New answers tagged

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This is physics, so the question should boil down to what you'd measure in an experiment. What would you see if you built a constrained mechanism and placed a load cell at the point where the constraining force is applied? The answers above give you theoretical approaches, noting that a load cell is actually a very stiff spring with a very sensitive strain ...


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In practice you're of course right, however allowing for infinite forces is the easiest way to handwave exact fulfillment of the constraints. As argued in Qmechanics answer, in practice the energy is limited, then the stiffer you make the spring constant the less max deviation you get from the exact constraints. The idea is that you can then compute the ...


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When you simplify your coordinate system with mechanical constraints, you are assuming that: The system will stay on the constrained path; and There will be no energy stored in the constraints. As a real mechanical constraint becomes stiffer, i.e., the magnitude of its restoring force as a function of deviation increases, it approaches this ideal more ...


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Your question reduces essentially to what conditions are necessary for one to be able to write $\mathbf{F}(\mathbf{r}, t) = - \nabla V(\mathbf{r},t)$ for some function $V(\mathbf{r},t)$. I'm assuming we are working in Euclidean space for simplicity, but in curved spaces the answer gets much more complicated. There are conditions necessary for this to be ...


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Classically, you don't need to interact with a particle at all to measure its position. The particle has a well-defined position value at all times. You just place a ruler on the ground and look down on it from above to measure its position in one dimension, for example - this may be practically infeasible for something as small as an electron, but the ...


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OP already seems to have thought long and hard about this and makes good points. In this answer we will review the argument for why constraint forces are said to be infinite. We will assume that OP talks about holonomic$^1$ constraints. To be concrete, let the constraint be that some generalized coordinate vanishes $$q~\approx~0.$$ We can implement the ...


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I think there is no single answer to this question; it depends on circumstances. If you have a three-dimensional region of space, with a sphere sitting in it, for example, then you might have a particle whose motion is constrained to lie on the surface of the sphere. In setting up Lagrangian mechanics for this case, you don't have to invoke any notion of ...


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Distances are measured via a metric and this is a symmetric bilinear form. One could ask can a useful distance could be constructed out of an antisymmetric blinear form, what we might call a cometric. In fact, they actually have been useful and are called symplectic forms. They are not used to measure distances but can be used to construct a volume form just ...


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As a general rule, if you know of multiple axes which can serve as principal axes, and all of these axes have the same principal moment, then any vector that lies in the linear span of these axes will also be a principal axis. In this case, we know of three axes (four, in fact) that can serve as principal axes: the axes connecting the central atom to each ...


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So you can choose one axis based on the symmetry (passing through one vertex and the center of the opposite base of the tetrahedron. You can choose the second axis based on the symmetry to pass through the center of mass of the tetrahedron and to intersect one of the edges. The third axis is chosen to pass through the center of mass and to be orthogonal to ...


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For simplicity assume a horizontal spring & mass system. With a force of $10\,\rm N$ applied assume the elongation of the spring is $1 \,\rm m$ and the energy stored in the spring is $\frac 12 k e^2 = \frac 12 \cdot\frac {10}{1}\cdot 1^2 = 5\,\rm J$. Now apply an extra force of $1\rm N$ in addition to the $10\, \rm N$, so the total force is now $11\,\rm ...


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Treat the body as made of tiny particles connected with springs to each other. You apply a force, the particles start to accelerate, the reaction force of the body rises as it deforms. Then forces are equlibrated, but there is some velocity, Newton's 1st law. However, remember the body is like a spring, so there is little more deformation, however the ...


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You have assumed that the coefficient of friction for the car is the same as that give for the trailer (0.15) so the frictional force acting on the car and the trailer is $1630 \times 9.8 \times 0.15 = 2396.1$ Newtons You have also assumed that friction acts in a backwards direction on the car. But you are told that the force exerted by the car on the ground ...


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The second equation is not correct. The amplitude refers to the motion along the arc, $A=L\theta$, or horizontally, $A=sin(\theta)$. These will all agree for small theta converging to $v_{max}=(T/2\pi)g\theta$. For large amplitudes the system is nonlinear and the motion is not simple harmonic motion (sinusoidal), so $v_{max}=\omega A$ does not apply.


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For small $\theta$, $\cos \theta \approx 1-\theta^2/2$. $v_{\rm max} = \dfrac{T}{2\pi}g\sqrt{2(1-\cos\theta))}\approx \dfrac {Tg\theta}{2\pi}$ and with $A \approx L\theta,\, v_{\rm max} \approx \dfrac{2\pi}{T}L\theta \approx \dfrac {Tg\theta}{2\pi}$.


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Gandalf's answer curiously ignores the fact that mass is being lost in their analysis of the pendulum and hence is entirely wrong on that part. For the leaky harmonic oscillator (spring or pendulum), the mass loss absolutely impacts the period of the oscillator. If you have a pendulum bob that is full of a liquid that is leaking out, then what actually ...


