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I'm not going to answer your question directly, but maybe can point you in the right direction. First assume you have some type of electromagnetic actuator that provides a linear force proportional to current. By shorting the windings of the actuator coil you will basically have a damper mechanism, a device that will resist changes in displacement. But such ...

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You are calculating an unreal force by using Newton's second law. Remember that Newton's laws are valid in inertial frames of reference. And Newton defined inertial frames as those frames where an object continues to be at rest or in constant motion unless acted upon by a real physical force.In your example the train or you is not an inertial frame. Hence, ...

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I think the real reason is because if you change your velocity then everything that used to be at rest should now continue just as they did before you changed your velocity. You can call it indifference, nothing cares how you move unless you interact with it. Or you can call it relativity. What this does is reduce the case 1) of things at constant motion ...

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It's not a force that would break the glass, it's an uneven force (i.e., a baseball smacking the center of the glass). A great illustration of this is air pressure. Air pressure has a force of about 100,000 $\frac{N}{m^2}$ - so a normal glass case big enough to hold a fire extinguisher or something would have about 10,000$N$ of force evenly distributed ...

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On an intuitive level if $a\ll R$ then nearly all of the deformation will occur close to the surface. Imagine for a bit that R is radius of the earth and you're pushing on some dirt with base ball so there's a circular contact patch with a radius of about 1/2 an inch. Now if earth were half as big would the forces/stresses/strain/contact area be any ...

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Consider this previous exchange on the Cosmic Microwave Background The crucial bit is this: However, the crucial assumption of Einstein's theory is not that there are no special frames, but that there are no special frames where the laws of physics are different. There clearly is a frame where the CMB is at rest, and so this is, in some sense, the ...

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You probably got voted down cause this can easily be google searched, but the simplest way to explain it is that a tide happens because the lunar tug on one side of the ocean is measurably more than on the other side of the ocean and as the earth rotates the tidal "bump" follows the moon so you get 2 high tides and 2 low tides a day. A tide is effectively ...

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At the risk of sounding like a broken record... it is a good idea to draw a diagram for all but the very simplest problems (and even then): You can immediately see that the normal force is made up of two components: $F_c \sin\alpha$ and $F_g\cos\alpha$. The friction results from the combination of both of these. In your approach, you ignore the normal ...

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We integrate $x dm$ to say "for every piece of mass $dm$, sum up the amount of mass times the coordinate $x$ of that mass". It makes sense to integrate over $m$, because every piece of mass has an $x$-coordinate. Likewise, in kinematics, it makes sense to integrate over $t$ because for every $t$ we have a velocity $v(t)$. However, calculus doesn't care ...

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The reason why a gyroscope does behave in this strange way is that if you try to rotate it's axis in some direction, the "endpoints" of this axis have to be pushed perpendicular to what our first intuition would say. In order to verify why the axis starts rotating in this strange way, let's make some simplifications: the gyroscope consists of two identical ...

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When the light hangs off the ceiling by means of the electrical cable, the cable actually acts as a very stiff spring. The light bulb provides a downward force due to its weight $mg$. The cable acts like a spring, slightly idealised here as a Hookean spring which provides the counter force (what you called the force of tension) of $k\Delta y$ with $k$ the ...

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The simple word is "damping". Initially when you hang an object from a string (spring, etc), it will move - side to side, and up and down. While it is moving, there will be a changing force on the object - after all it is accelerating / decelerating. This shows up as a force in the string that changes with time. In all "real" systems, there is also a ...

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This is a "good" model from a conceptual standpoint. It's the model Isaac Newton himself used when he tried to model the concept of aerodynamic lift. However, modeling air as a series of little balls bouncing off one side of the sail ignores the contribution of the air on the other side of the sail. Air consists of a bunch of "little balls", not just a ...

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Let's simplify this problem a bit more. Suppose there is no air resistance and the ground is flat and smooth. If we throw this bouncy ball (which I'm assuming will undergo elastic collisions) at some angle to the ground, $\theta$, then it will have two components of velocity; that perpendicular to the ground, $v_y$, and that horizontal to the ground, $v_x$. ...

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The fastest point of sail depends on the boat (both its hull shape and its sail plan), the wind strength, and the sea state. In general, a beam reach is not the fastest point of sail. For instance, in very light wind some boats will go fastest on a close reach due to the increased apparent wind from going toward the wind. For boats that sail faster than ...

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Consider the angle between the two vectors as $\theta$ and the following rules \begin{align} \vec{a}\cdot\vec{b} &= \|\vec{a}\| \|\vec{b}\| \cos\theta & \|\vec{a}\times\vec{b}\| & = \|\vec{a}\| \|\vec{b}\| \sin\theta \end{align} Now to construct the parallel vector use the direction of $\vec{b}$ and the adjacent side of the triangle ...

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When you calculate the "parallel" vector, you should not use the dot product of $a\cdot b$ but instead the normalized dot product $$\frac{a\cdot b}{|b|}$$ times the unit vector $b$. The projection of $a$ onto $b$ should always be independent of the length of $b$.

