# Tag Info

## New answers tagged classical-mechanics

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It's the same reason a knife cuts through an object so easily. Use a blunt object and you'll have a harder time. When a person is standing facing the waves or away from them, the force exerted on them by the wave covers a greater area all across them and therefore produces a greater net force on the object that would be their body. Standing on your side, ...

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Measure F at 3 (or more) different velocities and curve fit a parabola to the data.

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There are couple of methods that is widely used. incline plane method horizontal plane method There is a very good expriment suggested by Pellissippi State Community collage Below is the link that defines what and how to do it. Check out the below link ...

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Because your hand is softer than a hammer, absorving most of the impact power over the nail, reducing overall power needed to penetrate wood. To compensate this, is necessary to apply too much more power that your hand will simply be injuried by nail. If the resistence of your hand skin is lower than resistence of wood, only your hand will be perfured, even ...

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When a torque is applied to an object it begins to rotate with an acceleration inversely proportional to its moment of inertia. This relation can be thought of as Newton's Second Law for rotation. The moment of inertia is the rotational mass and the torque is rotational force. Angular motion obeys Newton's First Law. If no outside forces act on an object, ...

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Top above question regarding integration & derivatives... Integration of functions your pretty much screwed unless you program your own integral functions, or if you just need the area under a curve you can use the for example "SIMPSON'S RULE". For roots (x=0), can use the powerful "NEWTON'S METHOD"... x2=x1(assume) - f(x1) / f'(x1), then X2=X1... ...

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I've studied this in detail before way back in college days, After taking fluid mechanics course & modern physics (& mechanics physics, calculus, linear algebra, diff equations...) it made sense considering the air similar to a fluid. There is no simple equation like $$y_{x}=\frac{-gx^2}{2v^2\cdot \cos(T)^2}+\tan(T)x +y_\rm{initial}$$ You will have ...

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If you're strong enough, you can push the nail in by hand. You only need protection against the "equal and opposite reaction" of the nail against your soft hands. The nail can penetrate the wood because it is sharp, applying high pressure to the surface area it contacts. The nail's head is relatively much wider, so it will not penetrate your hand as easily ...

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Your solutions are wrong. As $$\frac{dq}{dt}=\frac{\partial H}{\partial p}\qquad \frac{dp}{dt}=-\frac{\partial H}{\partial q}$$ you get $$\frac{dq}{dt}=10\,p\qquad \frac{dp}{dt}=0$$ i.e. $$q(t)=10\,p_{0}\,t+q_{0}\qquad p=p_{0}$$ The $q$ coordinate flows in time in straight lines, while the $p$ coordinate doesn't change in time. So each phase ...

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Consider, the strongest forces will be at the end that was falling, with compressive force in the vertical direction in the center (from the impact) and tensile horizontal forces from the momentum of the sides that didn't hit the ground, pulling at the middle. So, assuming you're landing on an ideally flat, level surface, and assuming a column with ...

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TL;DR: In order for the nail to be driven into the wood, you need to apply sufficient force to split apart the wood fibers ("make a hole") and overcome the force of friction between nail and wood. The hammer can apply a much greater (instantaneous) force than the hand, because it is much harder (has a greater Young's modulus). If you assume a mass $m$ ...

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Your system has 2 degrees of freedom, but using $x_1$ and $x_2$ will not be helpful in determining the effective spring rate. To get the spring rate you need the extension $x$ of the connection point with mass $M$ and the tilt angle $\theta$. Do the substitution: \begin{align} x_1 & = x - a \theta \\ x_2 & = x + b \theta \end{align} The ...

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It is a method that is generally used for conservative unidimensional problems (problems with only one degree of freedom, here your angle $\theta$ or cartesian coordinate $x$). You'll notice that it is equivalent to using Newton's second law in this case : let us write the total energy $E = \frac{1}{2} m v^2 + V(x)$, $V$ being potential energy. The problem ...

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You can! But first, some precautions: Your hand is squishier than the wood (generally, for wood of a condition to have new nails driven into it and hands prepared to do some hammering). This means that it will take less pressure for the nail to damage your hand than for the nail to enter the wood. The pointy bit at the front of the nail and the broad ...

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There are no consequences. It's a convention that negative is for attraction forces and positive is for repulsion. This leads to forces tending to reduce the potential when attracting. You remove that sign, then attraction forces because positive, and forces will tend to increase the potential. The worst that would happen is a few inconsistencies within ...

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The gradient of a function $\nabla V$ is as you know a vector and this vector points to the (locally) steepest ascent. i.e. if you are on the hill the vector $\nabla V$ points to the tip of the hill and when you are on top of the hill the vector vanishes because there is no "steepest ascent" at that point. You can prove this fact mathematically but I'll omit ...

