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The equations of motion will be the same, except in your stated case, you can't assume that forces up and down the banked curve will cancel each other out. There will be a net force up the incline due to this, and the car will move in the direction of that net force (e.g., up the incline as it travels around the turn). Edit for additional comment: From the ...

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Firstly I don't understand how if the ball is initially rotating how it gains rotational energy from the top of the plane to the bottom, since there is no friction to provide torque. Your understanding is mostly correct. We can choose any point on the object to measure rotation around; let us choose the center of the ball. Both the gravitational force ...

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You are overcoming the static friction, which others have noted is generally greater than dynamic friction. But, the force needed at the point of contact is presumably the same when twisting versus when lifting, so why the difference in the behavior of the stand (lifting up, versus just sitting there)? The reason is because of the direction the force is ...

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For the same reason if you try to push a block or your coffee mug placed on the table it take some force before it move but when it starts moving you can lower the amount of force you applied but the block/mug keeps moving.

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Actually, spinning the pens bring a relative motion between cap and pen. When the surfaces of two objects are at rest with respect to each other static friction force acts between them and a kinetic friction force acts between then when they are in relative motion. Static frictional force $>$ kinetic frictional force. You may want to read about the ...

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This is not a complete answer, but by spinig the pen you give to your system the same energy as if you pull it out directly, in two diffrent forms.So the ''spinning'' energy overcomes the friction and the rest pulls out the pen.

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This is a straightforward application of the divergence theorem. First, split the integral on the left into its vectorial components: $$\int_{\partial \Omega}(P_0 - \rho g z)\left(-\mathrm d\mathbf{S}\right) = \sum_j \mathbf e_j\int_{\partial \Omega}(-P_0 + \rho g z)\mathbf e_j\cdot \mathrm d\mathbf{S} .$$ Then, apply the divergence theorem: \begin{align} ...

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Why is it that two carbon atoms fired at each other will bounce off and not stick together? It is because as the atoms move close together their orbital electrons begin to repel more than their nuclei attract each other's electrons. The result is greater potential energy as they approach and this leads to the tendency to move apart much like compressing a ...

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The math is almost trivial for someone beyond algebra 1. Write the kinetic energy of each particle as $p_n^2/2m_n$. Then converse momentum and kinetic energy in the center-of-momentum. You will see that the magnitude of the momentum each particle does not change.

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A big wheel is more expensive than a small wheel. Mopeds are usually "as cheap as possible" so will want a small wheel. But bigger wheels give you: better load bearing (good with heavy bike) Smoother ride (smooth over small bumps / holes in the road) Greater stability (inertia) More metal = Heavier = stronger (important as stresses are greater at high ...

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Motorcycles are traditionally designed for use on all types of roads and to an extent for off-road use. Mopeds are designed for urban use where the roads are better maintained. Larger tires see less likely to get stuck in ruts, potholes, mud etc. The basic question is more about engineering than physics in that regard.