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The point is that the ball gets a tangential hit by the ground. This changes the angular momentum of the ball. Consider a ball thrown with a horizontal speed v. It should also not rotate. Right before hitting the ground, the ball has an angular momentum of $$L=mvr$$ This is a result of $\vec{L}=\vec{v}\times\vec{p}$, which is also valid for linear ...

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Yes. One way to see this: if $S'$ and $S$ have coordinates $x'$ and $x$, then by the usual rule we know that $S'$ observes a distance of $\Delta x = x-x'$ between them. Differentiating on both sides, we get $\Delta v = v - v'$, $\Delta a = a - a'$, and so on. In other words, velocity, acceleration, and all higher derivatives behave like you think they ...

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Don't worry about relativistic corrections - they are insignificant for most planetary orbital motion at the level you try to model/understand it. You want to look at the vis viva equation which is well explained on Wikipedia: $$v^2 = GM\left(\frac{2}{r}-\frac{1}{a}\right)$$ where: $v$ is the relative speed of the two bodies $r$ is the distance between the ...

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Yes, you're correct. Since $S'$ is an non-inertial/accelerated frame of reference, all objects within this frame is acted upon by a pseudo force that is proportional to the mass of the object and whose direction is opposite to the direction of acceleration of $S'$. The fact that $S$ is accelerating(with respect to a rest frame on ground) with $a=5 m/s^2$ ...

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Yes, as viewed from S', S accelerates at 2 m/s2. One way to think of this is just imagining S' as a rest frame, meaning that the rest frame is travelling with an acceleration of 3m/s2 through 'absolute space'. Alternatively, just comparing the velocities of both frames after each second confirms that they are moving apart with an acceleration of 2m/s2.

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A wavefunction is a function that takes a configuration of all the particles in the universe (if just one particle is in a different place it is a different configuration) and assigns a complex number to that configuration. If you take the square of the length of that complex number you get something people like to call a probability-density. An X-density ...

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Kinetic friction is in the direction opposite the moving object's velocity relative to whatever surface it is sliding on. Static friction opposes a stationary object's tendency to slide relative to the surface it is resting on. In other words, it is opposite the direction in which the object would move—relative to the surface—if there were no friction. ...

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Friction is always tangential to the object's surface. If it's propulsive (causing the motion of the object) then it acts in the direction of motion. If it's resisting, then it acts opposite to the direction of motion.

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As @rmhleo pointed out in his answer, the frictional force doesn't depend on the surface area, because no matter which part of the object is in contact with the other surface, the total normal force (and thus the total frictional force) is unchanged. However, that assumes a couple of simplifying conditions: namely, that the two surfaces are consistent in ...

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Yes, frictional force does not depend on the area. This is clear from a simple mental experiment, think of a block resting on a surface. Assume it is a prism whose faces have different areas. Whichever face it is resting on, the friction force will be the same: when on the smaller surface, the contact is reduced compare with the case where the resting face ...

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$1)$ Let's make all the forces that would be acting on the blocks, in the first case. Now we apply newtons' laws of motion assuming that block of mass 2m accelerates downward with $a_1$ acceleration and the block of mass m accelerates upward with same magnitude of $a_1$ acceleration( because they are constrained to have same acceleration ...

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The idea here is that static friction is larger than dynamic friction. This is something that depends on the nature of the materials in contact and is not in general true because friction is not a fundamental force: it is a result of some very complicated phenomena at a microscopic level. An explanation that does make sense to me is to think of the two ...

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All of this is all right, but the problem is that I'm taking a course on electrodynamics and the teacher said that the work $W_{\mathrm{ext}}$ done by one force external to the system is $$W_{\mathrm{ext}} = \Delta K + \Delta U,$$ that is the change in the total energy of the system. I don't know where this comes from It follows from the work-energy ...

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Definitions and assumption: Mass of the hoop $m$, radius of the hoop $R$, force exerted by the string $F$, angle of rotation $\theta$ ("theta"), pivotal point P. It is assumed the string fits around the hoop without friction. Inertial Moment of the hoop: Around its centre axis the moment of inertia is $I_c=mR^2$ and with the Parallel Axis Theorem the ...

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No tricks needed. You can make it basically without a special design, as long as it follows the general guidelines: piece with hole tied with string to piece with rod. Of course the hole should be a fit loosely the rod. As for string length, it should allow movement of the dangling piece, and the shorter the harder it will be to master. As for the physics, ...

