# Tag Info

## New answers tagged classical-mechanics

0

Since the total mechanical energy is conserved, from which we get $KE_i+PE_i=KE_f+PE_f$, it should be more clear now that $\theta_0$ is the angle at which you raise the pendulum (i.e., that term represents the $PE_i$). The terms on the left side your first centered equation are the final kinetic and potential energies, after some time $t$. In effect, you are ...

0

Noether's theorem doesn't really discriminate between relativistic and non-relativistic theories. As long as there is an action formulation and a symmetry, she will provide a conservation law via the standard Noether procedure. However, the question formulation (v1) touches upon other issues that are far from trivial, such as, e.g., The status of an ...

0

If the Lagrangian is non-singular so that the Legendre transformation to pass from Lagrangian to Hamiltonian formalism is well defined, the answer is Yes. If a quantity is conserved in view of Noether's theorem in Lagrangian formulation, passing to the Hamiltonian formulation it turns out to be the generator of a canonical transformation that preserves in ...

2

Hints to the question (v1): Let us parametrize the problem wrt. an arbitrary world-line parameter $\tau$ (which does not have to be the proper time). The Lagrange multiplier $\lambda=\lambda(\tau)$ depends on $\tau$, but it does not depend on the canonical variables $x^{\mu}$ and $p_{\mu}$. Similarly, $x^{\mu}$ and $p_{\mu}$ depend only on $\tau$. The ...

3

Kepler's 3rd Law: The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit. If we take Jupiter's semimajor axis of 5.2 AU (avg value), then objects at 2.5 AU, 2.82 AU, 2.95 AU, 3.25 AU have orbital periods shorter than Jupiter by a factor of 3:1, 5:2, 7/3 etc. The fact that the ratios for the deepest ...

1

The Hamiltonian can be used to describe an evolution of the "density in phase" of a system of N bodies. The density in phase is a conserved quantity for a system in equilibrium by Liouville's Theorem. The position and momenta can describe any general intensive parameter. Gibbs used this approach to derive statistical mechanics. This approach of the concept ...

2

It took me quite some time to clearly understand the experiment you're describing. Actually, pouring a full bottle in a container is a quite intriguing thing. Consider the following starting configuration : This of course is an unstable situation, as the pressure $P_0$ cannot be at the same time the pressure of the air in the bottle, and the atmospheric ...

2

The definition of temperature in the kinetic theory of gases emerges from the notion of pressure. Fundamentally, the temperature of a gas comes from the amount, and the strength of the collisions between molecules or atoms of a gas. The first step considers an (elastic) impact between two particles, and writes $\Delta p = p_{i,x} - p_{f,x} = p_{i,x} - ( - ... 0 Conceptually, the reason they are attached to walls is to remove rigid body motion from consideration. When they are tethered to the wall, there is no translation or rotation about the center of mass of the system. So, when you break the connections to the walls and they are now free to drift through space, the number of admissible solutions increases. ... 3 Your question actually is one of the most important questions in analytic mechanics. This is because, when you explicitly write the Eulero-Lagrange equations for any constrained system with$n$degrees of freedom and Lagrangian of the form: $$L(t, {\bf q},\dot{\bf q}) = T(t, {\bf q},\dot{\bf q}) - U(t, {\bf q},\dot{\bf q})$$ where$T$is quadratic in ... 0 If C is the center of mass of the ball, and A is the contact point then the velocity of the ball at the contact point is $$\vec{v}_A = \vec{v}_C + \vec\omega \times (\vec{r}_A-\vec{r}_C)$$ If you have infinite friction then you have no slipping which means$\vec{v}_A =0$. If there was slipping then only the speed in vertical direction should be zero ... 0 The determinant is fairly easy to calculate. You know already, essentially, the eigenvalues of the stiffness matrix; more accurately, you know the eigenvalues of the matrix$\mathbf{m}^{-1}\mathbf{k}$, because the$\omega_i$are zeros of the equation $$0=\det(\mathbf{m}^{-1}\mathbf{k}-\omega^2).$$ (The more aesthetically minded would replace ... 1 Consider the positive quantity$X = (\omega - \omega_0)^2 (\omega + \omega_0)^2$. Let us make the approximation$\omega \approx \omega_0$: In the first factor: we get$X_1 = 0$and the relative error$\epsilon _X = |\frac{X_1-X}{X}| = 100 \%$. In the second factor: we get$X_2 = 4 \omega _0^2 (\omega - \omega_0)^2$and$\epsilon _X = | \frac{X_2-X}{X}| = ...

