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The problem is equivalent to 4 spheres colliding simultaneously, where top sphere center is at $60^o$ relative to the $x'x$ axes (same goes for bottom sphere): We'll name them: sphere A (dark blue), and spheres 1, 2, and 3. During the collision the spheres will behave like springs with an infinite hook constant. The forces on the spheres will be ...

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I) Disclaimer: In this answer we will use the (traditional) physicist's definition of tensors using indices and their transformation properties under coordinate transformations. Moreover, let us suppress time dependence $t$ for simplicity. II) Let the manifold $Q$ be the configuration space. The Lagrangian $L:TQ\to \mathbb{R}$ transforms as a scalar ...

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Here is how you deal this this problem as a system of equations. For each contact pair assign a normal direction $\hat{n}_k$ and and impulse $J_k$. The possible contacts are AB, AC, and AD. We can introduce symmetries and simplifications later. The initial velocity if body A is $v_A$ along the horizontal axis, and after the collision it is $v_A + \Delta ... 4 There are two primary factors that allow the cochlea to isolate frequencies. These are generally referred to as passive and active properties: tl;dr version: The passive properties are due to the mechnical properties of one of the membranes in the cochlea, the basilar membrane, primarily the width and stiffness at a given point. The active properties are ... 3 For gravitational systems one has to be careful making statements about entropy and the second law of thermodynamics. Your example is similar to the gravitational collapse of a gas cloud if you think carefully about it. In that case and in yours, the shrinking of the gas will raise it's heat. Now even though the increase of entropy due to the increased ... 2 You need to use the equations of motion in the expression for$dE/dt$. Lets take a simple example. For the Hamiltonian $$\mathcal{H} = \frac{1}{2}\dot{x}^2 + V(x)$$ (which is equivalent to the Lagrangian$\mathcal{L} = \frac{1}{2}\dot{x}^2 - V(x)$) we have that the time-derivative of the energy is $$\frac{d\mathcal{H}}{dt} = \dot{x}\ddot{x} + ... 2 The perfect head on collision is a special case where we don't need to worry about any relative angles. We can solve it using Physics 1 conservation of momentum and energy, all in the lab frame. For equal mass particles$$m u_1 = m v_1 + m v_2 \implies u_1 = v_1 + v_2,$$and$$\frac{1}{2} m {u_1}^2 = \frac{1}{2} m {v_1}^2 + \frac{1}{2} m {v_2}^2 ... 2 You can define the kinetic energy as you want. As long as they are consistent with each other, definitions are yours to choose. For example you can take $$T = \frac{1}{2}m v^2 \, ,$$ as a definition and use it to prove Newton's law by requiring energy to be conserved, $$\frac{dE}{dt} = \frac{dT}{dt} - \frac{dW}{dt} = m v a - F v = 0\, .$$ If you want a ... 1 Torque$\tau = \vec{r} \times \vec{f} = I \vec{\alpha} r$here is moment of arm, ie. the distance(& perpendicular distance) from the axis of rotation of the body. You can only have and define torque if you have an axis about which the subject will rotate. It will be an absolute force driven motion, with NO torques compelled to induce. 1 These problems are best solved with the Euler-Lagrange equations. Work in polar coordinates centred about the circle. Then the ring is at$(R,\theta(t))$. Its kinetic energy is, $$T = \frac{1}{2} m R^2 {\dot{\theta}}^2$$ From the spring there is potential energy. Let$A$be$(L,\phi)$in polar coordinates. The potential energy is proportional to the ... 1 I'm not familiar with "Modern Analytical Mechanics" by Pellegrini & Cooper so I can't comment on that one but I'm very familiar with the other two books you mentioned. Landau's books are generally excellent but tend to be shorter in length and sometimes very dense. Nearly every paragraph has some profound insight that you'll miss if you don't ponder ... 1 You can indeed!!!! Relativity is not a good answer since you are asking for classical (Newtonian) Mechanics. There are two approaches. The first one is that of differential geometry. In it you realize that the problem of evaluating Newton's equations in non-inertial frames is that acceleration introduces curvature to the coordinates systems used. This is ... 1 One of the most useful ways of describing SHM is obtained by associating it with the projection of uniform circular motion. Imagine that a disk of radius$\mathit{A}$rotates about a vertical axis at the rate of$\omega~\text{rad/s}$. Also imagine that a peg$\mathtt{P}\$ has been attached to the edge of the disk and that a horizontal beam of parallel light ...

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You could consider omega to be a pure indicator of periodicity in the cycle. Larger omega gives you more rads per second. Larger omega gives you shorter wavelength, and therefore more energy required to keep the cycle going. You can look at omega in the equation and get an idea of the amount of energy you are dealing with.

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