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3

When you have an impulse $F\Delta t$ (I prefer that notation over $m\Delta v$ because it allows impulse to be imparted without worrying about the mass of the thing giving the impulse), then The momentum of the center of mass changes as though the impulse was applied there, so $$m\Delta v = F\Delta t$$ The angular momentum changes according to the torque ...


1

In general if the imparted momentum vector $\vec{J}$ goes through a point $\vec{r}$ relative to the center of mass then the change in speed of the center of mass is $$ \begin{aligned} \Delta \vec{v} &= \frac{1}{m} \vec{J} \\ \Delta \vec{\omega} & = I^{-1} (\vec{r} \times \vec{J}) \end{aligned} $$ where $\times$ is the vector cross product. In ...


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I'd expect the motion of a shuttlecock to be quite hard to describe with great accuracy. A rough approximation is to assume a damping factor proportional to the velocity in the equation (so a force of the form $\mathbf F_v = - \gamma\mathbf v$). Some inaccuracy stems from the fact that I'd expect the coefficient $\gamma$ to depend on the orientation of the ...


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This video by Get Smarter Everyday is a quite informative one. He shoots the failure at a staggering 250000(quarter million) fps. And your assumption is right about the first fracture triggering the rest. PS:At some instances the spaghetti doesn't break into 3 pieces, but 4 or more! https://www.youtube.com/watch?v=ADD7QlQoFFI


1

The dynamic equations of classical mechanics are locally time-reversal invariant. You can replace $t$ with $-t$ in them and they won't change their form. A system with friction is NOT described by these equations, and that kind of system is not covered by the reversibility statement. Statistical mechanics and chaos theory give you the real arguments for the ...


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As it follows from the Mechanics by Landau-Lifshitz (which can be found here — page 28), the desired dependence can be found in the form of $$x(U)=\frac{1}{2\pi \sqrt{2m}}\int_{0}^{U} \frac{T(E)dE}{\sqrt{U-E}}$$ Where $T(E)$ from the law of conservation of energy and the initial statement $T_{½}=\alpha V^\beta$ $$T(E)=2\alpha ...


1

The terms trajectory and orbit both refer to the path of a body in space. Trajectory is commonly used in connection with projectiles and is often associated with paths of limited extent, i. e., paths having clearly identified initial and end points. Orbit is commonly used in connection with natural bodies (planets, moons, etc.) and is often associated with ...


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An orbit is a special case of a trajectory. Any ballistic flight path is a trajectory (throwing a ball or firing a cannon) and an orbit is a special case where the trajectory does not intersect the locally dominant gravitational body - at least in the short term.


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Is trajectory the same as an orbit? Yes and no. For example, you will see many textbooks and papers use the term hyperbolic orbit. I use the term myself. On the other hand, the ballistic part of a weapon's flight from the gun (or missile silo) is called a ballistic trajectory rather than a ballistic orbit.


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The angular velocity for the disk is given by $ \dot{x}_G^2 + \dot{y}_G^2 = v_G^{\, 2} = \omega_{disk}^2 r^2 $. Insert $\omega_{\text{disk}}^2$ into $$ T_A = \frac 12 m v_G^2 + \frac 12 I_A \omega_{\text{disk}}^2 $$ We can then add up $T_A$ and $T_G$ to form $T$, likewise for $V = V_A + V_G$. The Lagrange function is then given be $L = T - V$ and we need to ...


0

Nop! The definition of an tridimensional trajectory, is the set of 3D points $(x(t), y(t), z(t))$ determinted by the parameter $t$. Which means, it's a function $\mathbf r(t)$, where $\mathbf r: \mathbb{R}\mapsto\mathbb{R}^3$. Of course we can generalize to the $n$-dimensional case: $\mathbf r: \mathbb{R}\mapsto\mathbb{R}^n$, where $\mathbf r(t) = (x_1(t), ...


1

A basic understanding of what's going on gain be gained just using Kepler's laws and Newtonian mechanics. A simple way of dealing with multiple gravitational sources is to selectively ignore all but one of them. This is the patched conic approximation. Which gravitating body is in play? That depends on whether the spacecraft is inside the gravitational ...


1

Thermal expansion from an atomistic perspective: The energetic potential between two atoms can be approximated by two exponential functions, one for the attractive force between the atoms, one for the repulsive force. The superposition of these two force fields has a minimum at a certain distance. Examples for such empirical potentials are Stillinger-Weber, ...


2

Roughly speaking solid matter is on a lattice form, A three-dimensional lattice filled with two molecules A and B, here shown as black and white spheres. The molecules fit like LEGO , the forces tying them together are mainly the spill over electric field forces , attractive and repulsive forming the patterns of the lattice. In a single crystal one ...


0

 Why doesn't anything need the second derivative (acceleration)? Only Newton's gravity law does not use acceleration in the expression for force. In electromagnetic theory with retarded fields, forces are functions of past positions, velocities and accelerations of the charged particles.


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"Could we come up with a theory that only requires a snapshot of the positions? ...Does modern physics (e.g. relativity) have something to say about this curious thing?" Yes. The Theory of Special Relativity posits that there is no favored rest frame, and therefore there is no such thing as absolute velocity. In other words, velocity can only be measured ...


