrodynamics + classical mechanics could not explain a number of data. The datum that introduced the otherwise unknown and undefined h_bar was black body radiation. It was not possible to reconcile the fact that no excess ... 4 What would somebody take h-->0 to take the classical limit in the first place? Think for a minute about what the presence of Planck's constant in quantum theories does. It creates a granularity to the amount of energy that can be stored in a mode (but not in general, to the amount of energy that can be stored in most systems). Why does that matter? ... -1 Center or rotation isn't located where those two gears make contact, it's in a point closer to the smaller gear's center 1 Here we have a holonomic constraint: \theta-\omega t=\theta_0 (or \dot{\theta}=\omega). The question is where we should use it in solving the Lagrange's equations. At first you obtained equations from the Lagrangian for the free particle, solved them, and then used the constraint \dot{\theta}=\omega. But let us remember how to solve the classical ... 1 You're on the right track. "How do we know that the space shuttle passes throught the point P after losing speed ?" The assumption (in these types of problems) is that the thruster is applied for a very short time compared to the duration of the orbit. In that way, we can assume that applying the thruster is effectively instantaneous. So if the ... 1 It is assumed that the spacecraft fires changes its velocity in an instant, not over a period of time. Its velocity is decreased exactly at the point in time it passes through P. It is true that spaceships in lower circular orbits have greater orbital velocities, but in elliptical orbits the velocity changes with the distance between the two masses (since ... 1 Potential energy is the energy due to configuration of the system. If you keep three charges very, very far from each other, then the potential energy of the system is very effectively zero. But when you bring them close together to a specified coordinate, then the potential energy of the system increases from 0 to a positive value given by$$U= ... 0 Potential Energy is calculated of a system, ie, a system possesses potential energy and the capacity to do work with it. If you a raise a ball of weight mg to a height h above the surface of the earth, then the total potential energy of the ball will be PE = mgh as you have done work against the gravity of the earth and thus the ball possesses a certain PE. ... 3 If they rotate in the same direction at the same speed, there will be no effect. This is because the disks do not move relative to one another. If they rotate at different speeds or in different directions and there is friction between the disks, their speeds will gradually come closer to one another's until their rotation rates are the same. A good way to ... 0 HINT: Density is an intrinsic property of a body.It is independent of how much of the material is present and is independent of the form of the material, e.g., one large piece or a collection of small particles. Here the mass is based on area of the disc and not the volume. Let the density of the disc be equal tod$. Then, Mass of total disc initially ... 1 The problem is that you are equating too many things to$\dot{q_k}$. Usually$\dot{q_k} = \frac{dq_k}{dt}$, a total derivative, as opposed to a partial derivative. If$q_k$has no explicit time-dependence, i.e. it does not depend directly on$t$itself, then$\frac{\partial q_k}{\partial t} = 0.$In this case, the Poisson bracket reduces to:$ ...

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First notice that all forces are internal, so by the third newton law the object's center of mass cannot move. More explicitly. The horizontal component of the force on Y will create a reactive force on L in the opposite direction. This force will transfer through the red bar and push the green part to the right. The force over Y by the green bar will be ...

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No. It will only move if L is started spinning initially, or perhaps very slightly if L is dropped. Note that in the video, he starts it spinning before it moves, and at the end, once he stops the spinning motion and sets it down - it doesn't start moving again. That's also why springs are necessary to pull L towards Y (or the other wheel) -- so that ...

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The weight L needs to do work on Y to move the cart and therefore must itself move. The cart won't move at all if L is held rigid; otherwise it will be briefly accelerated while L falls.

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Parameters The surface gravity of Mars is ~0.376 g, where g ~ 9.81 $m/s^{2}$ for Earth. The surface pressure of the atmosphere on Mars is ~0.636 kPa, which is roughly 0.63% of Earth's atmospheric pressure (i.e., ~101.325 kPa at sea level). The density of air at STP on Earth is ~1.2 $kg/m^{3}$, compared to Mars at ~0.020 $kg/m^{3}$. Background Typical ...

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There are at least two generalizations of Noether's theorem. 1) Assume that the Hamiltonian system with Hamiltonian $H(z),\quad z=(p,q)$ has a one-parameter symmetry group $\{g^s_F(z)\}$ which is generated by a Hamiltonian system with Hamiltonian $F$. Then $F$ is a first integral for $H:\quad \{F,H\}=0$, moreover if $dF\ne 0$ then there are local ...

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simply no useful notion of 'force' in quantum systems in general. Yes you can do some calculations and sometimes handwave what a force is in a certain situation, but in most matters its of little particular use. Take the collisions at the LHC - there is no useful notion of 'force' you can ascribe to what is going on when all those particles collide. Yes, if ...

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With a cushion, the force is spread over a larger area, so the average pressure is lower. The total force is the same, it's the same as their weight obviously. Since the pressure is lower, none of your pressure receptors register a painful level.

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