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While neither the Lagrangian $\mathcal{L}$ nor the action $S$ are invariant under boosts of the form $$\dot{q}(t) \to \dot{q}(t) + v, \quad v \in \mathbb{R},$$ the Euler-Lagrange equations are. The dynamics of the systems are unchanged for any transformation that preserves $\delta S = 0$, i.e. a transformation of the form $$\mathcal{L}(q, \dot{q}, t) \to ... 1 It sounds like you are interested in symplectic reduction procedures. On of these methods is that of Routh's procedure to eliminate cyclic variables using a Legendre transform to a reduced-variable Hamiltonian called a Routhian. Forming a variational approach may be difficult for some reduction procedures, however we can view conserved quantities as ... 0 Yes, it is, but it is not easy to know the invariant quantities beforehand, so the usual way to use it is to write down the Lagrangian in terms of any (non-necessarily invariant) quantities, and then use the principle of least action to find the invariant quantities, Nother currents, etc. 0 This answer is to be read in conjunction with @ja72's answer. Assuming that you are modeling damped simple harmonic motion, where the drag force is proportional to the velocity, you can control 3 variables: the spring constant k, the mass m and the coefficient of damping \mu (where force of drag F = \mu v). This allows you to set the time for the ... 1 The general equation is$$x(t) = x_0+ {\bf e}^{-\beta t} \left( A \sin \omega t + B \cos \omega t\right)$$where the constants A and B depend on the initial conditions, and \beta and \omega on the mechanical properties of the system. Read http://hyperphysics.phy-astr.gsu.edu/hbase/oscda.html for more information For example if the initial ... 4 You usually cannot push your hand through the table, because it's a single solid. The atoms are held together by covalent bonds, which are electromagnetic in nature. Sand on the other hand is grainy - the SiO_2 grains do not interact with each other and are only held "in place" because of gravity. You can run your hand through sand similar to driving a ... 1 For this sort of problem it really helps to go back to the Riemann sum. You don't even need to write it down per se; you just need to think about what your \delta x is in the Riemann sum. So, in this case you're trying to find a "sum of forces" or we might say a "net" force. To find this, it helps a little to think about momentum conservation/Newton's ... 0 Well in this case, you could consider a really thin ring of elemental thickness say t, then proceed to form the integral as you have done to that point... Now, you might write dS as tdx , dx being an elemental strip along the circumference of the ring which subtends an angle d(theta) at the center. So you may now rewrite dS as t* R *d(theta) , and proceed ... 1 I guess the Alka Seltzer is an effervescent kind of tablet. When it is immersed in water, gas bubbles form on the tablet surface and surface tension effects prevent the bubbles from separating them from the tablet surface. By this effect the global density of the system tablet plus bubbles is going down overtime, until the global density approaches the ... 2 It's a complicated subject - but very well studied in the context of eddy current brakes, where the retarding force is used to create a braking force without mechanical friction / wear. For me, the starting point for finding out more was this post - in particular the posting by Jim Hardy contained lots of good links. It seems that some of the most ... 0 Some years later... I am reviewing this problem mostly for my own benefit, But it could be useful if somebody is still wanting to discuss the answers, particularly without any braking system. let me start with an alternative wrong solution, aka a variation: allow the system to drop the water without horizontal velocity, for instance using a periodic ... 0 It is easy to make a trap that is stable in some direction(s). Just not one that is stable against deviations in all directions. Your circular motion also encircles a region with charge, so you are not confining the circling charge within a charge free region. 1 As noted in some of the comments already, the answer to your question depends on what properties of quantum eigenstates you want your classical analogues to have. If you're thinking of eigenstates of arbitrary observables, quantum eigenstates have the property that the value of that observable is precisely defined. In classical mechanics, individual points ... 6 Your intuition was correct - the shaft will rotate in one direction and the housing/stator will rotate in the other. If you look up "moment of inertia" you will find that it is the rotational equivalent of mass. For almost any reasonable motor the moment of inertia of the shaft/rotor windings will be smaller than the moment of inertia of the housing/stator. ... 2 The astronauts working on the Hubble space telescope had to bring special low torque wrenches to counteract the effect of the torque of the motor spinning them around, due to conservation of angular momentum, although this meant far more use of muscular power to hold them in place. And also to avoid damaging the equipment they worked on, such as screws ... 1 I can give a back-of-the-envelope derivation of a drag force that ignores fringe effects and other complications. Say the conductor is a plate of thickness \Delta z traveling with velocity v in the x direction. Take the magnetic field to be constant in a rectangular area, with the the \Delta y side perpendicular to the velocity much longer than \Delta ... 1 I don't have the expertise you're looking for, but here's a crack at it: This is a really hard problem whose answer depends on the material properties. In particular, the answer depends on how big the 'eddies' are: if the current moves in roughly a big circle then the effect is large, but if there are lots of little eddies the effect is small. However, ... 0 The wheels use regenerative braking drive a propeller/turbine to provide thrust not the other way around. In a typical sailboat or other wind powered craft the wind provides thrust and the resistance given by the conveying mechanism (water or wheels) is typically minimized. In this scenario however, the wheels have a little bit of regenerative braking ... 0 Yes - rigid body motion in 3 space can be modeled by the Euler equations, which because of cross-axis coupling are nonlinear - but they can be integrated. A complete reference on how to model rigid body dynamics is given in these MIT lecture Notes. Solve equations (9), (10) and (11) for the rate of change of angular velocity around each principle axis, and ... 