# Tag Info

8

To apply Noether's theorem, which is what you are alluding to here, one needs to look at continuous symmetries of a Lagrangian description of a system's dynamics. The damped oscillation equation you have written, although it is invariant with respect to a time translation as you rightly say, is not a Lagrangian description. If you write the Lagrangian for ...

4

Conservation of energy is related to time translation invariance for systems that can be described by a Lagrangian. Dissipative systems in general are not describable by Lagrangian mechanics (without altering the formalism that is) and so Noether's theorem cannot be applied to check whether or not energy is conserved. EDIT: the dissipative system the OP ...

0

If you are still interested in this question (and hopefully still have uses for the answer), I will try to answer it in an efficient manner. First of all, The rolling resistance force is an interaction between the ground and the wheel, which is independent of speed ONLY when the ground surface is completely flat and rigid. If the terrain is bumpy/hilly, ...

0

This sounds like it's an example of ground-effect. When landing a standard fixed wing aircraft (from experience, a 2 seater Cessna) the aircraft descends at a constant rate towards the runway, assuming that the controls and engine power remain fixed. However, a few metres above the runway this descent slows dramatically - the point where you thought you'd ...

0

Imagine that friction between wheels and road disappeared in one second. Then the car would just go away moving straight along tangential direction, right? In the life it doesn't, consequently there is centripetal acceleration that keeps the car on the circle way. Consequently there is some force doing that. It is friction.

1

The force of friction acts both towards the centre of the circle and opposite the velocity vector of the car. Strictly speaking, the diagram you have does not show all forces acting on the car but it is enough for purposes of explaining the circular motion. As the text also explains, circular motion always requires a force pointed radially inwards because ...

1

I would look at this in a slightly different way. Rearranging it: $$m \ddot{x} = -(a|\dot{x}|+k) x = -k_{eff} x$$ If you look at it that way, it is really a variable, non-linear stiffness $k_{eff}$ that depends on the velocity, rather than a damping that depends on the position. In this respect (assuming $a > 0$), the stiffness coefficient has a lower ...

0

If by "solve" you mean to get an answer that is independent of θ, the answer is... Nope. You need θ, it doesn't cancel out. However, here's a slightly similar problem that you CAN solve: You've got a similar hillside, and a sled at the top of it. The height of the hillside is h, the friction coefficient is μ, and the angle is θ. You let the sled slide ...

1

As already explained in other answers and comments in General Relativity (GR) energy is not conserved. Some people and physicists say it is, it simply gets lost by the matter-energy and gained by the gravitational field, or viceversa; this is more pleasing to our sense of conservation of something, but it has problems in that the gravitational energy, is not ...

2

When the wheel comes in contact with the belt friction will act as there will be relative motion between the belt and the point of contact. Now friction will tend to act on the belt opposite to the velocity of the belt until slipping ceases. To find the work done by the external agency lets consider the energy changes : 1) The K.E of the wheel increases. 2) ...

0

If before putting the mass $m$ on the piston, in absence of external constraints, the piston is in equilibrium; since it is not mentioned in the problem description that gas has received some amount of heat; thus it is impossible that the piston moves upwards after putting the mass $m$ on it due to gas pressure. So, this question violates physics laws and we ...

0

To determine the coefficient of friction, (in an ideal situation, of course) push a known mass of the object with a given force F. Also the frictional force has a numerical value of (approximately) coeff. times normal force. So you can calculate the coeff. this way. There are momentary 'bonds' formed between the surfaces. These bonds arise due to Van der ...

1

Half of your questions are concerning Newton's law: Why is it non-uniform? Because the object would have different densities in different parts, so weight would be greater in some parts? Yes. Think of a car. It is in contact with the ground in four places and pushes down causing four normal forces. If the car is heavily loaded with bagage in the ...

0

Belive it or not, but I solved it in very funny and simple method. I made even lighter function, but it works better. Instead of using slip ratio percentage, I use slipping speed(non %, non ratio - a linear slip that depends on speed) and it works like a charm. $V_{slip} = \omega r - V_{longitudal}$ Where: $V_{slip}$ - wheel longitudal speed $\omega$ - ...