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It's because of conservation of energy. There are two masses of water involved: the water originally in the tub, and the water that you're adding. Because energy is conserved, the thermal energy that flows out of the water originally in the tub must be equal to the thermal energy that flows into the water you add. The equation you've written down just ...


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Pressure in a fluid (liquid or gas) acts equally in all directions, so the air pressure doesn't add to the weight of your metal. However, there is a small reduction in the weight of the plate due to buoyancy. To get some solid numbers, let's assume that your plate is made of iron. The density of iron is ~$7800\,\rm kg/m^3$, that of dry air at $20°$C is ~$1....


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Not the pressure but the density of air impacts the weight of objects due to buoyancy.


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I am not aware of "natural" superintegrable systems failing to translate across the quantization map and certainly not the classical limit. (Actually, these are all described by classical & quantum Nambu brackets which demolishes bogus "proofs" that NBs could not be quantized. This is why I barely listen to proofs for the non-...


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The answer to your main question is yes. There is a reduction in the weight of all objects due to their buoyancy in the air, not just helium balloons, as there would be if they were submerged in water, except that the air weighs only approximately $1.2kg/m^3$ compared with water which weighs $1000kg/m^2$ so the buoyancy is correspondingly less. But the ...


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No. Air can get under the sheets of metal you have described, so the atmospheric pressure, which pushes in all directions equally, pushes up as hard as it pushes down. There is no net force. If you had a perfectly sealed $1m^2$ suction cup and tried to pull it away from the surface, thereby creating a $1m^2$ cross section of vacuum under the suction cup, the ...


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From the Bernoulli equation for a constant-density fluid, $$ P + \frac12 \rho v^2 + \rho g h =\text{constant} $$ you can see that the “vertical pressure” $\rho g h$ at some point is the same as the weight of a vertical column of the fluid divided by the area of that column. A cubic meter of air at sea level has a mass of about a kilogram. If the atmosphere ...


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Everything works well with constraints which depend on time explicitly. The definition of holonomic system of constraints is that, for $N$ material points represented in Cartesian coordinates in the rest space of a reference frame (the definition does not depend on the choice of the reference frame) a set of $c< 3N$ conditions must hold $$f_j (t, \vec{x}...


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Let us start from the general form of Euler-Lagrange equations for a curve (I will discuss all this issue in local coordinates thoug global approaches are possible) $$\mathbb{R} \ni t \mapsto (t, q(t), \dot{q}(t)) \in \mathbb{R}^{2n+1}$$ $$\left.\frac{d}{dt}\left( \frac{\partial T|_R(t, q,\dot{q})}{\partial \dot{q}^k}\right)\right|_{(t,q(t),\dot{q}(t)} - \...


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1 & 2: Eq. (1) is not derived. It is a defining property of a generalized velocity-dependent potential. The form of eq. (1) mimics the Euler-Lagrange operator, so that we can bring Lagrange equations $$ \frac{d}{dt}\frac{\partial T}{\partial \dot{q}^j}-\frac{\partial T}{\partial q^j}~=~Q_j, \qquad j~\in \{1,\ldots, n\}, \tag{L}$$ in the form of Euler-...


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Yes, it is valid to claim that this condition would be comparable to the force required to push the chair over the same curb but only in case if the pushing force and pulling force (in pulling case) vectors are collinear to each other, in the same direction and having the same magnitude that means the pushing force should act on the same line as the pulling ...


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The momentum of the ship would be $m\cdot v=\frac{P\cdot t}{c}, c=3*10^8m/s$


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As @saumy points out, both A and B see equal and opposite horizontal forces and therefore the horizontal velocities are equal and opposite at all times. When B is directly above A, both bodies are instantaneously rotating around the mid-point of the string. The tension in the string due to the rotation is $mv^2/(L/2)$. When this force is greater than the ...


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Both would work. It is just a matter of convention. Notice that if $x_0$ is an extremum of the function $S(x)$, it is also an extremum of $\alpha \cdot S(x)$ for constant $\alpha \neq 0$.


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The answer depends on the nature of the force that is causing the simple harmonic motion (the "restoring force"). In the case of a simple pendulum, the restoring force is the weight of the object. This is proportional to the mass of the object. So if the mass of the object changes, the restoring force changes by the same proportion, and the period ...


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The flow of a classical Hamiltonian observable is the flow of its associated Hamiltonian vector field. The differential equation that an integral curve $(q(u),p(u))$ of this flow/vector field obeys is precisely your eq. (4). You should think of these integral curves as the motions you would get if the associated observable $f$ was your Hamiltonian - for $f=H$...


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In the case of solids, when you apply a force on the surface you have what is called a traction vector. This vector is defined locally as the force density over the surface. Inside the solid you have stress. This is a quantity that is of a tensor nature. The traction on the surface is the projection of the tensor in the direction of the unit normal. That ...


