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0

I agree with Peter Mortensen and Mark Eicheniaub. The differential equation is Ӫ=(g/h)sinӨ For small angles it should be Ӫ=(g/h)Ө The solution for this is t=√(h/g) *ln(Ө/ Өo) but only for small angles like 1°. We can calculate step for step (1°) 90 times, it implies a ln sum: ln((Өo +1)/ Өo) + ln((Өo +2)/ (Өo+1)) + ln((Өo +3)/ (Өo+2)) + ln((Өo ...


1

Ideal string cannot be described in terms of canonical ensemble (with Boltzmann probability distribution). The fact it is ideal means it has infinity of degrees of freedom and since each would have, according to the Boltzmann probability distribution, average energy $$\frac{1}{2}k_B T,$$ total energy of the string would be, on average, infinite. This is ...


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If you assume linearity and the string is finite, then the partition function is a sum of discrete energies: it is the same case as a string in 2D ( $y=f(x,t)$ ), but with two uncoupled dynamics: $(y,z)=(f(x,t),g(x,t))$. This can be shown by considering an oscillator in 2 dimensions with dynamics $\ddot{\vec{x}}=-k\vec{x}$: you will see you can separate it ...


4

The easiest way to calculate escape velocity, is neglicting Earths rotation and assuming the object takes of in a radial direction. Then, indeed, you start from $$E = K_1 + U_1 = K_2 + U_2$$ where $K_1=\frac{mv^2}{2}$ and $U_1=- \frac{GMm}{r}$. Since the range of gravitional forces is infinity, you say (theoretically, not practically) that an object has ...


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Building on the responses from ACuriousMind and Gennaro Tedesco, I will make an attempt to provide a satisfactory, though not mathematically rigorous, answer. Question: Does there exist a nontrival non-Legendre transformation T such that the function defined by F(q,p,t)=T[L(q,q˙,t)] contains the full dynamics of the system? Yes, any invertible ...


9

It's the same reason that most airplane propellers have only two or three blades. As the blade moves through the air (or the air moves over the blade), it leaves a wake. If the next blade encounters the wake, it will be moving through disturbed air and will be less efficient. The more blades you have, the more likely each blade will pass through air ...


9

I worked at the Wind Energy Project Group at the TU Delft in the Netherlands for a summer. One of the most interesting ideas was a wind turbine which only had one blade, with a massive ugly blob on the other side as a counterweight. The idea was, "if we spend three times as much money manufacturing just one really good blade, could we make the system better ...


1

The final equation turns out to be: $$ l \ddot{\phi} - a \omega^2 \sin{\omega t} \cos\phi + g\sin\phi = 0$$ Now, as Aleksandar has suggested, all that remains is to put in the condition of small oscillations i.e. small values of $\phi$. Therefore \begin{align} \sin\phi &\approx \phi \\ \cos\phi &\approx 1 \end{align} Your equation reduces to $$ ...


2

This is the picture of your problem So point M movement is described by $x_M = a \sin(\omega t)$. This is a system with one degree of freedom and for the coordinate that completely describes this system we will use angle $\theta$. Let us describe $x$ and $y$ position of a pendulum at any moment $$x_m = x_M + r \sin(\theta)$$ and $$y_m = r \cos(\theta)$$. ...


2

Question: Does there exist a nontrival non-Legendre transformation T such that the function defined by F(q,p,t)=T[L(q,q˙,t)] contains the full dynamics of the system? Answer: any function that produces the equations of motion under some sort of rules that you state is an allowed function to describe the dynamics. In particular any function that you can ...


2

I don't have that text, but I can find the table of contents on the internet. Somewhere in that text (most likely chapter 5 on non-inertial reference systems), there should be a derivation that for any vector quantity $\boldsymbol q$, the time derivative of that vector in an inertial frame and a rotating frame that share the same origin are related by $$ ...