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because we want to find the trajectory of the particle and if we know it's position in different times we can calculate it's motion. Also, the acceleration can be obtained when we have the X(t). then acceleration is not a new information to use and it is embedded in the X(t) (acceleration=$d^{2}X/dt^{2}$)


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I am trying to answer your questions one by one. By the way, I saw your comment. 1. You say in the question: "Why doesn't anything need the second derivative (acceleration)?" Yes the acceleration is needed for obtaining the velocity. I some situation we are given the force and the mass, as in an electrical or gravitational field, not the velocities. Then ...


1

The reason that you only need to specify initial position and velocity to exactly solve the equations of motion for a system is simply because Newton's Second Law (which is the equation governing motion in Classical Mechanics) is a second-order differential equation. The upshot is that to solve a 2nd-order ODE, you basically need to take 2 integrals. Each ...


1

Perhaps you could use the expansion of the pressurized container to measure the internal pressure? Even a metal container will expand by a few microns if it pressurized - this expansion could be measured using lasers. Does that meet your definition of 'not altering'? Although I suppose even lasers would heat the container by a miniscule amount, which ...


0

No, I don't think that you proceed correctly. You need the relationship between force and distance, and this is what you should integrate. So, please follow my formulas. So, I understand that $V$ is the velocity of the ion (not of the box). From your formula $T=aV^b$, I deduce the acceleration $A$ acquired at the end of the trip, by assuming something that ...


1

If I throw a small rock (m = 1kg) at a big rock (100kg) the small rock rebounds. Let's say my weight is 80kg, if I would jump into a big rock instead of bouncing back I would move in the same direction as a big rock. The big rock is heavier but it is not reflecting me. Why is that? There are two conditions which are to be met if a body A ...


1

Two laws govern collisions: conservation of momentum and conservation of energy. Momentum is the product of mass and velocity, so we can write $$\sum m_i\cdot \vec{v_i} = const$$ Conservation or energy is a little bit trickier, since energy can be converted from one type to another. In an elastic collision, the kinetic energy is conserved, so $$\sum ...


2

How accurate do you need to be? The problem with these calculations is that massless, frictionless pulleys are usually out of stock at Acme Mail Order. High school physics will give you the tension in the rope, a different set of high school equations will tell you how much extra tension is needed for a certain acceleration. A rough metric is add 10% for ...


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Read Lagrange's Mécanique analytique (English translation: Analytical Mechanics). The book is split up into two parts: statics and dynamics. The first chapter, "The Various Principles of Statics," is a beautiful historical overview. Lagrange works out many problems; for example, he has a chapter entitled "The Solution of Various Problems of Statics." But, ...


1

Yes, you absolutely need to count the back pressure. Otherwise the force required would be essentially independent of flow rate or the properties of the fluid. For a given fluid, you need to assess the pressure needed inside the syringe to make the desired flow. The force on the plunger needs to overcome that. In many cases that will be the dominant ...


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The force you have to exert on the plunger is $F_1 - F_2 + F_3$. If you didn't include the back pressure term it would be just as easy to squirt treacle as it would be to squirt water, which obviously isn't the case.


1

To quantize a classical system, start from the Poisson bracket $$\{x_i, p_j\} = \delta_{ij}.$$ This relation defines $p_i$ as the momentum canonically conjugate to $x_i$ and is equivalent to Hamilton's equations. Quantize by letting $x_i, p_j$ be Hermitian operators on a Hilbert space, with commutator $$[\hat x_i, \hat p_j] = i\delta_{ij} $$ (identity ...


-1

Yes, there is a limitation, and it comes from mathematics itself. Gödel's first incompleteness theorem states that no consistent system of axioms whose theorems can be listed by an "effective procedure" (e.g., a computer program, but it could be any sort of algorithm) is capable of proving all truths about the relations of the natural numbers (arithmetic). ...


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What's the mathematical process and physical logic? The Fourier transform of position space ($\vec x$ domain) is wave number space ($\vec k$ domain). This is an unambiguous, well understood mathematical result. By the De Broglie hypothesis, the momentum is $\vec p = \hbar \vec k$. This is physical hypothesis with experimental confirmation. Although ...


0

A Fourier transform is the decomposition of a position space function into a basis of plane waves, each of which has a well defined momentum. $$ f(x) \sim \int \text{d}p\; F(p) e^{\text{i}px} $$ This relies on the quantum mechanical idea that waves can have a well defined momentum.


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If you look up centripetal force i think you will find your answer. I wont post any links because i dont think i'm allowed to at the minute but a diagram will help you understand it better than i can explain it


1

You might be best off creating a bunch of trial systems and evolving each one independently using the (much easier) Hamilton's equations, which are good old ordinary differential equations. After all this is how the Liouville equation was constructed in the first place by people like Gibbs---in the limit of an infinite number of trial systems, you get the ...


0

Escape for a moment from the definition "work is equal to the force times the distance ('times' meaning scalar product of course)” The meaning of "work" may be interpreted as a transfer of energy, or specifically, the transfer in or out of a system's kinetic energy [1]. Your Question: "If I lift some object from a ground, the force to be put in above ...