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The gradient of a scalar function always points in the direction of greatest ascent. That is, a scalar function has the most rapid increase in that direction. If you imagine walking around a mountain, the gradient of the height function of the mountain at each point will point towards the steepest direction. On the other hand, forces in a conservative field ...

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Take a simple 1D example. We'll plot gravitational potential energy as a function of horizontal position for the side of a hill: We normally define $F = -\nabla V$ and that means the force points downhill i.e. if we let the particle go it will roll downhill. If instead we defined $F = \nabla V$ the force would point uphill i.e. if we let the particle go ...

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One way of looking at the change is that you are changing the definition of potential from The potential at a point is the work done on a unit mass in going from an arbitrarily chosen zero of potential to the point to The potential at a point is the work done by a unit mass in going from an arbitrarily chosen zero of potential to the point So ...

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This the same as changing $V$ to $-V$. For example, if $V$ described a potential well then changing it to $-V$ would describe a bump.

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Basics, Newton law and acceleration integration: sum(F) = m.a v = a.t The force exerted by the hammer is dissipated in about no time, because the hammer is very though. The acceleration is therefore great. Now, considering that your hammer is .2kg, going at 10m.s^-1, and hitting the nail during 1ms, it is pushing by (0.2kg*10m/s)/0.001s = 2000N That's ...

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The "should have been proof" relies in the extensions of the springs being the same. If you put a=b that condition is not satisfied. Find a spot where the two extensions are the same and you will get "should have been proof" answer.

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It depends what you mean by vibrational motion. A point mass indeed does not have internal vibrational degrees of freedom, it can just have translational motion, as you pointed out. However, if you attach this point mass to a spring you create a harmonic oscillator. It has a translational motion, but this oscillating motion can describe vibrations such as ...

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There's many different things at work here. First, there's the issue of acceleration. Hammers are very hard and solid, so when you hit the nail head with the hammer, the energy and force of the blow is delivered at almost an instant. Hands, on the other hand, are rather soft, and will spread out the same amount of energy and acceleration over a longer time ...

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The inertia matrix for a thin rectangular foil (laying along the xy plane) in body coordinates is $$I_{body} = \begin{vmatrix} \frac{m}{12} b^2 & 0 & 0 \\ 0 & \frac{m}{12} a^2 & 0 \\ 0& 0 & \frac{m}{12}(a^2+b^2) \end{vmatrix}$$ where $a$ and $b$ are the side dimensions. The inertia matrix in world coordinates, while rotated by ...

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This is in fact possible, as evidenced by the number of videos online of someone hammering a nail in with their bare hands. More to the point, the human hand is plenty hard enough to hammer in a nail into some pretty hard if it is swung fast enough. It's just that the damage to the hand would be rather devastating, and so our pesky brains keep us from ...

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Your hands are soft whereas the hammer head is very hard. Therefore your hand will come to rest upon striking a nail in a much greater time as compared to hammer. Now, force is the rate of change of momentum. The hammer is heavier so momentum is more initially even if hand and hammer hit at same velocity. As well as the time taken for the momentum to change ...

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It is just because when you hammer a nail, you apply a force to the nail by changing the momentum of the hammer $F=\dot p$ This force helps in overcoming friction and thus pushing the nail inside. This isn't the case with your hand. Moreover, if you try to develop the same amount of force with your hand, you will get hurt because of the pressure applied to ...

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While I have never conducted such an experiment (yet), I conjecture it will be the energy of the projectile what will determine the results. It will also bring momentum, but this would be comparable roughly to three A380 jets at cruise speed, so it probably would not change much on the result. The projectile would start evaporating already when passing ...

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Taylor series of the space function $s(t)$: $$s(t) = s_0 + s'(t_0)(t - t_0) + \frac{1}{2}s''(t_0)(t - t_0)^2$$ Now, having $s'(t_0) = v_0$ and $s'' = a$ you get: $$s(t) = S_0 + v_0t + \frac{1}{2}at^2$$ assuming $t_0 = 0$

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A simple intuitive explanation is that distance equals average speed times time. The average speed is $1/2(0+aT)$, and time is $T$ (assuming starting from rest, and constant acceleration).

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Looking at the graph you can also see that the displacement is equal to the average velocity $\times$ time.

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yes, the 1/2 is due to the double integral. How $at$ evolves during the interval $[0,T]$ ? It's value is 0 at the begining, and $aT$ at the end. So the cumulated value in $v$ cannot be $aT^2$, or it would mean that the value was constantly equat to $aT$ during the interval. When you sum-up varying quantities, you really have to compute the integral, not ...

