# Tag Info

## New answers tagged classical-mechanics

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$y=\frac 12gt^2$ This is the best hint I can give since it is not a meta exchange site.

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First of all the distance traveled in $n$-th second is the distance traveled in one second between $t = n$ and $t = n-1$ seconds lets first calculate the distance traveled in 3 seconds. $S=ut+0.5 g t^2$ (here initial velocity is zero) $S=0.5 \times 10 \times 3^2 = 45 \, \mathrm m$ now lets calculate the distance traveled in 2 seconds $S=ut+0.5 g ... 0 In order to solve this question, you have to calculate distance travelled in 4 seconds as well as in 3 seconds. And then subtract these two to get the distance travelled from 3rd to 4th second (which is 3rd second). What you did was that you calculated the total distance travelled from$t=0$to$t=3$seconds, instead you have to find the distance travelled ... 0 It helps to remember that invariant quantities are seen as scalars to the transformation (they have no indices in the target space). In the other hand, covariant quantities are objects that transform in a certain way. Example: Vectors in$R^{2}$, under rotation$R_{ij}$, transform covariantly since$v'_{i}=R_{ij}v_{j}$, but it's length is invariant since ... 1 Whatever you are doing, I think you are making this too complicated. The normal way to solve this problem (approximately) is to ignore all the terms in the Fourier expansion, except the one at the resonant frequency of the swing. Then the solution is just the normal time-dependent solution for a damped oscillator forced on resonance. 3 The issue is that the underlying classical physics is determined by equations of motion (EOMs) (i.e. Newton's 2nd law), which are common for initial value problems (IVPs) and boundary value problems (BVPs). For BVPs , the EOMs can often alternatively be formulated as Euler-Lagrange (EL) equations of a stationary action principle. The latter approach does ... 9 Indeed the problem with boundary conditions, generally speaking, is not well-posed. There are boundary conditions admitting no curves or admitting many curves, satisfying both these conditions and Euler-Lagrange equations. Examples. (1) Think of a particle constrained to stay on a smooth sphere where it can freely move. If you assign the North and the ... 0 About second part of your question, I should say that I couldn't understand it because it may the polygon like a star has no side contacted with the ground. About first part of your question, I should say "It is not possible". “Let us say a polygon shaped object is stable on a side when the center of mass "falls" inside the base”. If you accept this phrase ... 0 Supposing these vehicles are very solid, have both of them crash into identical parked train wagons, from the back. If after the crashes both vehicles are still intact and not moving any more, then you will see both train wagons moving away at the same speed. The extra supposition (solid vehicles still intact after the crash) is there to assume no momentum ... 0 From those we can get$f(x(t),t) = x + Ct + D\,$after integration. How can I interpret this$f(x(t),t)$? The constants of integration$D$and$C$are, respectively, the displacement between the origins of the two frames at time$t=0$and the constant velocity with which the origin of the primed frame moves with respect to the origin of the unprimed ... 0 Lagrangian mechanics can be derived directly from Newton's second law using only algebraic manipulation and a some calculus. This includes both the general form of the Euler-Lagrange equation and the specific form Langangian$L = T - V$. No assumptions of stationarity, use of the calculus of variations, or even any reference to the concept of action are ... 0 There is nothing wrong with looking for plane-wave like solutions of the form$A \exp (i (\omega t - k x) )$. Given the linearity of the equations, and as @ignacio pointed out the fact that the$\exp (i k x_n)$form a basis of solutions, you can write a more general solution as a combination of these plane waves. This solution isn't necessarily periodic ... 3 The oldest work on this preceeds quantum mechanics by more than 100 years. it was done by Malus in 1809 about experiments with polarized light. See http://www.mat.univie.ac.at/~neum/papers/physpapers.html#CQlightslides 0 The obvious difference is that Newton's equations retain their form for all inertial reference frames when Galileo's Principle of Relativity is used, but Maxwell's equations are not invariant under this transformation. Instead one must use the Lorentz transform, which recognizes that there is a fixed speed for light,$c$. This limit was recognized by ... 0 Two of Maxwell's equations combine to yield a wave equation with a fixed wave velocity, the speed of light$c$, for both of two observers in relative motion to one another, contrary to the behavior of waves in Newtonian mechanics. 1 Yes, the viscosity of a dilatant suspension will decrease if the viscosity of the solvent decreases. Surprisingly I struggled to find experimental data to back this up. Perhaps everyone thinks it's too obvious to be worth publishing. The best I could do is this school exeriment report. The authors timed the fall of a ball through the suspension, so lower ... 4 This is the setup described in the equation: The acceleration is defined in terms os the displacement of the bow$x$by: $$a = 6000 \left(1 - \tfrac{4}{3}x\right) \tag{1}$$ So initially$x=0$and when we substitute this into equation (1) we get$a = 6000 \text{ms}^{-2}$. When the arrow leaves the bow so$x=\tfrac{3}{4}$and we get$a=0$. So far so ... -1$a=6000(1-\dfrac{4}{3}x)$is a differential equation in$x$. Solve it then substitute$x=0.75m$to find out$t$. 0 Firstly, given a differentiable Lagrangian$L(q,\dot{q},t)$, we can always form the Lagrangian energy function $$\tag{1} h ~:=~\sum_ip_i \dot{q}^i-L ,\qquad p_i ~:=~\frac{\partial L }{\partial \dot{q}^i }.$$ Secondly, make the assumption that $$\tag{2} \text{The Lagrangian } L=L(q,\dot{q}) \text{ has no }{\it explicit} \text{ time dependence.}$$ ... 0 Let a$d$-dimensional Hamiltonian system (i.e.$2d$-dimensional phase space) be given. Then the existence of$d-1$observables$c_1,\dots,c_{d-1}$that are in involution with each other and with the Hamiltonian means that there are$d$constants of motion along each orbit - the Hamiltonian itself and the value of$c_1,\dots,c_{d-1}$. Systems for which such ... 1 You forgot about the free body diagram for the cupboard! It is pushing on the earth too. So you can't apply a net force to the earth by pushing between the floor and the cupboard which are both fastened to the earth. 2 The question is, what do you mean by "force"? :) I mean, what confuses you is probably: Why do we call $$F=ma$$ a physical law, if this is the only definition of force? This is a good thought. But fortunately Newton has not just this one axiom, but two (the historical "first" is a special case of$F=ma$, but there is a third), which are connected. So I ... 0 I know what you say, that's How I did initially, and it gave me a wrong solution. I will post the exercise to you solve by yourself. If I got the same system, without the rope, considering there is friction between the blocks, and A and the ground, is there friction between the blocks while I'm pulling block B but the system doesnt start moving? 0 The definition${\bf F} = \frac{d{\bf p}}{dt}$is valid in all inertial frames (assuming not considering relativity). If you have another inertial frame you can replace${\bf v}\to {\bf v' = {\bf v} + {\bf v}_0}$, so that${\bf p} = m {\bf v}\to {\bf p' = {\bf p} + {\bf p}_0}$and since${\bf v}_0$is constant, have$\frac{d{\bf p'}}{dt} = \frac{d{\bf ...