3

The statement is really about the transformation between inertial co-ordinates and co-ordinates fixed to the body. This is expressed by: $$D_t = d_t + \omega(t)\times\tag{1}$$ where $D_t$ is the "total" derivative, i.e. the time derivative in the inertial frame, $d_t$ is the time derivative in the frame fixed to the body. Since there there are no torques ...


0

You will want to plot the exponent on a linear scale: the log scale is more appropriate for the difference graph should you choose to use it. The graphs for the exponent from differences in q and p individually may be misleading. You may also want to increase the time scale 100-fold so the phase space parameters of the single oscillators sync up a couple ...


1

Comment to the post (v2): It is true that if $t\mapsto q(t)$ is a solution to the Euler-Lagrange (EL) eqs. for the action $$S[q]~=~\int_{t_i}^{t_f} \! dt L(q^1,q^2,q^3, \ldots; (\dot{q}^1)^2,(\dot{q}^2)^2,(\dot{q}^3)^2, \ldots)$$ of a time-symmetric Lagrangian, then the time-reversed path $t\mapsto q(-t)$ would also be a solution to the EL eqs. In fact, ...


0

I believe the correct approach in a go-kart is leaning outwards. I think this is useful because go-karts do not have a differential between the left and right wheels. So the wheels are forced to turn at the same rate. Because the turn radius is not the same for the inner and outer wheels there must be some slip. Leaning outwards facilitates this without the ...


1

When there is a pressure difference across a rope (or a "uniform load" of some kind) then the tension in the rope has to be such that the net force perpendicular to the rope exactly cancels the force. For this, the rope needs to be bent, and the combination of the radius of curvature and the tension gives you the net force according to this diagram (from the ...


2

In today's understanding of Nature, there is nothing completely isolated. So technically there will always be interaction with the surrounding, at least from a quantum physical perspective. Here vacuum is not empty i.e. it does allow for electromagnetic interaction and there will be heat loss due to these vacuum effects. Furthermore also the other concepts ...


1

First, I'm going to define some notation. I prefer the direct notation of modern continuum mechanics as presented in this book by Gurtin et al., so I won't be carrying around the indices that you use. Lastly, the following holds for small deformation, rate-independent plasticity with isotropic hardening. $\mathbf{T}$ = Cauchy stress tensor $\mathbf{T}_0 = ...


1

In mechanics, a mass $m$ experiences a force $\textbf{F}$ along some path $C$. The work done on the mass is given by $$ W = \int_C \textbf{F} \cdot d\textbf{r},$$ such that the energy of the mass increases by $W$. Positive work corresponds to energy being added to the system in question (which is inevitably taken from the surroundings). Edit: To answer ...


0

Due to the amount of momentum (and energy) transferred to you. Assuming a completely inelastic process (and an idealized one dimensional setup) you get from the conservation of momentum ($v$ is the initial velocty, $p$ the initial momentum, $v'$ is the final velocity, $M$ is the mass of the object hitting you, $m$ is your mass): $$ p = M v = (M + m) v' $$ $$ ...


0

In your last equation $U$ is a function or $n$ variables. Which of these does your $x$, $y$, and $z$ in that equation represent? To find the contribution to the potential energy due to the action of forces on a particular particle, one has to take the partial of the potential with respect to the variables representing the position of that particle. I ...


2

It's not taking partial derivatives with respect to an observed particle's position, but rather the space of all possible positions of that particle. Think of the potential energy as being defined prior to the particle having an actual path. Really, at heart, these things are defined on a phase space not on ordinary physical space.


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Not to mention, there are cases where the local extremum of the action isn't a physically realiziable path. Consider the plane with all the points satisfying $y > |x|$ removed. now, consider a start point of $(-2,1)$, and an end point of $(2,1)$ on this manifold, with the Lagrangian $\frac{1}{2}m\left({\dot x}^{2} + {\dot y}^{2}\right)$. Clearly, the ...


3

As demonstrated in this paper, the trajectory can never maximise the action but can in fact lie on a saddle point in cases where the potential has the appropriate spatial variation (at least partially repulsive) and where the final state is taken sufficiently far 'downstream' (beyond what these authors call the 'kinetic focus').


