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## Hot answers tagged newtonian-mechanics

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Yes it is, if you can throw it hard enough. Not bothering with things like air resistance etc. (I think this is the least of the plausibility problems) you need to put the spear into a low-Earth orbit, such that the centripetal force is provided by gravitational acceleration. $$\frac{v^2}{R} = \frac{GM}{R^2}$$ Using $R=6400\ km$ and $M= 6\times10^{24}\ ... 9 No. No matter how hard you throw. Since orbits are ellipses, all trajectories meeting the criteria must pass through the ground at one point except for the surface-grazers. Air resistance will not be negligible so there's no point in assuming it will be. The effect of air resistance on any shape other than a lifting body is a drag force straight backwards ... 7 Shouldn't there be many psuedo-forces to account for planetary motion? In theory, yes. In practice, no. Consider the third body perturbations induced by Alpha Centauri (a two solar mass star system at a distance of 4.37 light years) on Voyager 1, which is currently about 130 astronomical units from the solar system barycenter. This is on the order of ... 5 TL,DR: it cannot be done because there is no spear strong enough to withstand the acceleration. Long answer: There are two orbits that would give you this result. The first is one where you throw the spear "slightly up" - it would rise out of the atmosphere as drag slows it down, and slowly descend at ever-steeper angle upon re-entry. With the right launch ... 4 No, you cannot throw a spear hard enough to circle the earth. The impact depth$D$of a wood spear of length$L=2.5m$(density$d_1$below$1000\frac{\mathrm{kg}}{\mathrm{m}^3}$) in air (density$d_2$about$1.2\frac{\mathrm{kg}}{\mathrm{m}^3}$) is about 2km (using Newton's approximation$D = L \frac{d_1}{d_2}$). Note that the velocity does not enter here, ... 4 You really have two questions here How do we identify inertial frames? How is it that acceleration is not relative when position and velocity are? The first one is harder than the second. Identifying inertial frame We define an inertial frame as one in which the laws of physics take on their usual (simple) form, and identify non-inertial frame by the ... 3 According to this paper (note: there's probably a paywall), I can draw three important reasons why you can not actually throw a spear around Earth. The paper, for those who can't access it, is titled "Effect of Vibrations on Javelin Lift and Drag". In it, they show that the amplitude of induced vibrations in a javelin is greater with greater throwing ... 3 Suppose you are an observer stationary in this infinite gravitational field, so you feel a gravitational force$g$just as you do standing on the surface of the Earth. The only difference is that this acceleration$g$is constant and doesn't change as you go higher or lower. In that case your spacetime is described by the Rindler metric (as Phoenix87 ... 3 From the perspective of Lagrangian mechanics, the tension$T$is a constraint force that does no virtual work. Can you see why? Hence it can the be ignored in the Lagrangian formulation, cf. D'Alembert's principle. See also e.g. this Phys.SE post. The only remaining force in the Lagrangian formulation is gravity, which we encode via its corresponding ... 2 It doesn't. It's only when the projectile and freely falling body are released from rest at the exact same time, and the effects of air resistance are roughly equal on both of them (feather aimed at a cannonball, not so much). And that's simply because gravity affects them the same. Its analogous to how a car going 70mph looks like a car going 0 mph when ... 2 The force$mg$that you reference is the force due to gravity on a test mass$m$. That is the mass you drop. Now,$g$is a function of distance, $$g=-G\frac{M}{r^2}$$ If a mass were at the end of a tube, infinitely far from the mass$M$then$g=0$, $$\lim_{r\rightarrow \infty}-G\frac{M}{r^2} =0$$ So the object would not fall when you drop it! Thus ... 2 You have assumed that the entire energy dissipated by friction is the KE of the hammer on impact. But the problem details that besides the energy on impact, the hammer gains energy by dropping further. It loses PE corresponding to dropping an additional$0.02m$. That energy has to go somewhere, and it goes into work done against friction. So you could ... 2 A slightly more fundamental way of looking at elastic behavior is in terms of the elastic moduli. These still assume a (locally) linear relationship between the extent or shape of an object and the applied force, but they are defined in such a way that they only depend on the material and you have to put it the dependence on length and area. To use the ... 2 The Lagrangian boundary value problem is ill defined in general (it is well defined when the initial and final configurations are sufficiently close to each other). There is no existence and uniqueness property of the solutions. Conversely, the local and global problem with initial data, if the kinetic energy of the Lagrangian is positive-defined is always ... 2 If it's going too fast, it'll wind up escaping Earth altogether. Obelix might be able to throw it hard enough it spiraled around at least once (for sufficiently large values of force imparted by magic potion). Sadly, once the spear has made its first lap, we know its speed has fallen so much it is now back at head height. It won't make another lap at the ... 