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I'm a bit late to the party, but for future reference of other students stuck in this problem, you can easily find a solution if you assume a solution for $S$ in the form of $$S(x,t) = x\cdot f'(t) + g(t)$$


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Let's consider a sligthly more simple example. Let's say I have $h(k)=f(k)g(k)$. I want to calculate $h(k)$ to first order. I can come up with two ways to do this. One way is to just Taylor expand $h(k)$ in the usual way and the second way is to first expand $f$ and $g$, then multiply them as polynomials and then truncate. If I know the expansion of $f$ and $...


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I don’t have that textbook but I guess what authors want to derive is your last equation, \begin{equation}T = \frac{2V}{g}+(\frac{T^2}{3}-\frac{2V^2}{g^2})k+O(k^2).\ \ \ (1) \end{equation} This equation is equivalent to your first self-consistent equation, \begin{equation} T = \frac{kV + g}{gk}(1-e^{-kT}) \ \ \ (2)\end{equation} at the lowest order about $k$...


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at each point on the path you know the tangential vector $~\vec t~$ the normal vector $~\vec n~$ from here you can obtain the transformation matrix between the moving frame and inertial frame $$\mathbf S=\left[\vec t~,\vec t\times\vec n~,\vec n\right]$$ the magnitude of the columns is one $\Rightarrow~\mathbf S^T\,\mathbf S= \mathbf I_3$ so your vector $~\...


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First of all if you want to throw the ball after thirty degree of rotation of arm then according to me you should focus on the tangential velocity on that instant at which you have to release the ball. I'm going to assume that the torque on the Catapult is going to remain constant all the time because you haven't told that if the arm is attached to the ...


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This is in part a question about Biology. The structure of a trunk and a branch is shown below. Drilling holes $A$ and $C$ will impact on the growth and health of the branch more than drilling hole $B$ because more of the growth and transportation part of the tree is affected. Drilling hole $A$ might well weaken the branch as this is equivalent to cutting a ...


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A strange question since drilling a hole in a tree is going to damage the tree no matter where you drill it. If this is in a physics/engineering context, my guess is you are supposed to treat the tree branch as a cantilever beam and think about where it is in tension and where it is in compression.


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Ignoring air resistance, gravity, and a bunch of other variables here's an answer that will hopefully suffice: Note: I use surface area interchangeably with volume as sheets of paper are very thin. Why is a single sheet of paper easy to go through for a bullet? Well this is due to that large difference between the ability for the sheet of paper to resist the ...


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Suppose you have a cart connected to a wall with an ideal spring oscillating frictionless back and towards the wall. Now on that cart, you mount a pendulum that can only oscillate orthogonally to the cart’s direction of motion (with a different frequency). This way, the motion of the cart and pendulum are completely independent of each other. Now, consider ...


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In the context of dynamical systems, quasiperiodic motion is characterized by two (or more) incommensurate frequencies (i.e., frequencies which are not rational multiple of each other). As an example, if you consider a point jumping along a circumference of length $1$ at fixed irrational (say, $1/\sqrt{2}$-long) steps, you also get quasiperiodic motion — ...


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In this context quasi-periodic should be what in mathematics is called almost periodicity. Essentially, a function $f(t)$ is quasi-periodic if there is a quasi-period $T$ such that \begin{equation} |f(t+T)-f(t)|<\varepsilon, \forall x \in \mathbb{R} \end{equation} for some small $\varepsilon>0$. Notice that if $f$ was periodic, above we should have $=0$...


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Mathematically (as on the linked wiki), a quasiperiodic function is one which satisfies $f(x + \omega) = g(x, f(x))$, where $\omega$ is some constant, the quasiperiod. Quasiperiodic motion is motion which is describable by a quasiperiodic function. In common language, which is almost entirely unrelated to the mathematical definition: a system is quasi-...


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either Einstein's hypothesis is false or Newton was wrong that there is attraction between 2 objects. As you continue studying General Relativity, you should soon see that the physics community has long ago come to terms with the idea that General Relativity is a reinterpretation of gravity that replaces Newton's picture. Newton's simplified picture of ...


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I believe a proof can be given closer to the spirit of classical mechanics, at least if assuming the constraint is holonomic in nature. The unconstrained system is a particle moving in a central potential having the Lagrangian $L(q,\dot{q},t)$, obeying the usual Euler-Lagrange equations and having an energy function $h(q,\dot{q},t)=\sum_{j}\dot{q}_{j}(\frac{\...


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If the CCW torque about point (A) produce by the force (P) is greater than the CW torque produced by gravity, then the rod will tip up from the surface of the incline.


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Yes, if the coefficient of friction $\mu$ is greater than $\tan\theta$, the body won't slide, but could easily tip over, if the line of the weight falls outside the base, as in the diagram below. Even for the $20^{\circ}$ angle, usually a body wouldn't slide, but if it were tall the tipping could still occur.


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There is a great way to show that the Lagrangian (the thing you want to minimize) is actually equal to $T-V$. The "proof" comes from the book "Quantum Field Theory for Amateurs". To begin the proof we first of all should consider what the average kinetic and potential energy is as a functional $$T_{avg}[x(t)]=\frac{1}{t_2 -t_1}\int_{t_i}^{...


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