0

For a system of point particles, the definition $$\vec{L}_i=\vec{r}_i\times\vec{p}_i$$ is always true; it's just a definition. I see no reason why that won't work here. The only choice you have to make is where to measure the position vectors $\vec{r}_i$ from. A particularly convenient position from which to measure $\vec{r}_i$ is the rotation axis. One ...

2

Let there be $n$ coordinates $q^j$. Ref. 1 is discussing in Section 2.4 a type of non-holonomic constraints that is known as semi-holonomic constraints. However we interpret OP's question (v2) as mostly being about counting independent degrees of freedom in constrained systems, and not so much about semi-holonomic constraints per se. Therefore, to gain ...

1

This is a fact about the hamiltonian compared to the lagrangian which I find not trivial (and worth to keep in mind). Suppose that the lagrangian $L$ and hamiltonian $H$ are cyclic with respect to some coordinate $q_1$. Then we have a theorem (cfr. [1]): The evolution of the other coordinates $q_2,...,q_n$ is the one of a system with $n-2$ independent ...

2

I) The restricted$^1$ transformation (RT) $$\tag{1} (q,p)~\longrightarrow~ (Q,P) ~:=~(q, \sqrt{p} - \sqrt{q})$$ of OP's professor with inverse RT $$\tag{2} (Q,P)~\longrightarrow~ (q,p) ~:=~(Q, (P+ \sqrt{Q})^2) ,$$ and with Hamiltonian $H=\frac{p^2}{2}$ and Kamiltonian $K=\frac{p^3}{3}$ is indeed interesting. Apparently we should assume that $p,q,Q\geq ... 1 If$\vec{p}$the vector connecting the center of mass of b1 to the center of mass of b2 then you must have $$\vec{v}_2 = \vec{v}_1 + \vec{\omega}_1 \times \vec{p} \\ \vec{\omega}_2 = \vec{\omega}_1$$ $$\vec{a}_2 = \vec{a}_1 + \vec{\alpha}_1 \times \vec{p} + \vec{\omega}_1 \times \vec{\omega}_1 \times \vec{p} \\ \vec{\alpha}_2 = \vec{\alpha}_1$$ 1 Your solution looks fine to me. Yes: the angular momentum is preserved in the horizontal plane (the weight is vertical and the reaction of the sphere surface is a central force) so your first relation is fine, just remember that$\theta$is not the vertical angle, but lies on the plane tangent to the sphere at point B. There are two kinds of rotational ... 2 One way to see the relationship of Hamilotian classical mechanics and Quantum mechanics is not to look for a direct translation of Hamiltionian -> Quantum Hamiltionian (which exists: Geometric Quantization), but consider the reverse relationship. Given a Hamiltion operator and evaluating it on wave functions of the form$e^{\frac{i}{\hbar} \phi}$(which can ... 0 First, you are right: in this problem angular momentum around the vertical axis is conserved. This is because all forces acting on the particle have no azimuthal component. Note, that even if we assume that the particle is rolling on the bowl surface without slippage, the angular momentum of its own rotation would be of negligible compared with the angular ... 0 I do not have a solution, just some steps to get there. I have parametrized the problem with spherical coordinates,$\varphi$is the azimuthal angle (around the hoop),$\psi$is the nutation angle (drop from horizontal plane) for a position vector $$\vec{r} = \begin{pmatrix} r \cos \varphi \cos \psi \\ -r \sin \psi \\ -r \sin \varphi \cos \psi ... 2 The canonical (Hamiltonian) formalism offers one of the main paths for quantizing gravity. General Relativity can be expressed in terms of the ADM 3+1 decomposition of spacetime: http://en.wikipedia.org/wiki/ADM_formalism And Hamiltonian's underlie quantum mechanics: http://en.wikipedia.org/wiki/Hamiltonian_(quantum_mechanics) Not only does this provide ... 6 First of all, Lagrangian is a mathematical quantity which has no physical meaning but Hamiltonian is physical (for example, it is total energy of the system, in some case) and all quantities in Hamiltonian mechanics has physical meanings which makes easier to have physical intuition. In Hamiltonian mechanics you have canonical transformations which allows ... 