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53

No, a car cannot steer on a frictionless surface. This has little to do with gyroscopic action and more to do with conservation of momentum: to turn, even when conserving its speed, the car needs to accelerate at right angles to its motion, which changes the total momentum of the motion. This change in momentum requires a force which, in normal roads, is ...


47

If the wheels had spun fast enough for a gyroscopic effect to become noticeable, the only result on a frictionless surface (which would be the same without a surface at all) is that when you turn the wheels, the rest of the car would rotate instead of just the front wheels :) You need some reaction force to alter the trajectory, like a sail or surface ...


26

Yes you can It is actually possible with a real car, but you would have to be very patient to steer a little bit. Suppose you have built a car with power on the big front wheels to induce a gyroscopic effect. If you rotate the wheels, the direction in which the center of mass is going will not change directly, but the angle in which the rest of the body ...


20

Newton's third law is pretty near to the mark. All of the phenomena you cite stem from the principle of conservation of momentum in an isolated system, itself ultimately a result (through Noether's theorem) of the fact the physical description of that isolated system is unchanged if we shift the spatial origin of our co-ordinate system. So, if you're in ...


16

In your problem, "Earth" is not an isolated system. The combined "Sun-Earth" system, however, is, so we can know that the angular momentum of the Sun-Earth system is conserved. As the earth's mass is accelerating the sun, you have to take its angular momentum into account as well. While the mass and size of the sun mean that we can ignore its motion with ...


11

You are a point in the circle. The torque is: $$ \mathbf\tau = \mathbf r\times\mathbf F $$ Where $\mathbf r$ is the position of Earth and $\mathbf F$ is the force (radially towards the sun). Notice when your reference point is somewhere in Earth's orbit, as you said, and your object is earth, the force will not be parallel to the position. Therefore, the ...


10

Since there is no friction, then it will not affect any other forces that may act on the car. The direction of wind blowing on the car may change its trajectory, as any driver will attest when driving in high winds. Turning the car wheels may have a slight affect on the resultant direction of the force. If the car has curved roof, then it may acts as ...


6

If I'm not wrong, it's basically the same principle in which an astronaut would throw something in empty space and, with so, move in the opposite direction. It's not thrusting against something but throwing energy and power by burning fuel according to the law of inertia... I could be wrong so I'd like someone more knowledgeable to double-check, please :)


6

I'm a little confused why you ask about GR at the very end of your question. If your question is simply how a rocket is able to accelerate in space without having anything to "push" off, then we can tackle your question pretty well with classical mechanics. Newton's Third Law has its limitations at times, but for this question it will work just fine. ...


5

A model I encountered as a kid, back when the Space Race was in full swing, is still the simplest explanation I've found: Think about an inflated balloon with its neck closed. It doesn't go anywhere, because the pressure in all directions is equal. (Which is what keeps it inflated, too.) Now open the neck. What makes the balloon fly around isn't actually ...


5

No need to call the general relativity. You can understand it using only classical mecanics laws. If you are on a skateboard, and throw a mass in front of you, you will move backwards. As long as the mass does not hit the floor, the center of mass of {you+the mass} is the same, however you did move. The rocket engine ejects gaz from combustion. The ...


5

Possibly your confusion arises from not considering that KE is a scalar, whilst momentum is a vector. Yes, there is of course a connection between KE and momentum: $K = p^2/2m$ (for non-relativistic bodies). Thus, for two equal-mass particles heading directly towards each other at equal speeds $v$, their velocities are $\pm {\bf v}$ and their momenta $\pm ...


4

I'm 27 and since I was about 15 I had the same doubt you do. Only a couple of years ago I realized why momentum is always conserved in a collision, whereas the same is not enforced for energy. (They must have told me this at some point -- or points --, but I guess sometimes I just don't pay much attention) First of all, let's make it empirically clear that ...


4

Friction is the only force that would cause the car to move along a different path. On a frictionless surface, the gyroscopic effect could change the orientation of the car a bit, but not the trajectory of the car. In other words, the front car would no longer point along the direction of travel, but would "skid". (That is, if you could call frictionless ...


4

Is there no relation between the momentum and kinetic energy? Are they not linked with each other? No, they're not. Conservation of momentum only says that $$m_1 v_{1_-} + m_2 v_{2_-} = m_1 v_{1_+} + m_2 v_{2_+}$$ I used subscripts 1 and 2 denote the particles involved in the collision and the subscripts - and + denote the pre- and post-collision ...


