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I assume you know the equation of motion$^{\dagger}$ of a simple (small angle) pendulum, call it $\theta(t)$. The angular acceleration $\alpha$ is given by: $$\alpha=\frac{{\rm d}^2\theta}{{\rm d}t^2}$$ Let the period of oscillation be $T$, then the first 'end' of the oscillation is reached at $t=T/4$, plugging this in the equation of motion will then give ...

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$$\det\left[\frac{df_h}{dq_k}\right] \neq 0 \tag{1}$$ because we are dealing with a bijective and bi-differentiable transformation of coordinates (see below). Hence, if $$\left[\frac{\partial L'}{\partial \mathbb q'} (\circ) - \frac{\mathrm d}{\mathrm dt} \frac{\partial L'}{\partial\mathbb{\dot q'}} (\circ)\right]\cdot\frac{\partial\mathbb f}{\partial\... 4 I cannot imagine what discrepancy you are talking about. The EOM yields$$ \partial_ r \left (\frac{p_r^2}{2m} + \frac{p_\theta^2}{2mr^2}\right )=-\dot p_r, $$which amounts to$$ \frac{p_\theta^2}{mr^3} = \dot p_r. $$It is quite misleading to think of it as "the" radial force, as it willfully skips the centripetal acceleration. The radial ... 0 In quantum mechanics the volume integrated momentum of a free particle is always h/\lambda. Light consists of massless particles and Maxwell's equations are what the Schrödinger equation is for massive particles. The formula p=m\gamma v holds for any mass except zero but in all cases, p=\sqrt{E^2-m^2c^4}/c. 7 I would agree with John's suggestion that the most "fundamental" definition of momentum is the quantity that is conserved when the Lagrangian is invariant under spatial translations \mathbf{x}\rightarrow\mathbf{x}+\mathbf{s}. However, do terms like Lagrangian even make sense when we try to include massless objects? Yes, absolutely. In fact the ... 3 Would the normal force then equal the net downward force on the block due to both its weight as well as the atmospheric pressure? There's still air underneath the block, so air pressure acts on the block in all directions, cancelling out. It's different for a suction cup. In that case a (partial) vacuum is created between the cup and a smooth surface and a ... 3 A treatment in which m\ne0 doesn't need a separate discussion: The four-momentum p^\mu and four-force f^\mu satisfy f^\mu=\frac{d}{d\tau}p^\mu (which reduces to a conservation law in the absence of an external four-force) and p^\mu p_\mu=m^2c^2 if we keep dimensions consistent in the convention p^0=E/c. (Personally, I'd recommend setting c=1 to ... 0 Just adding to the previous answers, you take a harmonic oscillator potential energy and differentiate it twice, and you get the spring constant. So it only makes sense to take the second-order term of the Taylor expansion of the Lennard Jones potential evaluated around the potential minimum. Here's a post that shows how to arrive at the minimum of the ... 0 I've described the method to calculate the frequency of oscillations for arbitrary potentials in my answer to this question: Calculating Frequency of Oscillations About a Stable Equilibrium Point. I've adapted my answer here from the one there. The basic idea is this: First, find the (local) minimum of the potential you are interested in, using the standard ... 0 In the n-exp notation we write for the Lennard-Jones potential:$$U_{LJ}(r)=\varepsilon\Big[\Big(\frac{r_0}{r}\Big)^{2n}-2\Big(\frac{r_0}{r}\Big)^n\Big]$$where n=6 and \varepsilon is the bonding energy. Applying an harmonic approximation at the potential minimum (at {\displaystyle U(r_{m})=-\varepsilon }{\displaystyle U(r_{m})=-\varepsilon }), the ... 1 If you have a general oscillator with restoring forces and damping forces:$$F_{rest}=-kx,\text{ } F_{damp}=-b\frac{dx}{dt}$$Such that the equation of motion would be:$$m \frac{d^2x}{dt^2}=-b\frac{dx}{dt}-kxm \frac{d^2x}{dt^2}+b\frac{dx}{dt}+kx=0$$This equation has solution of the form :$$Acos(\omega x + \theta)e^{t/\tau}$$Where you can ... 0 It is a rather fundamental mathematic concept. The roots of first derivative are points of local extrema, where forces are vanishes. But these extrema could be local minimum (a stable equilibriium point), or local maximum (a unstable equlibrium point). The criterion of checking local maximum or minimum or saddle point (in higher dimension space) is to check ... 