14 Some more comments to add to user1504's response: For a system with configuration space of dimension n, Hamilton's equations are a set of 2n, coupled, first-order ODEs while the Euler-Lagrange equations are a set of n uncoupled, second-order ODEs. In a given problem it might be easier to solve the first order Hamilton's equations (although sadly, I ... 4 Cool question! Thanks to user lionelbrits for his answer that prompted me to pull out my mechanics books and check the definitions of "canonical transformation" given by different authors. If you look in Goldstein's classical mechanics texts in the section on canonical transformations, then you'll find that canonical transformations are essentially defined ... 21 There are several reasons for using the Hamiltonian formalism: 1) Statistical physics. The standard thermal states weight pure states according to Prob(state) \propto e^{-H(state)/k_BT}. So you need to understand Hamiltonians to do stat mech in real generality. 2) Geometrical prettiness. Hamilton's equations say that flowing in time is equivalent to ... 2 I) In case of a point particle with mass m (and no moment of inertia), the best one can do seems to be to model the friction/drag via a Rayleigh dissipation function {\cal F}(v^2) with a friction/drag force$$\tag{1} {\bf F}_f~:=~-\frac{\partial {\cal F}(v^2)}{\partial {\bf v}} ~=~-2{\cal F}^{\prime}(v^2){\bf v},$$i.e. the Lagrange equations read ... 1 The original coordinates satisfy the equations of motion when the integral of p\, \dot{q} - H(p,q) is minimized, and the new coordinates satisfy the equations of motion when the integral of P\, \dot{Q} - K(P,Q) is minimized. There is no requirement that H and K be numerically equal. The transformation is canonical if the Poisson bracket remains ... 0 The translational accelleration will be the force divided by the mass. The cross product of the force vector with the vector from the touch point to the center of mass is the torque applied to the object. 2 The problem with the phase space flow in Hamiltonian mechanics is that the flow itself is non-dynamical, that is, the flow is immediately defined for a given Hamiltonian, so there is no independent equation governing its evolution. Thus, Liouville equation is simply a transport of a scalar variable in a given flow. So, dimensional analysis of the flow ... 2 I) At least three different quantities in physics are customary called an action and denoted with the letter S: The off-shell action S[q;t_i,t_f], The (Dirichlet) on-shell action S(q_f,t_f;q_i,t_i), and Hamilton's principal function S(q,\alpha, t). For their definitions and how they are related, see e.g. this Phys.SE answer. II) OP's ... 4 Your mistake is to assume that the water will stop "no matter how long a bottle you take". It will not - you just need a longer bottle than you expect. To be precise, you need a column of water 10 meters high to counteract atmospheric pressure. 1 Both approaches are equally correct in this case. F = mv^2/R is just a consequence of the law for rotational motion, which says \tau = I\alpha (Torque = Moment of Inertia * Angular acceleration). The former formula may be used in case the objects in consideration are point masses. But the latter, more general version of the formula is applicable for ... 0 One classic book along these lines is Mathematical Methods of Classical Mechanics. V. I. Arnold. Graduate Texts in Mathematics vol. 60, Springer, New York, 2000. Available e.g. here. This book is mathematically very formal and very clear; I loved it when I took analytical mechanics because it avoids the phycisists' smudges of rigour and presents one ... 0 This answer contains some additional resources that may be useful. Please note that answers which simply list resources but provide no details are strongly discouraged by the site's policy on resource recommendation questions. This answer is left here to contain additional links that do not yet have commentary. W. M. Oliva. Geometric mechanics (Lecture ... 