2 Since a system must obey the law of momentum conservation, the center of mass of a system (which can be made of one or many bodies) must have constant velocity if no external force is applied. Hence, a body can rotate around its center of mass, or it can rotate around any other point, but only if under the influence of an external force. Therefore one can ... 2 @ChrisDrost's answer is correct, but we can actually remove the assumption that the friction is constant by considering conservation of angular momentum instead. If we put our origin at a point along the ground, then there is no net torque on the sphere: The frictional force always points directly towards (or away from) the origin, and the normal force ... 1 So this is a phenomenon which is known in billiards as "backspin": you hit a ball off-center and it simultaneously has a motion "forwards" but a spin that imparts a force on the ground to send it "backwards". Trick shots where you induce extreme amounts of backspin by hitting the ball almost vertically downwards are known sometimes as "massé shots", if you ... 1 Try to think of this problem using a polar coordinate system. x is essentially the radius r or \rho, measured from pivotal. w is simply the angular velocity. So the position vector of the object is x\hat{\vec r}+\theta\hat{\vec \theta} So the velocity vector is \dot x\hat{\vec r}+w\hat{\vec \theta} The hatted vectors are unit. So the ... 3 It is well known that adding a total time derivative to the Lagrangian does not change equations of motion. The Lagrangian above adds a term$$-q\dot q=-\frac{1}{2}\frac{\mathrm{d}q^2}{\mathrm{d}t}$$(a total time derivative) to the free particle Lagrangian \dot q^2. It is thus fully equivalent to to the standard free particle Lagrangian (up to an ... 1 $$\mathcal{L}\left( q,\dot{q},t\right)= \dot{q}^{2} - q\dot{q} \tag{01}$$ $$\dfrac{d}{dt}\left(\dfrac{\partial \mathcal{L}}{\partial \dot{q}}\right)-\dfrac{\partial \mathcal{L}}{\partial q}=0 \tag{02}$$ \dfrac{d}{dt}\left[\dfrac{\partial \left(\dot{q}^{2} - q\dot{q}\right)}{\partial ... 1 The eddy currents are caused when the metal plate which is subjected to a magnetic field is take out rapidly. This causes current to flow within the metal plate and thus opposing the already existing magnetic field (Lenz's law). To calculate the current I think the approach would be as follows. We have F_d, external force B, magnetic field ... 0 You may be interpreting the explanation of the coordinates wrong. "The angle between the top of the bigger cylinder and the position of the smaller one" does not tell us where the center of the angle is measured from, but if it is measured from the center of the larger cylinder, which perhaps is itself at a height y, then the proper formula would be m g ... 2 The variables involved here are classical and you resolve them classically. They enter into the operator because they are parameters of the wavefunction. So let's do this a little more broadly. For continuous systems, we want a family of solutions based on some parameters which I'll collectively identify as \alpha \in A; the solutions are then labeled ... 3 When calculating gravitational potential energy the only thing that matters is the position of the centre of gravity. So if the vertical position of the centre of gravity changes by \Delta h the gravitational potential energy changes by \Delta V = mg\Delta h. 2 If, by "Newtonian Mechanics" you mean what Newton derived, then yes, by definition. But if you mean "classical mechanics" including rigid body dynamics then the answer is a resounding "no"[1] and the main reason is that Newton's three laws by themselves are not enough to imply conservation of angular momentum. For conservation of angular momentum, you need ... 1 If "for every action there is an equal and opposite reaction" (F12=-F21) allows friction as a legitimate reaction, then no. Non-conservative forces like friction are not fundamental and would allow non-conserved energy and momentum if additional constraints are not included (like "heat" from thermo). They are not time-reversible. You can't derive ... 0 Control moment gyros , like those used on the space station can store tremendous amounts of energy and are used to create torque to change spacecraft attitude (rotation). The energy is stored by the wheel's high angular velocities, and the torque is created by tilting the wheel (pushing against the wheel's angular momentum). 1 There are four ways that I can think of to store mechanical (elastic) energy. Axial Stretch {\rm d}W = \frac{F^2}{2 E A} {\rm d}x Torsion {\rm d}W = \frac{T^2}{2 G J} {\rm d} x Bending {\rm d}W = \frac{M^2}{2 E I} {\rm d} x Shearing {\rm d}W = \frac{c F^2}{2 G A} {\rm d} x See https://en.wikipedia.org/wiki/Castigliano%27s_method and last page in ... 2 We may construct a system with Hamiltonian not T+V but energy still conserved from any system where energy is conserved by making the phase space description generally covaraint: Starting from an unconstrained Hamiltonian H_0(p,q) with Hamiltonian action$$ S_0 = \int \left(p_i \frac{q^i}{\mathrm{d}t} - H_0(p,q)\right)\mathrm{d}t we may turn it into a ...

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In general you can write the kinetic energy of a free particle as: $$T = \frac{1}{2} m \,\vec{v}\cdot\vec{v}$$ which holds whatever coordinate system you choose (could a physical quantity such the trajectory of a particle depend on the coordinate system that you choose?). We can rewrite this: T = \frac{1}{2} m ...

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I'm not sure why you're talking about an $x$ and $y$ component of velocity when you're working in a polar coordinate system. Maybe you're confusing $x(t)$ (the extension of the spring as a function of time) with the Cartesian coordinate $x$. These are very different things. To understand what the radial component of the velocity is, assume the pendulum isn't ...

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Velocities in the kinetic part of Lagrangian The variable $\;x\;$, that represents the displacement of the string from its position at rest, has been replaced by the variable $\;s\;$ in order not to be confused with the coordinate $\;x\;$ of a Cartesian system. The velocity of the particle $\:\mathbf{v}\:$ is analysed as follows ...

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The volume element is $(dr)*(rd \phi)*(dz)$. Hence, the extra r in your integrand should be eliminated.

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