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With a strong grasp of Lie Algebra and Calculus of variations, "Invariante Variationsprobleme" should provide all the foundation one needs to build Newtonian Mechanics (and so much more). The deeper reason that we use either of these formalism is that they agree with experiment; that either formalism predicts the other is far less valuable than that they ...


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No. The easiest way to see this without invoking rotating reference frames is to write out Newton's Law's in polar coordinates, which work out to be: \begin{align*} F_r &= m \ddot{r} - m r \dot{\phi}^2 \\ F_\phi &= m r \ddot{\phi} + 2 m \dot{r} \dot{\phi} \end{align*} From these, it's pretty easy to see that if we have $F_\phi = 0$ and $\dot{r} ...


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In a rotating frame of reference you usually consider the centrifugal force, one of the three possible fictitious forces. This is because you usually consider it as rotating at constant angular speed. There are other forces however: the Coriolis force that appear when the object moves in the rotating frame of reference. The Euler force when the angular ...


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The rule is simple: regardless of which forces are acting, if the motion is accelerated then there is a net force, otherwise, there not net force. The only kind of non-accelerated motion is motion in a straight line at uniform speed. In particular you options: a)wrong, at it is accelerated in the curved part. We do not know in the straight part. b)wrong, ...


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David Z, excellent answer - You comment that "... it does require that space defines some sort of absolute rotational reference frame". I believe that the existence of the reference frame is provable - In a nutshell - The proof is based on the relationship between angular velocity and the centripetal force it produces. It assumes that the same angular ...


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Your question is very confusing, so I will first attempt to answer the spirit of your question with a cleaner scenario. It is possible to "move" from one place to another if there is minimal friction. You can do so with yourself, a large box, and a bag full of baseballs. But it isn't as cool as it sounds (that is why I had to put quotes around move). If ...


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Your profile lists your age as 17, so I assume you're still at school. At this stage in your physics education you'll only have been exposed to differential equations that have relatively straightforward solutions. I assume the education system does this to avoid putting you off. If you continue studying physics you'll quickly learn that the vast majority of ...


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You're correct that (B) is incorrect, but for the wrong reason. Kinetic energy doesn't have to be conserved since, in any closed system with only conservative forces, mechanical energy is conserved. There are many physical situations where kinetic energy isn't conserved (inelastic collisions for example), hence we do not know kinetic energy is conserved. ...


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You can calculate the maximum height above the launch point with $$y_c = \frac{v_y^2}{2 g}$$ and draw a horizontal line at this height. Now you must know the initial direction of travel you can fit a parabola to this slope while being tangent to the height line at $$x_c=\frac{v_x v_y}{g}$$. The general shape of the curve is $$y = y_c - K (x-x_c)^2 $$ By ...


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As a side note skidding is not a yes or no state with tires. See this answer for more details. The sum of the normal force and friction force that act on the car is the reaction force to the sum of the weight and centrifugal force of the car on the road. We can equate them in the coordinate frame parallel and perpendicular to the road. $$N=m\,g\, ...


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No. W=o for internal forces is valid for all bodies irrespective of their rigidity. Because a non rigid body can b deformed only due to external forces. Not due to internal forces. Because internal forces acts along the same line. With equal and opposite direction causes net force to b 0


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Answering your three questions: He knows the relationship between radius and terminal velocity, but the drops are too small to measure their radius with any accuracy (1 ┬Ám is tiny - looking at such a drop from a distance makes it no more than a speck of light, even with a "micro telescope"). Meauring the terminal velocity, he can deduce the size ...


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With this answer I'll try to note the mistake committed in the original formulation "Landau implies that he has not assumed incompressibility at this point so how is it that one can choose the first form for Newton's second law without loss of generality?" Well the problem is originated when you assert: "If one instead uses $F=\dot p$ so that $$f = ...


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Can Lagrangian Mechanics be justified without referring back to Newtonian Mechanics? Sure; one can deduce Newtons Laws from it. The question is should one? By deducing Newtons Laws one is missing the crucial aspect of induction; the reverse procedure and in a sense more difficult; that is the discovery and invention of a theory that covers a wider ...


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The answer to this question is very much analogous to the answer to how aeroplanes fly. See Physics SE Question "What Really Allows Airplanes to Fly?" and the best (IMO) answer is this one here. But basically the airfoils, sails or vanes - whatever they may be called - deflect the flow of air. They do this by pushing on the air and changing the latter's ...


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Since the Lagrangian results in exactly the same equations of motion as Newton's laws, I'd say that based on their agreement with experiment both are on equal footing. Of course, to get the right equations of motion from Lagrange's equation you have to pick the right Lagrangian, so then you ask how we systematically pick the right Lagrangian. The recipe in ...