2 The dissipative contact force during the collision itself is not described by an 1/r potential, so there's no reason why the virial theorem \langle T \rangle=-\frac{1}{2}\langle V \rangle for 1/r potentials should hold. 1 However, my teacher says that in classical mechanics light cannot be both and for quantum mechanics we can. Does this mean we need more trust in the validity of quantum mechanics because there is a sense of ambiguity? We only trust that we know more about nature than a hundred and fifty years ago, not because of ambiguity. We have gone to much smaller ... 4 Doesn't look too good to me.$$$$0 Yes, trajectories of non-autonomous systems can cross in phase space. One way to understand why is to consider that the system's explicit time dependence means that the time t is needed to unambiguously determine the state of the system, i.e., the "complete" phase space includes an extra dimension t besides, say, (x,y): so what appears as a ... 0 As the research paper rightly points that for non-autonomous system trajectories may intersect. The gist of this concept lies in the definition of autonomous and non-autonomous system. For a non-autonomous system: And for an autonomous system: Thus for a non-autonomous system, the equation governing the evolution of system changes with time due to the ... 0 Canonical momentum is the momentum. Kinetic momentum is only part of the momentum. The other part is the potential momentum. For example, the magnetic part of the Lorentz force does not conserve kinetic momentum. Only the total, canonical, momentum mv+qA is conserved. A problem could be that the total momentum is not gauge invariant. There are two ways to ... 0 This is a question I've been playing with myself a lot lately! But the simplest way to describe it is that they are simply different quantities. The canonical momentum is the derivative of the Lagrangian function with respect to the generalized velocities. The kinetic momentum is essentially just the "normal" momentum of a free particle with no ... 1 TL;DR: Stepping function:$$ \text{Theta}_{n+1} = \theta_{n+1} = \theta_n + \dot{\theta_n}\Delta t  \text{Angular acceleration}_n = \alpha = \ddot{\theta}_n = - \frac{mg}{I} ( \frac{x}{2} \cos \theta_n - \frac{y}{2} \sin \theta_n )  \text{Angular velocity}_{n+1} = \omega = \dot{\theta}_{n+1} = \dot{\theta_n} + \ddot{\theta_n}\Delta t $$Free body ... 2 The key when changing discrete sums to integrals is to ask oneself what things in the integral need to be small and which ones can be large. In the case of the moment of inertia around an axis, the formula$$I=\sum_{i=1}^Nm_i r_i^2$$is valid for the case of N point particles, with m_i the mass of the ith particle and r_i its distance to the axis. If ... 0 Because the ball is exerting a different force on the two ropes. Assume the ball has mass m and the two ropes are attached at two angles \theta_1 and \theta_2 with respect to the vertical line connecting the ball to the ceiling. Then, the forces acting on the ball are three: gravity (m\textbf{g}), directed towards the bottom and the two tensions \... 0 The tension in each "suspender" depends on the final, equilibrium geometry. For ropes, that depends on the lengths of the ropes and the geometric relationship between the attachment points on the ceiling. If you have two springs rather than non-stretching ropes, the lengths will change until the the two horizontal force magnitudes are equal. The ... 2 OP's sought-for derivation of the worldline formalism essentially boils down to the observation that the point-mechanical propagator integrated over a Schwinger proper-time parameter reproduces the corresponding field-theoretic propagator, even in the presence of various background fields, see e.g. Ref. 1. References: O. Corradini & C. Schubert, ... 1 I know that the friction is μmg Not always true. \mu mg corresponds to the maximum friction, f_{\rm max}. The static friction force f can be anywhere between 0 and f_{\rm max}:$$0\leq f\leq f_{\rm max} = \mu N$$To solve the problem, first, write down the net force and torque equations$$ma=F-f \qquad \text{and}\qquad I\alpha = fR$$Using v=\... 2 The answer is randomness and symmetry, as in the post you linked. In a container without a lid, air acts much like a lid would have. When air molecules bounce off the walls of the container, molecules in the wall vibrates some, but they are held in place by atomic bonds. At the open "lid", air molecules from the container bounce off air molecules ... 