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If you are in a flat space(time), i.e. without any source of gravitation, in a spaceship, and you emit a ray of light across the spaceship, both the spaceship and the light will be in the same frame of reference. The frame will be inertial - not accelerated - and therefore the light will follow a straight path. Yet if a boost is applied to the spaceship, it ...

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The Lagrangian is defined in the most simple case as a function of $q$, $\dot{q}$ and $t$: $L(q,\dot{q},t)$. This notation implies that $q$, $\dot{q}$ and $t$ are by definition independent variables. This is how you have to interpret the partial derivatives to $q$ and $\dot{q}$, it doesn't make sense to write: $q = f(\dot{q})$, because both are considered ...

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I found the (shamefully simple) answer myself: The simple double pendulum is a conservative system, hence due to Liouville’s theorem the phase-space volume given by a given ensemble of trajectories is constant over time. However, if the system exhibited transients and thus attractors, all trajectories starting within the basin of attraction of a given ...

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On a vortex mixer the liquid is swirling inside a test tube. In a centrifuge, the test tubes (containing the liquid) are held in a fixed position inside a rotor, and the rotor is spinning inside the centrifuge.

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Like tides on the ocean, it is a consequence of the tidal forces (if the claim is correct at all which I am verifying now). Tidal forces result from inhomogeneities of the gravitational acceleration. If we approximated the external gravitational acceleration $\vec g$ caused by the Moon or the Sun to be constant all over the Earth and its vicinity, the Earth ...

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A vortex is a local disturbance caused by turbulent flow. It carries fluid and particles inward toward a central axis of rotation. The central axis is not fixed. It moves within a larger body of fluid. The speed of the vortex is greater near the axis of rotation and decreases toward the edge of the vortex. Unlike a vortex, a centrifuge creates ...

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First of all, if you are at either pole, you are not traveling in a circle. Any place else, you would be traveling in a circle, with the largest being at the equator. For you to understand what is happening, I will use two cases: 1 - No spin generated force. When you are at a pole, you don't "fly out" into space, nor do you sink into the ground, therefore, ...

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You've taken a wrong turn with the cross product business. A cross product is a vector operation - it maps two vectors to a third vector. However, you seem to be imagining that your three equations for $F_x$, $F_y$, and $F_z$ have a cross product in them, but they can't, because those equations contain only scalars. The Lorentz equation as written like this ...

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You need to remember that there are two forces acting on the pendulum: (1) Gravity toward the earth, and (2) Tension toward the center of the circle formed by the arc that the pendulum describes. If gravity is resolved into a vector perpendicular to the arc, and a vector tangential to the arc, the tangential component is a restoring force that returns the ...

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The velocity is always tangent to the arc of the pendulum, but the acceleration is not. This is because of the centripetal acceleration, which is always directed along the pendulum toward the center of rotation. So the acceleration has components both tangent and normal to the arc of the pendulum. When the speed of the pendulum is increasing, the ...

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It looks qualitatively correct. The acceleration and velocity are tangentially in the same direction when the pendulum is speeding up, and tangentially in opposite directions whenever the pendulum is slowing down, as it should be. Right at the middle point, when the pendulum is momentarily moving at a constant speed, the acceleration is purely in the radial ...

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Well the usage of the potential is not purely classical: in quantum mechanics we promote $\hat x$ to an observable and then reinterpret $V$ as acting on square matrices via its Taylor expansion about some point. We can interpret these things by using Ehrenfest's theorem prescriptively, to obtain the rough equivalence. Some definitions will be helpful: an ...

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To justify this in theory, you must start from quantum electrodynamics which treats the electromagnetic field quantum mechanically. So, there is no classical potential at the fundamental level (although, in practice, one does make use the notion of classical external fields a lot in quantum field theory, but this notion is in principle flawed as a ...

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In classical mechanics calculate the evolution of a particle means to know its position and its velocity for any time. In general if I say that a particle is in position $x$ and has velocity $v$ and ask you about the kinetic energy. That is $\frac{1}{2}mv^2$ and this is independent of the system you have. The answer is always correct. Solving the equations ...

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Part of your trouble is the incorrect belief that QM doesn't contain force. Let's review how QM works. First, we take classical equations in however many variables we may have, including derivatives of variables with respect to others. We then replace every variable, except time and the variable in a derivative's "denominator", with an operator on the ...

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For your particular example, the kinetic energy of the ball does indeed depend on the height of the ball. The total energy of the ball is constant and the potential energy of the ball is different at every different height y. As a consequence, the kinetic energy of the ball must adjust at every different y to keep the total energy constant. You can see this ...

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