3

Newton's third law states that if object A acts on object B with force $\mathbf{F}_{AB}$, then object B must act on object A with force: $$\mathbf{F}_{BA}=-\mathbf{F}_{AB}$$ When expressed in terms of A and B's momentum, the same equation can be written as: $$\frac{\mathrm{d} \mathbf{p}_A}{\mathrm{d}t} = -\frac{\mathrm{d} \mathbf{p}_B}{\mathrm{d} t}$$ ...


3

Since the early 2000s, Matthew Bate and collaborators have been producing smoothed particle hydrodynamic (SPH) simulations of collapsing clouds. The clouds have an initial uniform density, no net angular momentum, but a turbulent velocity field. These clouds begin from rest and collapse under their own gravity to form hundreds of stars including many with ...


3

On a completely frictionless floor, with the absence of other external forces, the centre of mass of the car will continue in the same trajectory for ever. Hence no steering is possible. However, irrespective of whether the front wheels are rotating or not, turning of the front wheels will produce a counter torque changing the orientation of the car, albeit ...


2

I asked the question because I did not believe in the accepted answer that has been sitting for more than 3 years. I have my own understanding, but since it is not good practice to put it with the question, I am posting it as one possible answer. My problem is that I do not believe the first statement quoted in the question which is contradicted by the ...


2

There is an angular momentum problem with regard to star formation, but you have the sense of the problem completely backwards. The problem is not where the angular momentum arises. The problem is where does it go. Gas clouds a tenth of a parsec across have been routinely been observed to rotate at about one revolution every five or ten million years or so ...


2

You could start from the premise that there was not net angular momentum in the universe at all; but it would still be the case that everything of interest was spinning. On the scales of stars and planets there are (at least) two important mechanisms that result in individual systems having angular momentum. The first is turbulence. If you take a parcel of ...


2

In this scenario, the dust in over-dense regions will fall on radial trajectories to the gravitational centre of their over-density. Assuming density inhomogeneities are continuous (meaning no abrupt change in density), we can always model this in a symmetric way local to each over-density. This means that in the frame of reference of the centre of gravity ...


2

Start with the force felt while holding weights and spinning with arms at full extension. Ask if it is easier or harder than when not spinning. Here you are forcing the weights to move from a straight line and to go in a circle, the force has to be felt all the time to keep pulling the weights into a circle. To make the circle smaller requires even more ...


2

If the (entire) system does not change if you displace it in space, then the total momentum (of all particles) will be conserved. Momentum is the generator of translation, which is a specific case of Noether's theorem. To relate it to your example of living creatures: imagine cat on very slippery ice and neglect air resistance and friction. If you slide ...


2

Conservation of linear momentum, for a physical system whose particles are initially at rest in a given inertial reference frame is equivalent to the fact that the center of mass of the system remains fixed at its initial position. You see that this constraint is quite week for a system made of a large number of particles as a human body. Starting form a ...


2

Two things. First of all, in general, if you make simplifying assumptions that aren't actually true, then why would you expect actual momentum conservation? You should only expect your calculation to yield approximate momentum conservation. Secondly (and more importantly), is that if you are applying a constant force $f$ to this system, then you should ...


2

The simplest way to get this intuitively is to consider a rocket where the exhaust gasses escape in two opposite directions. So, there is a nozzle at one side and also the opposite side. In this case, the rocket will go nowhere. The chemical reactions produce gasses at high temperature and pressure and they then accelerate and escape in both directions. If ...


1

You assume that the angular momentum of the right hand rod will be zero after the rods have separated, but this is not so. Consider this diagram showing the rods before and after: Angular momentum is always measured about some reference point. Any reference point can be used, but for convenience I've chosen the centre of mass of the two rods and I've ...


1

In canonical quantization one constructs the Hamiltonian formalism. Energy conservation is therefore manifest (since Hamiltonian is time-independent and commutes with itself). Quantum-mechanically, the Hamiltonian of the system can be expressed via particle creation-annihilation operators. So, the total energy of the field is also the total energy of all ...


1

It is unclear what you mean by transformation involving space-time, well, at least, I found two possibilities. The mentioned transformations are not gauge, since $\Lambda$ does not depend on space-time. So one way to make it involve space-time is to set $\Lambda \rightarrow \Lambda(x)$. The given Lagrangian is not invariant under these gauge ...



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