1 One has not to confuse forces and effects of the forces. The resultant force's effect is the motion of the system to which the force has been applied. This motion is certainly different if a given force acts for a longer or shorter time. But the second Newton's law says that the acceleration at a given time t is given once one knows the force at the same ... 0 Since the movement is in a plane, you can indeed assume the motion to be in the xy-plane i.e. the \theta=\pi/2-plane. The Lagrangian for the system constrained to the \theta=\pi/2-plane is given by setting \dot\theta=0 and \theta=\pi/2 in the given Lagrangian. 0 Yes, the tension in the spring is changing continuously. At any short duration of time the increment in the stored energy in the spring can be written as: $$\Delta{W} = Tdx = {\frac{\lambda x}{L}}dx$$ Now, if the limit is from 0 to x. then, W= \int_{0}^{x}\Delta{W} = \int_{0}^{x} {\frac{\lambda x}{L}}dx = {\frac{\... 1 A rigid material or object (the limit of stiffness → infinity) can have no hysteretic losses because deformation work can't be performed on it anyway. However, all real materials with finite stiffness exhibit internal friction/damping. It's therefore reasonable to assume that—all else being equal—hysteretic losses decrease with increasing stiffness. 1 The elastic potential energy of a spring is$$\mathrm{EPE} = \frac{1}{2}kx^2$$where k is the spring stiffness (and I believe is related to your \lambda by \lambda = kL) and x is the spring extension. Let y_1 be the initial height of the mass, and y_2 be the final height. Since we know that the mass has no kinetic energy at either of these ... 1 Comments to the post (v3): It seems that OP's principle (4) only reproduces the dynamical side \sum_i{\bf F}_i of Newton's 2nd law, not the kinematic side m{\bf a}, so it only applies to creeping/overdamped motion. Also OP's principle (4) treats velocity \dot{x} as independent of position x. In a conventional variational action principle, they are ... 1 Let's forget about non-inertial reference frames and fictitious forces. The key thing to note is that \frac {v^2} R is not a horizontal force, it is a horizontal acceleration. So when we apply Newton’s second law the \frac {v^2} R terms are on the opposite side of the F=ma equation to the forces. Because the acceleration is horizontal, it has zero ... 0 You can use the constraint that they should reach that point simultaneously for collision to happen. that is the time taken will be equal. 4 Quantization is a huge topic, so we will only sketch the basic idea. Let the classical phase space be a symplectic manifold (M,\omega). As OP already knows, Poisson brackets should, roughly speaking, be replaced by commutators, cf. e.g. this Phys.SE post. Also we know from the HUP that we can only measure commuting observables simultaneously. A basic idea ... 0 That is one constraint that you got now. In order to find a relation between u and v, \theta has to be eliminated. This can be done with a second constraint which uses the range. Since ball 1 reaches the highest point at half the range and given that D=10, the next constraint can be$$\frac{u^2\sin(2\theta)}{2g} = 10.$$Using this relation \theta ... 0 the only practical way to prevent gas loss through the balloon material is with a thin layer of metallization applied to the balloon wall, and by making the wall nonelastic so it will not shed the metallization when it stretches. This solution is already used commercially for balloons by making them out of mylar film to which a thin (typically 0.0003") ... 1 The term P in your expression is an integral of Force by d\dot{x} (by treating x as a constant):$$ -\int d\dot{x} F_{total} = - \int d\dot{x} \{-bx - \frac{dU}{dx} \} $$Treat x as a constant ans carry out this integral over d\dot{x}$$ =\frac{1}{2} b\dot{x}^2 + \frac{dU}{dx} \dot{x} = P $$Thus, it recovers the force if your partial derives ... 2 The \mathbf p which appears in Newton's second law is a function of one variable, so the derivative which appears is just the derivative \mathbf p'(t). There is no notion of partial derivatives because there is only one variable, and there is no such thing as the "total derivative" of a function all by itself. This is how it works: If you ... 1 Newton second law$$\frac{d}{dt}(m\,\vec v)=m\,\frac{d\vec v}{dt}=\vec{F}\tag 1$$the velocity ~\vec v is in general depending on the generalized coordinates ~\vec q~ the velocity of the generalized coordinates ~\vec{\dot{q}}~ and the time ~t \Rightarrow the total derivative is$$d\vec v=\frac{\partial \vec v}{\partial \vec q}\,d\vec q+ \frac{\...