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The parachute can not be opened instantaneously. The parachute is fully opened at the "vinicity" of the point $C$. The acceleration due to gravity is constant but there is another force one needs to take into account : the air resistance.


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In the case of radial freefall is from rest at some initial $r=R$, the motion will be periodic if you treat the gravitating body as a point-mass and ignore collisions. Since the radius is strictly positive, it makes sense to substitute $$r = R\cos^2\left(\frac{\eta}{2}\right) = \frac{R}{2}\left(1+\cos\eta\right)\text{.}$$ while conservation of specific ...


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You are correct that centripetal force does not affect speed. So let us consider tangential speed, which is the speed the turning car would have if centripetal force were removed. Linear speed = tangential speed = distance / time = (2 * pi * radius) / time. (To simplify, assume the car goes through one full circle at uniform speed.) The linear speed of the ...


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I would say that indeed the center of the car still move at 50km/h. The angular velocity is $\omega = v/r$, with v the tangantial velocity and r the radius. Therefore the wheels on the inside will go slower (smaller v) and the wheel on the outside will go faster (larger v) to keep the same angular velocity $\omega$. Best, Samuel


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To answer your question directly, when the direction of the velocity of the front of the car does not match the front wheels' direction, the tire is deformed, acting like a spring which exerts a lateral force on the car. The deviation between these directions is called the slip angle, and to a first order approximation the lateral force on the tire is ...


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Think about what the parameters are that determine the period of the pendulum $T = 2 \pi \sqrt{ \frac{l}{g}}$. And is there a change in any of these variables.


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There is an implicit assumption in this problem that the chain is perfectly flexible. In this limit, the part of the chain that just hit the table cannot transmit any force back up the chain. It just bends out of the way when it stops. If you consider the falling part of the chain as your "system", its surroundings include the Earth (acting via a distance ...


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You've inserted the scale between yourself and the planet Earth. The scale's spring is compressed by the force exerted on you by Earth's gravitation. The compressed spring tries to uncompress itself and pushes back on you with linear restoring force. Linear restoring force is exactly equal to the force exerted on you by the Earth's gravitation. By ...


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There is an omission in input data: glider's "fuel" is a sum of its potential and kinetic energies. So heavier glider (towed to same height and velocity) requires more work from the towing plane and starts with more energy than lighter glider.


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Let's say the small piston has inner surface 1 centimeter^2, and the large piston has inner surface 5 centimeter^2. Let's say the pressure in the system is 1 Pascal (1 Newton/meter^2). The pressure per cubic centimeter over ALL surfaces inside the system is the same. The pressure of 1 Pascal at the small piston is also 1 Pascal at the large piston. BUT, ...


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when you accelerate the vehicle and you turn, the center of gravity shifts from different parts of the car body. Maybe I am misinterpreting your question though..


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Let's call the rightward horizontal direction $+x$, and the upward vertical direction $+y$. Both balls reach point 1 at the same time, going the same speed. They both have the same $x$-component of velocity. At the beginning of the dip in B's path, ball A remains at constant velocity, $v_1\hat{i}$, but ball B gains in $v_x$ until the bottom of the dip. It ...


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Energy is required to push something over a distance. In the case of the elevator, when it's not moving a brake can be engaged and the power removed and the elevator will sit just fine. That's because it doesn't take any energy to keep something still. But wait, if all I want to do is keep the helicopter still, then it doesn't require any energy? Sort of. ...


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Assume a person is falling towards the earth. We know that there is a force and thus an acceleration acting on the person. The opposing force is the gravitational force exerted by the person onto the earth equal in magnitude (Newton's Law of Gravity). This force produces an acceleration (Newton's Second Law) but because the mass of the earth is massive as ...


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The equation from the link is an approximation. In your analysis you do not include the mass of the car when considering linear acceleration. Nor do you consider the reaction force of the wheel/axle accelerating the car. You can split the driving torque into two portions, the portion that rotationally accelerates the wheel, and the portion that is ballanced ...


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Lets consider a situation which is in equilibrium, which consists of two pistons with different surface area which are connected to each other by an incompressible fluid. I will also assume that gravity does not play role, so no increase in pressure inside the fluid do to gravity. In this case the situation can be represented with the following drawing: ...


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I realize this thread is three years old. In real life, you measure the angle of trajectory with MOA. Or minute of angle, rifle scopes mostly come standard with these MOA adjustments. Minute of angle adjustments equate to 1/60th of each degree out of 360 degrees. On average one MOA is ~1 in. of bullet drop per 100 yds. Out to 2000 yards you will most likely ...


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Ernie is close to the correct answer, but the fundamental thing that needs to be considered is how the internal energy of the body flows. I researched this in a very interesting book I''m still reading, Principles of Animal Locomotion . Chapter 7 addresses running and section 7.5 discusses Internal Kinetic Energy . Limb accelerations can store kinetic energy ...