0

Rather than a book, consider the Susskind Physics lectures made available on Youtube as well as Stanford on iTunes via iTunes-University. You want to watch the latest version of the course lectures on Classical Mechanics. After a small bit of introductory material, Susskind gets right into the Lagrangian and Hamiltonian approach to mechanics using a simple ...


0

The best way to describe a rotating reference system is via a simple change of coordinates. If you have a description of a phenomenon in some set of inertial coordinates $\{t, x, y, z \}$, then you can obtain a description of its motion in a rotating reference frame with coordinates $\{t', x', y', z'\} $ by making an appropriate substitution. For example, ...


1

If your system is rotating about a pivot then the equation of motion is: $$ \frac{d^2\theta}{dt^2} = \frac{T}{I} $$ where $T$ is the torque and $I$ is the moment of inertia. The torque may or may not be a function of $\theta$ depending on your system. Simply calculate the total torque by adding the torques created by your two masses then solve the ...


1

Okay, that was the physics part. If your answer was that the windmill should be built as large as possible, then what are the common engineering problems that occur as we scale up? As large as possible is definitively the correct answer. The common engineering problems which comes, is that you don't have a big-enough crane to build it. It's practically ...


0

Practically also we can explain it.……Consider a plane or a table top.An object is placed at one of its edge.Now when we push the object perpendicular to the edge it will move in the direction of applied force i.e perpendicular to the edge of table top with velocity v0 for example.…………………………………………………… Now we take another force which is parallel to that ...


0

Tl;dr: Vector components act like sides of a triangle, where the length of the vector coincides with the triangle's hypotenuse. Mathematically, this follows from the fact that velocity in the plane is an element $v \in \mathbb{R}^2$. Consider a Euclidean coordinate system whose origin coincides with the spatial location of the particle in motion$^1$. Let ...


0

To record ACuriousMind's comment: Many "angular quantities" are actually pseudovectors rather than vectors, for the difference, see this question. Actually, he's being too modest: go straight to his answer to the question "When is it useful to distinguish between vectors and pseudovectors in experimental & theoretical physics?". The only thing ...


1

It seems that the exact exit conditions of a Vortex doesn't influence too much on the Vortex functionality. For example a Vater Vortex; it doesnt matter if you take the water out from the top, or from the bottom. The basical funcionality of the Vortex remains very much the same. Please look this film Secondary flows. in 1:20-2:55 and 11:00-12:00 as a ...


1

This is going to take a bit, so be prepared. Also, some of what I say will not be exactly correct, in order to get the larger point across, but bear with me. First, the Rutherford model really did not speak to the question of emission lines. It simply noted that, on the basis of Rutherford's scattering experiments, all of the protons in an atom had to be ...


2

The main idea. It makes sense that the "redundant" constraint modifies the result because when you introduce the Lagrange multiplier, you not only restrict your attention to the paths that satisfy the constraint, but you also restrict your attention to variations that preserve the constraint. In more detail. For clarity, let $\mathscr P$ be the set of all ...


0

Surely this just depends on the potential well in which the particle finds itself. If the function you describe is the solution of the equation of motion, then that will indeed be the motion of the particle. The function you wrote will give rise to a path that "never ends" - the particle keeps slowing down, but never stops. Of course in practice there are ...


0

Let us suppress explicit time dependence $t$ from the notation in the following. Hamilton's eqs. are the Euler-Lagrange (EL) eqs. for the so-called Hamiltonian Lagrangian $$\tag{1} L_H(q,\dot{q},p)~:=~ p_i\dot{q}^i-H(q,p).$$ In other words, the solutions to Hamilton's eqs. are stationary points for the Hamiltonian action $$\tag{2} S_H[q,p]~:=~\int \! ...


0

Hydrogen is a stable atom and its spectrum is discontinuous and was fitted mathematically by the Rydberg series. Classical electromagnetic theory predicts that the electron, due to its continuous acceleration on the orbit, would emit photons continuously and would drop to a lower energy until it fell on the proton and neutralized it. To explain the ...