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Assuming the scale is attached to the pulley then it is measuring the force $F$ that is supporting the pulley. Considering the weights and pulley as a single system, the external forces acting on the system are $F$ and its total weight $W$ - the tension in the rope $T$ is an internal force and can be ignored. But note that the system is not in equilibrium. ...

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There is a second factor, the nature of the string. Consider a "U" shaped section of the string. We move the top left end of the "U" - perhaps push it in some direction or another. The surface is frictionless, but the string has elasticity and its present shape. Will the top left part of the "U" (and oerhaps left side) move but ...

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Starting with a lower cylinder (and assuming that the given v and 2v are measured relative to the base), it is clear that the center of mass is moving to the left with a speed of v/2. Combine that with a tangential velocity of v/2 to get v at the top and an angular velocity of v/(2R). For an upper cylinder $v_c$ + $v_t$ = 2v and $v_c$ - $v_t$ = -v. (Where ...

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As a rule of thumb, physically meaningful quantities use total derivatives. Partial derivatives are mostly mathematical tools that help us express total derivatives, since the implicit functional dependencies that even allow us to speak of a difference between partial derivatives and total ones aren't inherent features of the physical system. Rather, they ...

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It is total derivative of momentum $p=mv$ w.r.t time: i.e. $F=d(mv)/dt=mdv/dt=ma$

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The Hamiltonian is: $$H=\boldsymbol p^T\,\dot{\boldsymbol{q}}-L(\boldsymbol q~,\boldsymbol p)$$ with $$L=L_0(\boldsymbol q)+\dot{\boldsymbol{q}}^T\,\boldsymbol a+\frac 12 \dot{\boldsymbol{q}}^T\,\boldsymbol T\,\dot{\boldsymbol{q}}$$ Step I $$\boldsymbol p=\frac{\partial L}{\partial \dot{\boldsymbol{q}}}=\boldsymbol T\,\dot{\boldsymbol{q}}+\boldsymbol a$$ $\... 2 Qmechanic is correct, eq. (1) has a typo and should have a factor of$1/2$in front of the$T$. If you don't find this satisfactory, then you might also argue that the$T$in (1) has not been properly defined yet, so that multiplicative factors are irrelevent. In eq. (2), we then choose$\mathbf{T}$to have a coefficient of$1/2$. This is handy for ... 0 Friction is a non conservative force. A conservative force can be written as the negative gradient of some potential, namely $$\vec F_{\rm conservative} = -\nabla U$$ Since friction is non-conservative, therefore it cannot be described as a gradient of some potential. Gravity on the other hand is a conservative force. So$V\$ can indeed be a gravitational ...

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The question arises, if at the slackened state there is no tension in the string, what causes the free end to move? The answer is in the statement preceding this one, "Now we give the puck some velocity" Giving the puck velocity means there was a force applied to the puck, even if briefly. When the force is removed, the puck continues at constant ...

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