0 Here there is an error in your angular momentum calculation.L=mvr is valid only if that is a point size particle.Here it is a cylinder,so angular momentum L about your point of contact should be$$L=L_{cm} +I_{cm}\omega$$where L_{cm}=angular momentum of CM about your orgin$$and$$I_{cm}=moment of inertia about CM So applying angular momentum ... 0 Actually, if you know that your current state is N(n)=16, there's no way for you to discern under that law whether N(n-1)=4 or N(n-1)=-4. There isn't a constraint stating that N(k)\ge0 in the problem statement. 1 Is it because there are two arrows into each state (which happen to not be shown in the solution)? Both N(n) and -N(n) go to N(n+1)=N^2(n) under time advancement. Thus, under time reversal, you're not sure one to go back to. 0 It looks like the friction force will decelerate the cylinder until (at some time t) there is no sliding, just pure rolling on the surface. Until then the friction force will equal \mu mg (the sign depends on the direction of the axis). Therefore, \mu m g t=m(v_0-v_f), where v_f is the final velocity of the center of mass. On the other hand, we can ... 1 The Euler Lagrange equations just give the differential equations that determine the motion of the object. The end points are boundary conditions for the differential equation. The differential equation which the brachistochrone curve satisfies will have its constants fixed so that it reduces to a line when you give it the boundary conditions of the object ... -3 Moment is bending due to linear force and the distance from the axis is perpendicular whereas in torque rotation takes place beyond 360 degrees. 0 When you throw a ball into the air, there are two times it will have a certain height s. Once going up and once coming down. That is what the two times represent. Both are valid. 0 A scale has two options to work. a) It either acts as a pendulum with gravity counteracting the weight imbalance, or b) there is a return spring that tries to level the scale if it is not horizontal. Either way the effect is the same. The angle is proportional to the imbalance. In your case the details do not matter, and so you get to pick the design (the ... 2 If we have some coordinates q_i and some momenta p_i, then a generator of a transformation is defined as a function g(q_i, p_i). By definition, this generates the transformation$$q_i \to q_i + \epsilon \frac{\partial g}{\partial p_i}p_i \to p_i + \epsilon \frac{\partial g}{\partial q_i}$$So if we want the generator of translations, we want ... 2 This is similar, but not quite identical to Newton's cradle, with the difference being the heavy objects placed on the middle coins. To explain things, first consider the simpler case where there is no heavy object on top of the coins, and suppose the 5 nonmoving coins in "frame 1" are separated by a distance L. When two objects of mass m and ... 1 There is a thresh-hold below which the ball will not bounce determined by the mass of the ball and gravitational acceleration (force of gravity on the ball in other words) In order for the ball to bounce (meaning physically lose contact with the surface of the ground) the compressive force created by the impulse of impact (the point at which velocity is ... 1 It's not missing, it's in the \ddot{R}(t) matrix. It doesn't show up on its own when you do the calculation with matrices instead just vectors however. Vector equations The first equation should be written$$\mathbf{x}_A(t) = \mathbf{\phi}(t) \times \mathbf{x}_B(t)$$(just reverse where you have A and B) since given the position in frame B you ... 1 When the (inertial) mass is zero, then the acceleration can be non-zero for zero force. This is similar, conceptually, to what has been discussed recently regarding an ideal conductor. Consider Ohm's Law:$$V = IR$$Now, what if$R = 0\$ as is the case with an ideal conductor? Clearly, the voltage must be zero for any current. The current through the ...

1

I agree with Brandon Enright's comment. But even if there were massless ropes, if m=0 and F=0, then F=ma still would hold for any finite a.

Top 50 recent answers are included