1

Distinct spectral lines (rather than an infinite range of energies and hence white light) are a result of only specific transitions being allowed, which implies a set of distinct and unvarying energy levels. Classical models of the atom before quantum mechanics couldn't explain why there should only be certain electron energy levels.


0

Your presentation is a little confused. I would read your book again very carefully, taking care to understand the meaning of each variable. The integral that you write for potential energy is the definition of potential energy, if $F$ is taken to be an internal force, that is, a force between two objects within the system. For example for free fall at ...


1

The whole energy-concept is a reformulation of Newtons laws. Starting from $\vec{F}=m.\vec{a}$, you could wonder about the effect of a force during a displacement. You call the concept 'work' and do the math $$W=\int_{\vec{x}_1}^{\vec{x}_2} \vec{F} d\vec{x} = \ldots = \frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2$$ To save yourself some work you define $$T = ...


4

Yes, there exists a Legendre transformation from $g(p)$ to $f(x)$: $$ f(x)=p(x)x-g(p(x)) $$ with $x=dg/dp$. Here the notation $p(x)$ means $p$ written in terms of $x$. In your case, the Hamiltonian is a function of $p$ and you are transforming it to a function of $\dot{q}$, so you must use Hamilton's equation to get the velocity: $$ \dot{q}_i=\frac{\partial ...


2

First of all, the hamiltonian contains the coordinates $q_i$ and their momenta $p_i$. You have to calculate the velocities $\dot{q}_i$. For that, you'll need the Hamilton-Jacobi equations $$\dot{q}_i = \frac{\partial H}{\partial p_i}$$The Legendre transform, as noted in the comments, is involutive, so the lagrangian is just the Legendre transform of the ...


7

That you get two answers for time $t$ is perfectly fine! How the is it possible that an item with a constant acceleration can displace over the same distance in two different time periods? Remember, that the $17$ meters you put in is not distance, but a position. The object passes this position at $t1$, then it decelerates, speeds down, gets a negative ...


2

I haven't checked your calculations, but the reason you can have 2 solutions is as follows. After 4.3 seconds, the object has travelled 17m from its start point, say heading along positive x axis. After 10 seconds it reaches a halt and starts to go backwards, and after 15.6 seconds, it's back at the point 17m from the start.


2

There's nothing wrong here about the equation giving you two instants. Say, your object is moving in the positive $x$ direction, it started with $v=5$ , but it's slowing down. "Slowing down" basically means that the direction of the acceleration is opposite to your velocity, that is, $a$ is in the negative $x$ direction. The object will cover a certain ...


1

I believe that you must study (not simply reading but trying to prove) that beautiful story of angular momentum quantization starting from the non-commutativity of the coordinate and momentum of a particle along any axis. The following is a starting point (found in any introductory book on Quantum Mechanics). In Classical Mechanics the angular momentum ...


0

Secondly, if we consider Maxwell's equations, which are not invariant under Galilean transformations, but we require them to hold in all inertial frames, doesn't it immediately follow that the speed of light has the same constant value in all inertial frames from this assumption (given that Maxwell's equations imply a constant speed of light). Why is it ...


1

Why is it given as an axiom of special relativity? In a nutshell: because of the Michelson-Morley experiment. This ought to have detected a variation in the speed of light caused by our motion through space. But it didn't. So Einstein reasoned that speed = distance / time, and that if the speed didn't change, your the time did. And then if your time ...


0

Other answers are right, but let me put it without math: Water can't come out of the bottle if air can't go in. (Except: if the water in the bottle is so tall that water can come out even if air can't go in.)


2

Water stops draining from the jar into the dispenser once it forms an interface as draining of more water would result into the formation of a vacuum in the jar because no air can rush into the jar to displace the water as it has an interfacial-lock. Consider the water level above interface $= h$, water level below interface $= x$ now $$P_{surface}= ...



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