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I think we must be able to accommodate a definition of a force on some particle which is independent of the motion of the particle, \emph{for all kinds of forces}, to surely verify the statement like 'force on a particle equals the time derivative of the momentum of the particle'. Is that true? The quoted statement is not the best formulation to be ...

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Great question. I believe that yes, Liouville's theorem is the key part of the justification for this in classical stat mech. The reason for this is that it leads to a time-invariant equilibrium measure. If you used a volume measure that wasn't time-invariant in this way then it would be very strange, because on the one hand you would say that you had no ...

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You can use these free body diagrams:

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As the problem is initially described, the nozzle is located on the left bottom side of the tank with the nozzle exit facing downward. if this is the case, there will be no horizontal force to act as a thrust to start the tank in a horizontal motion. Any thrust that may be developed by the water exiting the nozzle will be in the opposite direction of the ...

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In order to maintain the constraint of the pivot during the impact, a reaction impulse is needed. See the figure below for what I mean. At the center of mass the velocity is $v = a\,\omega$. This is a result of the two impulses $$(F-R) \Delta t = m\, a\, \omega$$ If the angular velocity is $\omega$ then the net impulsive moments at the center of mass are ...

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You can "derive" the Lagrangian formulation from Shannon entropy arguing Liouville's theorem in reverse.

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As you have hypothesized in the comments, this will be a time integral of a force at the pivot point. As I read the problem, it would actually be the force exerted by the rod on the pivot that you're integrating, not the other way around. More formally, the "impulse imparted on the pivot point" is a vector representing the total linear momentum transferred ...

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Going from action to EOMs is simple: it is just (functional) differentiation. Going the other way from EOMs to the action is hard: It is (functional) integration, and sometimes impossible! OP is now essentially asking: Can we integrate one more time? Well, not the action itself. But if we replace the EOMs and the Lagrangian $L$ with their dynamical ...