1

If you let $x_i$ be the position of the $m_i$, you can write a set of coupled equations $$\begin {pmatrix} m_1\ddot{x_1}\\ m_2\ddot{x_2}\\ m_3\ddot{x_3} \end {pmatrix}=A\begin {pmatrix} x_1\\ x_2\\ x_3 \end {pmatrix}$$ where $A$ gives the forces from the springs. With more masses and springs you have more lines in the equation. If all the masses are the ...


0

I don't know how this is supposed to simulate planetary motion, but we can still look at what data you would expect. The force on the stopper is $\frac {mv^2}r$, which if there is no friction (good luck!) equals the gravitational force on the masses at the bottom. We would therefore expect that for given mass on the bottom, $r \propto v^2$, which doesn't ...


1

I hope this is useful as an answer to your question, because although this is not about your exact system it gives a helpful interpretation of what normal modes are. For the water molecule we can consider it as three masses linked by two identical springs. Similar to your system there are three normal modes, which can be represented as the following motions ...


2

Step 1 The mass is dropped ... The rotational energy of the torus does not change. Going to stop you right there. Although total energy is indeed conserved, rotational energy does not need to be conserved. Remember that the potential energy of dropped objects is generally converted to heat when they go thud on the ground. Angular momentum is however ...


5

Actually, if the mass comes to rest relative to the torus after landing, the energy of the system goes down. Let $I$ be the moment of inertia of the torus, $r$ be the radius of the mass from the axis of rotation before it is dropped, $R$ be its radius after it lands, $\omega_1$ be the angular velocity before the mass is dropped, and $\omega_2$ be the ...


1

It probably doesn't help to look at the system from above. I suggest you draw a free-body diagram from the side-view like:                                           ...


1

I believe that one could rephrase the question as "if the limit of the speed of sound in a medium must be the speed of light in vacuum, what does that mean for the limit on rigidity of an object?" Speed of sound is given by $$c=\sqrt{\frac{E}{\rho}}$$ - it depends on both density and Young's modulus. I would consider "rigidity" to be just the modulus, and ...


1

Moving the mass inside the rotating object takes energy - when it is "lowered" against the artificial gravity it does work against the force resisting this motion so the final kinetic energy of the system with "lowered" mass is not the same as before. The equation: $$E=\frac12I\omega^2=\frac{L^2}{2I}$$ is the clearest way to see why this is so - when ...


14

In step 1 you lower the mass and this generates energy. Let's say you store this energy is a spring, and for the sake of argument let's say the energy stored is 1J. The energy has to come from somewhere, and of course it comes from the rotational energy of the torus so the rotational energy of the torus is now 99J. In step 2 you slow the torus, perhaps by ...


3

Step 1: the rotational energy $E=1/2 \omega L$ does increase because $\omega$ increases (bacuse I decreases) and L is a constant ( $L=\omega I$). This is wrong again in step 3: The rotational energy of the torus does not change So the potential energies of the lifted objects is not the same in both situations, because they will also convert rotational ...


1

Yes, I agree it looks like the second equation is missing an overall factor of $\Delta t$ on the right. The action is a function of all the $x$s for the whole trajectory, but the Lagrangian is not. It is only a function of a position and a velocity. So it makes perfect sense to evaluate it using the position and velocity for a single time. Points 8 and 9 ...


0

The moment of inertia, $I$, is, of course, given by $$I = \sum m_i r_i^2$$ where the sum is over every bit, $i$, of the object with mass, $m_i$, and at a distance, $r_i$, from the axis of rotation. You might need to calculate this by integration. $$I = \int_x ~r(x)^2 \rho dx$$ where $$m = \int_x \rho dx$$ You integrate along the pendulum in ...


0

Let's do it in 1D for simplicity: you consider a portion of thread of length $dL$ and section $S$, with a net body force density $f$, say $f=\rho S g$ where $\rho S$ is the lineic mass density. On $dL$, you also have stress from the rest of the thread, which are $T_+ = \sigma_{zz} (z+dL) S$ at the $z+dL$ end and $T_- = -\sigma_{zz}(z) S$ at the other end. ...


0

From user27118, I was able to grasp a better understanding of what needed to be done to set up the system of ODEs: \begin{alignat}{2} m_{eq}\ddot{y}_1 &= k_{eq}(z - y_1) + c(\dot{z} - \dot{y}_1) - m_{eq}gy_1 + F_1(t)\\ ...


4

The distribution of force on the ball bearings, and therefore inner, and outer races of the bearing depend on the Adjustable tension. This tension preloads the bearings. First consider the case where the bearing is supporting no weight but the cones have been tensioned. The tension from the nut compresses the bearing axially, compressing the balls. Viewed ...


1

Measurable quantities are described by time evolution of observables classically. Think about this for a bit. In Classical Mechanics what we see evolving in time are quantities like momentum, position, etc, i.e observable quantities. This is why the Heisenberg picture is useful. Exactly the same expectation value is obtained when we leave the state fixed as ...



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