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Speculating here... I suspect that for light weights the answer is yes - with the right technique. Your center of mass moves up and down which requires energy being absorbed and expended by your legs. Moving your arms with small weights should allow you to even out the motion, lowering the peak stress on your legs so they tire more slowly. In a sense you ...


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Consider your arms as pendulums. The period of a pendulum is determined by its length and by gravity. Though the period is affected neither by the weights nor by the amplitude of your arms, these two quantities affect your balance and efficiency. When you run you get into a rhythm of arms and legs. Your legs are do the work; your arms are along for the ...


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If we don't consider air friction with the ball, then you can see two different effects: an increase in length of the path and an increase in velocity of the ball. Turns out that the second effect more than compensate the first. An exact calculation isn't simple for arbitrary shapes, but we can see a simplification: let's say from (1) to (2) the length is ...


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It appears that you are only considering the system consisting of the vehicle, i.e. car+wheels+brakes, as @user31782 has pointed out. Examining only this system, there is no way for the car to stop: it is traveling through space at a certain speed, and that speed is independent of the spinning wheels; if you apply the brakes the wheels stop spinning (and ...


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You can distinguish from vertical and horizontal velocity. Both balls have the same horizontal velocity, the difference lies in the vertical component. Up to the second trough there is no difference, but then the second ball accelerates downwards. It can't go straight down, which means that the gravity partially accelerates it horizontally. This difference ...


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Without defining what a thing is, it makes little sense to discuss the ontology of a thing. Does an apple exist? First, one must say what an apple is; once we agree on that, it's straightforward to show (by example) that apples exist. Given a definition of force, force certainly does exist; we can point to time derivatives of momentum that we observe, and ...


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Force is a concept which describes, and can be used to manipulate, real phenomena that exist regardless of the existence of the human race. Newton's second law (Acceleration = Force / Mass) is a definition of force. Mass certainly exists, as do velocity and its time derivative, acceleration. There is no reason to suppose that force does not exist. Force ...


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Other answers and the referenced paper assume a constant radius turn. This path would require a discontinuous steering angle, which is not only non-physical but a somewhat poor approximation for how human drivers drive. The reason drivers don't approximate this technique (besides that it would take very rapid steering wheel movement) is that it would incite ...


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If the ruler has uniform mass(mass acts in the centre), and the rope is light and inextensible, then yes, the tension is equal throughout. Take moments about A, to find the tension of the rope at B, or take moments about B, to find the tension of the rope at A. Either way you should get the same result, since the ruler is in equilibrium, so no resultant ...


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When 1686 Newton writes "Principia...", the inertial frame concept does not exist yet. However, we can find in it Corollary IV (introducing the center of mass CM concept for any interacting body set), Corollary V (Galileo's Principle of Relativity, applied to any limited body set with CM at any uniform velocity), and the today almost forgot Corollary VI (a ...


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By definition you should use total force on the system. In this case total force acting upon system is load, F. What $mx''+kx$ is, a response of the system, not some random force from nowhere, it is just Newton's law. By intuition it looks like that. There is an impedance due to the spring and there is an impedance due to the mass at the end of the spring. ...


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But when I look at it from an inertial frame, I cannot intuitively understand how does the spinning of the Earth makes the mass free falling more slowly than when the Earth is not rotating? In the inertial frame, the mass will have the same radial acceleration whether rotating or not. But on a rotating earth, the mass also has a tangential speed. ...


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I think you are overcomplicating this. Consider an arbitrary point P moving with linear speed $\mathbf{v}_A$. Linear momentum is $$\mathbf{P} = m \mathbf{v}_{cm}$$ Angular momentum at the center of mass is $$\mathbf{L}_{cm} = I_{cm} \mathbf{\omega}$$ Linear velocity of the center of mass is $$\mathbf{v}_{cm} = \mathbf{v}_A + \mathbf{\omega} \times ...


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(I've read your question yesterday and could not find peace because of it, since I could not answer it to myself satisfactorily. ^^ But I figured it out and I hope the following helps... ) They key point is that, if the gravitational force acts as a centripetal force, the amount of centripetal force needed to let an object go round a circle with angular ...


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Now, when reading about Special Relativity, some books says that prior to Einstein there was one "principle of relativity" that could be stated as follows: The laws of Mechanics are invariant in every inertial reference frame That is not the best formulation, since constancy of laws is already assumed implicitly. What is meant by the ...


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You seem to be asking for a higher order approximation to the acceleration due to gravity based on its variation with height. So if the Earth's mass is $M$, and we're at a radius $R$ from its centre, then for a small radial displacement $h$ from this, the acceleration due to gravity is: $$g = \frac{GM}{(R+h)^2} \approx \frac{GM}{R^2}\left(1 - \frac{2h}{R} + ...



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