0

I think the book answer is correct. The small angle approximation is not necessary : $sin\theta$ does not appear in the analysis. $x = r\theta$ is exact. The motion is a rotation about the point of contact with the horizontal plane. Using the Parallel Axes Theorem, the moment of inertia about this point is $\frac{1}{2}Mr^2 + Mr^2 = \frac{3}{2}Mr^2$. ...

0

This is a linear and homogeneous differential equation. That means you can produce new solutions by adding other solutions together and multiplying solutions by a constant. That means they form a vector space. I'll try to avoid over complicating things for you, but the idea is that there is an infinite number of solutions, you can make new solutions out of ...

1

The differential equation you quote is fairly standard in university physics/engineering course but definitely requires some calculus to solve. As a first step, if you know how to differentiate products and chains, you can substitute the given solution into the differential equation and verify that it is indeed a solution. It contains two arbitrary constants ...

0

Friction between two surfaces is determined by the roughness of the two surfaces. This sounds obvious, and a perhaps equally obvious-sounding question is "is a polished surface smooth?" You could make the argument that a polished concrete surface is smoother than an unpolished one, but then again if I gave you a polished concrete ball you could probably ...

1

It depends also on the slope of the surface, for example, during WW2: "Among the features of the Soviet tank [T34] considered most significant were the sloping armour, which gave much improved shot deflection ...", sloping amour helps deflect incoming projectiles (look here).

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The primary mechanism by which explosions cause damage to materials is through the momentum transferred by the shock wave. I don't see how a smoother surface would mitigate this. However, there are plenty of other ways to strengthen materials, an ancient example being work hardening.

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The nature of frictional forces is primarily electromagnetic forces, in case of both rough surfaces and in case of adhesive surfaces. A rough surface acts like it has teeth on it, which goes into the grooves of another surface sliding on it, and the teeth apply a contact force (which is fundamentally electromagnetic) on the groove walls, which we see as ...

0

The gravitational force of the first mass is $$F_1 = m_1\cdot g/ r\cdot sin(\varphi)$$ and on the second mass $$F_2 = m_2\cdot g/ r\cdot sin(\varphi+\pi)$$ which is in total $$F_g=F_1+F_2=m_1\cdot g/ r\cdot sin(\varphi)-m_2\cdot g/ r\cdot sin(\varphi)$$ which cancels to $0$ if $m_1=m_2$. So if your disk is aligned vertically and the two masses are ...

0

Easiest way to think about it is as an electrostatic force between the atoms/molecules of two different materials causing repulsion. This is based purely on the EM field of the atoms/object, and their proximity to one another. This is the case for both "simple mechanics and fluid mechanics" so for both a "tyre on a road" and "fluid flowing in pipe." Is ...

0

The displacement in x direction will be a straight line all the time including the travel on the incline plane. There is no reason it will change its value. The displacement in y direction will be zero on the horizontal plane and change in parabolic because of gravity.

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The straight line should be followed by a continuous, downward parabolic curve. There is a gradual change between the straight line and the parabola. Because while the block is on the plane, the x coordinate is still increasing with time but now at a decreasing rate because of the acceleration $gsin(\theta)$ along the plane, opposite to the velocity of the ...

1

The only way that friction can appear is for there to be tension, since it is tension that will give rise to the normal force needed for friction. Now if we note that the friction must result in a difference in tension between the left and right strings (if the masses are different) then there will be a continuous change in tension. The normal force at ...

0

If the pulley is free to rotate and perfectly frictionless then the tension in the string is the same throughout, from A to D. So the tension in BC is not zero. You are correct in your comment stating that the net force on BC is zero, because the tensions in AB pulling right and in CD pulling left are equal. However, the net force acting on section BC is ...

0

The discussion which has ensued from the question here and in A conceptual doubt regarding forced oscillations and resonance hinges on how resonance is defined for particular situations and what is meant by the natural frequency of the driven system. An often used definition of resonance is: Resonance is the maximum steady state response of a driven ...

0

Friction can be recognized in a statistical description of a microscopic view, not only in macroscopic view. It is when energy (or any other conserved quantity) that was previously organized becomes disorganized. In this case, when the atoms that were flowing along the surface, after interaction with that surface, lose that flow, and the energy associated ...