2 You just need a different, more fundamental, definition of inertial frame. Landau and Lifshitz define it as "a frame of reference that describes time and space homogeneously, isotropically, and in a time-independent manner", or within Newtonian dynamics, a frame of reference in which objects not acted by any force stay at rest or move with constant velocity. ... 2 Any trajectory, on a sufficiently small scale, can be thought of as being "instantaneously" part of a circular trajectory by dint of the fact that you can talk about the instantaneous radius of curvature of the trajectory,$r$. When you look at the direction of the particle on a (small piece of a) circular orbit with radius of curvature$r$, then after a ... 2 Is there some flaw in my reasoning or is the work kinetic energy theorem valid really only for particles? It does not matter whether the system is modeled as number of particles or rigid body. What matters is whether you want the work-energy theorem to refer to macroscopic kinetic energy only or allow other kinds of energy. In your example, when you ... 2 I'd just like to add to the answer given by dmckee♦ regarding identifying inertial reference frames. The confusion you're having appears to be due to using a naive definition of an inertial reference frame. The first thing you need to understand is that gravity is another one of those pseudo-forces that dmckee♦ mentioned. Even if you ignore the Coriolis and ... 2 What follows is a version of the statement you want to prove which assumes that any two frames are related by a spacetime transformation that leaves time invariant up to translation and that preserves Euclidean distances. Because of these hypotheses, the statement below is a Newtonian answer to the question. I'm confident, however, that a similar ... 1 Your first line is actually wrong. The corrected one should be $$\ dU = - F\cos\theta ds$$ . Since you're not adding/removing any energy, the force (whatever your potential energy is coming from) should increase the kinetic energy and decrease the potential energy. 1 The easiest way, though it may not be what you want: make the sphere and cube out of rubber (so it doesn't slide) and choose a large enough sphere. The constraint of rolling without slipping can get rather complicated in 3D, but we can simplify things to the 2D case. First consider a circle radius$1$fixed at the origin. A square side length$2a$moves ... 1 A physical scenario will help you visualise some of the other answers, particularly Muphrid's and nonagon's. Imagine an aeroplane coming in to land, as a wheel of its landing gear touches the ground. If the tyre isn't spinning before contact, the point of contact on the tyre is moving at the same speed as the aeroplane, approximately$70{\rm m\,s^{-1}}$, ... 1 Suppose you surround yourself with a sphere of test masses that are too small to have any significant gravitational field. You are in an inertial frame if the masses remain as a sphere and do not accelerate away from you. You do not need to refer to any other frames - this measurement is done entirely in your own frame and works even if you are in a sealed ... 1 Yes - if the plate is stiffer, then the deflection will be smaller and so the amount of work done (which is force times distance) will be less. For a typical linear elastic situation, the work done will be$\frac12 F x$where$F$is the final force and$x$is the displacement. The factor$\frac12$comes about from the fact that the initial force needed for ... 1 ABS is not intended to shorten the braking distance, this system is for prevention of sliding when it occurs between roar and tire due to the exerted force exceeds the friction force. ABS triggered by sensors' signals to control panel and cosequently control panel commands to decrease the braking fluid thus slightly allowing acceleration. Then it again ... 1 Let me put it this way,all forces that we know arise from interactions between two particles/systems. This is true either for fundamental forces such as gravitation, electromagnetism as well as derived forces as friction,viscosity,etc. The potential or the potential energy as a concept is introduced to take advantage of the conservative character of most ... 1 In case of tomato falling down, kinetic energy is gained both by the tomato and the earth. However since$E_K = \dfrac{\mathtt{P^2}}{2M}$(Same$\mathtt{P}$for both) the heavy mass of earth shows very small change of kinetic energy. The kinetic energy of the system would therefore include both, the earth and the tomato. The$E_K\$ of earth is neglected and ...

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Any body which is free falling in a viscous medium will reach a terminal velocity. Although mathematically this is usually attained after an infinite amount of time, from a physical point of view you can assume that such velocity, up to uncertainties, is actually reached by the body after a finite transient. The forces that come into play are the fluid ...

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Suppose you raise your book to a height. Here as you said the potential energy of book will increase. now what happens if I let the book go down? The potential energy will now convert into kinetic energy and hence the potential energy of the book will decrease. Now what If I stop the book in the midway by catching it before it touches the ground? When ...

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