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Considering momentum: $$m_1u_1+m_2u_2=m_1v_1+m_2v_2$$ Therefore, $$m_1u_1-m_1v_1=m_2v_2-m_2u_2$$ Factorising gives us: $$m_1(u_1-v_1)=m_2(v_2-u_2)$$ Allowing us to rearrange to: $$\frac{m_{1}}{m_{2}}=\frac{v_{2}-u_{2}}{u_{1}-v_{1}}$$ Using the fact that you say the velocities can be interchanged, we obtain a final answer of: ...

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You ask: could the gravitational force of an average asteroid be extrapolated to the rest of the asteroids in its family? But the question is what properties the members of a family have in common. You could argue that all the asteroids in a family will be made up from the same material and will have the same density. That means as long as you can ...

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From considering the De Broglie Wavelength, $h/mv$, we can show that every wave has momentum if it is considered as a particle derived by its Wavelength.

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You can approach this in a couple of different ways. 1) The motion of the center of mass of the particles will follow the same path the grenade would have followed if it had not exploded. 2) The total momentum of the two pieces immediately after the explosion will be equal to the momentum of the grenade immediately before the explosion. Enjoy the algebra ...

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For the second part to travel down the path of the full grenade, it must have a velocity $-V_x$. By taking momentum into account, it can then be shown that the other part must have a horizontal velocity $3V_x$. Momentum is conserved, even though kinetic energy is not. You can then use these new velocities to calculate the distance between the two points ...

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Mechanical waves are those that need a specific medium for their propagation. Of course, the medium has to be physical. There are two types of mechanical waves: transverse and longitudinal. But, first of all, electromagnetic waves that show transverse property are not mechanical waves since they don't essentially need any medium for their propagation. The ...

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No. It is an equation of a moving function. Take any general function dependent on position and time such that the function is f(x+ct). Now, partially differentiate it by applying chain rule. You will see that you end up with the same differential equation as above.

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Solutions to the wave equation need not be periodic. In fact, any profile with the argument $x \pm c t$ will satisfy the wave equation.

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Your wording is a little off. SMH doesn't care about the size of displacement. SHM is the lowest order approximation to a general oscillation. A general oscillation does care about the size of the displacement, and when you ignore the higher order effects you will miss various phenomena. Examples of things you will not be able to describe are harmonic ...

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When you ignore higher orders in the force, assuming it is linear in the position, you are assuming the potential energy is quadratic at most (quadratic, parabolic or harmonic approximation). Then you loose global information about the potential. We are not able to say for example whether the movement of the particle is bounded or unbounded as we increase ...

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For linear momentum to be a constant, its value should not change. Being linear momentum a vector quantity, it's value is completely described by specifying both magnitude and direction. In a circular motion, even though the speed remains a constant, the direction of velocity (which is tangential to the point on a circle) is changing throughout the motion as ...

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Does this mean that without any external force the kinetic energy of any thing can be increased? No, of course not. Just because there is no torque that doesn't mean that there isn't a force. In this particular case, you have an inwards radial force that is performing work by countering and exceeding the centrifugal force. This work is what causes the ...

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No, more force is required because of the higher spring constant. If the spring with the higher spring constant happens to have more mass, then inertia will come into play, but consider the following you can have two springs with different spring constant and same mass the expansion of a spring can be done arbitrarily slowly so that inertia (which opposes ...

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Firstly what you did wrong was conserve energy. You see, if we ignore rebound then that means the collision is inelastic. Thus there will be net loss of energy whose amount we do not know. Anyways, to work around the problem, you have to conserve angular momentum about the point of collision. The reason being that about this point, resulting torque due to ...

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The thickening of oobleck and similar materials is due to a phenomenon called dilatancy. This happens because shearing the suspension forces the water to flow at very high shear rates through the restricted gaps between the solid particles, and that requires a very high shear stress. However the water itself is not thickened in any way, and its electrical ...

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The main reason for this difference in stability is the orientation of the feet of the clones. If you were to do this experiment on dry land by simply putting sandbags on someone's feet (the water makes it hard to move your legs fast enough to catch yourself, and so do the sandbags) and pushing on them from different angles, you would find that they fall ...

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The real world is messy and annoying, so maybe. But it shouldn't. Magnetic fields are created by currents, and the direction of a current depends on both the direction in which charge carriers move and the sign of the charge that is moving. When you spin a whole capacitor, in any orientation, you move both positive and negative charge carriers in the same ...

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The acceleration of an object by fictitious forces is by the definition of what a fictitious force is due to the acceleration of the non inertial reference frame only and does not depend on its mass. In order to make Newtonian mechanics work in those non inertial frames these fictitious forces are introduced and defined to be equal to the mass times the ...

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