1

If there is no friction the energy you put in initially will be conserved and the flywheel will rotate forever, then you don't need to put in any extra power to keep it running.

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For First Case: the rotation produced by the torque at the centre of wheel will rotate the wheel in the clockwise direction, but here friction is present, as friction opposes the motion of particle that's why it acts in anticlockwise direction and helps the body to move. For Second Case: here $mg\sin\theta$ will act along the line of centre of mass, which ...

0

Imagine there are two objects instead of one. One object has a certain mass, and experiences friction. The other object has another mass, and no friction. If those two objects were joined together, you would have no difficulty figuring out the equation of motion. But actually the problem is harder than it looks: the way you have drawn it, the object has ...

1

I agree that friction in the drive mechanism reduces thrust, rather than opposing the motion of the car. However, this is not the case for wheels which are not in the drivetrain - ie where there is front/rear wheel drive instead of 4-wheel drive. Friction in non-drivetrain wheel mechanisms are then sources of resistance to motion. If the car has rear-wheel ...

0

A free body diagram will show a car in motion has air drag force, gravity force and friction force on it. The net force keeps the car accelerated, decelerated, or moving at constant speed. Friction force is due to relative motion between wheel and ground. Engine output spins the wheels (torque from power train system balances the torque produced from the ...

0

PSI does not change with pipe size, only the surface area it is pushing on. Static head pressure is .433 per vertical foot. Water cannot flow over the top of a vessel which is gravity fed. It will only flow as high as the water level in it, because the water in the hose is also pushing back with gravity. A lot of these answers you are reading are ...

1

what's the backward force to balance $F_{fs}$ so as to keep the car moving uniformly? Is that the Force produced by engines? The forwards force comes from the torque produced by the engine of the car and is transferred into the ground via static friction. The retarding force that keeps the car moving at the same speed is mostly air resistance (as well ...

1

I would suggest that the major reason for the ball becoming stationary is its inevitable interaction with the air - i.e. friction, resistance to being parted and eddy currents.

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The reasoning you have stated is correct, as the ball maintains a constant velocity in those conditions. However, in reality, the ball squishes a little bit in the direction which it is moving, so a greater frictional force acts (opposing linear motion instead of "assisting" rotary motion), causing a deceleration and hence the ball stops. It is also ...

1

You have made a model of a viscous fluid coupling which was used in a number of four wheel drive vehicles to transfer torque. The system relies in the fact that adjacent planes of moving liquid experience a viscous force between them.

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It is due to the viscous nature of any liquid. When you stir, the liquid starts spinning and this causes the liquid (the part which is in contact with the pot) to "drag" the pot(due to friction) along with it in the path of its motion.Hope this answers your question.

2

A probable explanation for this effect is simply that the bottom of the pot might be a bit bulged out, as to form only one point of contact around which the pot then can rotate relatively freely (with little friction). As you stir the water inside the pot, the moving water molecules exert a frictional force on the walls of the pot, dragging it in the same ...

1

The force of friction is defined as $F_f = \mu N$, where $N$ is the normal force. In the case of a flat surface free of external forces, you can use Newton's laws to determine that $N = mg$, where $m$ is the mass of the object. Notice that we have made no reference to the objects size, or area of contact. This is because in these examples we have ...

2

So given a spring with spring constant $k$ can one predict what dissipative force the spring will exert when extended? The answer is "No" because they depend on different things. The stiffness depends on the elasticity of the bonds between the atoms/molecules which make up the spring and the damping depends on the permanent distortion of the bonds which are ...

1

Spring stiffness is not responsible for energy loss. Consider a spring with stiffness $k$ but no damping. The work done in compressing the spring by a distance $x$ will be stored as the potential energy(PE) of the spring. For a spring compressed by a distance $x$, the PE is given by $\frac12 k x^2$. This energy is not lost and can be used by letting the ...

-1

Why do you think that It doesn't make sense to me that the normal force would be smaller than the perpendicular component of mg Yes, it is, because the ground is bending away below the car. If it would bend quicker the car would be flying and there would be zero normal force. No, it's okay. The normal force would be $mg\cos\alpha$ without the ...

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