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TL;DR: Substitution inside the delta function yields a Jacobian factor $$\tag{1} \delta(f(v))~=~ \sum_{v_{(0)},f(v_{(0)})=0 }\frac{1}{| f^{\prime}(v_{(0)})|} \delta(v-v_{(0)}).$$ Here the sum is over the zeroes $v_{(0)}$ of the function $f(v)$. Let us for simplicity consider velocity $v$ rather than momentum $p=mv$. So energy conservation $$\tag{2} ... 1 You can actually think of the "little rain drops" that you are picking up as providing some resistance. When you travel at velocity v, and have area A, you are "picking up" all the material in a cylinder with volume V=vA per unit time. That volume of material needs to be accelerated to velocity v, requiring a force F\Delta t \propto m \Delta v = ... 0 The only factor of \pi in the problem comes from the final integral you have to do, which in dimensionless form is$$\int_0^1 \sqrt{\frac{u}{1-u}} du = \frac{\pi}{2}$$This integral is pretty hard, so if you did it quickly (which I think you did, since you didn't mention any painful integrals here), then you made a mistake here. 0 Once the constraining force is broken normal parabolic motion takes over in the direction perpendicular to the tension force to the center of the circle. Depending on the angular velocity at that moment will give you the shape of parabola and mg straight downwards of the ball will be a factor. Draw a force diagram before and after the moment of release. ... 0 What is the configuration needed for a circular-motion? The answer is, there must be an inward(towards the center) force perpendicular to the instantaneous velocity. The minimum velocity required for the bob initially to loop the whole loop is \sqrt{5gR}. In this situation, the tension at the topmost point becomes zero, but since there is velocity, which ... 1 For the second case you want to divide by the force, not the velocity. You are basically computing what fraction of the time you spend at a particular point in phase space. However what you have is a probability density. So P(z,p) is something you multiply by a volume in phase space to get a probability. A correct way to get it would be to consider a ... 1$$ m \ddot{\vec{r}} = \vec{F} $$multiply by \dot{\vec{r}} and integrat over t:$$ m \int_{t_0}^t \ddot{\vec{r}} \cdot \dot{\vec{r}}~ dt = \int_{t_0}^t \vec{F} \cdot \dot{\vec{r}}~ dt$$With \frac{1}{2} \frac{d}{dt} (\dot{\vec{r}}^2) = \ddot{\vec{r}} \cdot \dot{\vec{r}} it follows:$$ \frac{1}{2} m v^2 + \left( - \int_{t_0}^t \vec{F} \cdot d\vec{r} ...

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Here is a solution for a spring force that varies directly with displacement. It thus varies with time implicitly, but has no explicit dependence on time or any other variable. Givens and Assumptions oscillator with mass $m$ amplitude of oscillation $A$ oscillator displacement, $x$, varies with time, but $x(t)$ is unknown spring applies force varying with ...

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What about using Galilean free-fall? From $S= \frac 12 g t^2$ and $v = g t$ you get that velocity after a fall $h$ follows $$h= \frac 1{2g} v^2$$ We conclude that if the ball is consuming some essence to get velocity from the line of fall, this essence must be "stored" in space as the square of the velocity. The idea works because if we have already a body ...

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When you write: Derivations (or at least, convincing arguments) of the kinetic energy formula that didn't require the work formula required relativity to make sense, which is unbelievable considering that Newtonian mechanics were established well before relativity. I assume you are referring to arguments like Ron's argument. Although such ...

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To obtain some kind of practical answer, you have to determine how k varies. For example, if k varies with temperature, I would determine its value at -50, 0, and 50 degrees, then use those values and calculate T (which varies inversely as the square root of k). I would Use more points if a higher accuracy is required. If a formula is required, I would ...

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When you talk about g's, you are talking about acceleration, not force. So, we use the formula for centripetal acceleration of the outside rim of the tire: $$a = \frac{v^2}{r}$$ where $a$ is the acceleration, $v$ is the velocity of the outside edge of the tire, and $r$ is the radius of the tire. $$a = \frac{v^2}{r} = \frac{27.778^2}{0.1909} = 4,042~m/s^2$$ ...

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When you jump from any given height, the force pulling you down is gravity with $F=mg$. This makes you accelerate to faster speeds as you fall farther, obviously. When you hit the ground, you do not experience the same acceleration. Otherwise, it would take just as long to stop falling as it took to get up to that speed. Hitting the ground imparts a much ...

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You're looking at the net force acting on him at the instant he begins falling. And you are correct that, in the absence of air resistance, the net force that is acting on the man the instant after he begins falling in the same, given by $F_g=mg$. However, there's an old saying that says, "it's not the fall that kills you; it's the sudden stop at the end." ...

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Recall that force is equivalent to, $$F=\frac{dp}{dt} \sim \frac{\Delta p}{\Delta t}$$ Where $p$ is the momentum, and $t$ is time. Momentum is given by, $$p=m \cdot v$$ Where $m$ is mass, and $v$ is velocity. When you hit the ground falling from 2 meters. The change in velocity is very small, and the change in time is small. When you hit the ground ...

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To find the minimum velocity at the bottom-most point, we find the minimum velocity at the uppermost point. This minimum velocity is the one such that the centripetal force is equal to the force of gravity. Any lower and gravity will pull the object down and out of the loop, any higher and a faster velocity would be required to generate it (thus, not a ...

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A tension less than zero isn't really physical; this is the point where the string stops being taught and the object doesn't make a complete circle. Therefore, when finding the minimum velocity for an object to make it around a loop, we solve for when the tension at the top is zero as it is the minimum possible tension for the object to keep going in a ...

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