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44

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 ...


42

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 ...


25

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 ...


24

In start-up and hover each blade produces more or less constant sound. But the sound is attenuated by distance and may not be the same in all directions. Therefore you hear it differently depending on the blade's position relative to you. So as the blades rotate, the sound you hear pulsates because the blades alternately get to positions where you hear them ...


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 ...


9

Ultimately, what's special about angular momentum is this: Look up in the sky. A certain set of physical laws pertain in that direction. Look to the north. A certain set of physical laws pertain in that direction. Look to the west. A certain set of physical laws pertain in that direction. Those physical laws: They're the same in all directions. There's ...


8

I physically understand it as the momentum of an object rotating around something given a certain position. However, I can't give a physical explanation to the formula. Why do we multiply the linear momentum by the position? Why does the angular momentum is a function of the position? Angular momentum $L = mv * r$ (This is a late answer, but I ...


7

It is great that you "think differently" about problems - that is at the heart of all innovation. When it comes to the rotation of planets, you have to go back to the origins of the solar system: Planets are formed by accretion: a large cloud of debris starts to experience some gravitational pull, and as one "lump" becomes bigger than the others, it starts ...


6

The answer depends on what the symbols mean. The question does not make it clear how the symbols are defined. The most confusing quantity is $\omega_2$. How is this defined? Is it the angular velocity of the disc relative to the fixed lab axes or relative to the axle about which it is rotating (where this axle itself will be rotating at $\omega_1$)? Also ...


6

Actually, your book is correct. Even if the most usual uses of angular momentum involve circular or rotating motion, this is not the general case. An object moving in a straight line has angular momentum in a reference frame in which the origin does not fall on the the line. To see this simply remember the definition of angular momentum ...


4

First, assume a spherical helicopter... A helicopter isn't a sphere, or even close to one. Consequently, you get complicated acoustic effects where the pressure waves from the blades reflect off other parts of the helicopter, such as the tail boom. You also get intermittent reflection of the tail rotor sound off the main rotor blades, intermittent ...


4

Consider something like a door. A piece of wood with a hinge on one edge. Maybe it is one meter tall and three meters long. Now say that you're trying to hold the door in place, at the position half a meter from the hinge, while someone else throws a baseball at the other side of the door. If the baseball hits the hinge, you don't have to push at all. If ...


3

There are several ways to describe a particle's motion. For example, in 2 dimensions, you could use cartesian $x,y$ coordinates or polar $r,\varphi$ coordinates. To each coordinate, we can associate a 'quantity of motion' or 'generalized momentum'. If a given coordinate corresponds to a symmetry of the system, the corresponding quantity is conserved by ...


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

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 ...


2

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

That's what they wrote. But I am really confused why they wrote so. In fact I can't imagine torque & angular momentum without circular motion. Why did they tell so? What is the cause?? Please explain. A planet in an elliptical orbit has angular momentum alright, there is no problem. The magnitude of angular momentum does not change, is conserved, ...


2

I think the Hamiltonian is not necessarily the energy for the following reason: you can demonstrate that the Lagrangian may be deduced from the D'alembert principle which is linked to the concept of force, etc. but it may be also deduced from the Hamilton's principle which is a pure mathematical concept applied to physics (a certain quantity has to be an ...


2

This is an abstract answer, but I find it extremely helpful to the kind of "basic nature" question you seem to be groping for. Think of two things: Noether's theorem and a thought experiment "what if we had evolved as unsighted but clever beings?". As in David Hammen's Answer, it is Noether's theorem that would tell us that if our physical laws are ...


2

Why does the angular momentum depends on the position? Angular momentum is always defined relative to a reference point, say $\mathbf r_0$, (which is often, but not necessarily the origin). If the system is invariant under rotation around this reference point the quantity that we call "angular momentum with respect to $\mathbf r_0$" is conserved. (Note, ...


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

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

The key here is the phrase in the original question: $\omega_2$ is relative to the black arm This means that in the lab frame of reference, the disk is spinning with angular velocity $\omega_2-\omega_1$. And with that piece of knowledge, the answer follows what you already know: the total energy is the energy of rotation of the center of mass plus the ...


2

It is generally caused by poor piloting technique. Blade-slap occurs when the helicopter is allowed to slowly drift downward while the pilot is still applying significant power. When blade-slap occurs, the pilot should either stop the descent, or lower the collective to enter a more positive descent.


2

For a ring: $Ke = \frac{1}{2}m ( \int_0^{2\pi} (Rw_1-rw_1cos(\theta)+rw_2cos(\theta))² +(rw_2abs(sin(\theta))² d\theta) $ $Ke = \frac{1}{2}m \int_0^{2\pi} R²w_1²+r²w_1²cos²(\theta)+r²w_2²cos²(\theta)-2Rw_1²rcos(\theta)+2Rw_1rw_2cos(\theta)) -2rw1cos(\theta)rw_2cos(\theta)+r²w_2²sin²(\theta) d\theta$ $Ke = \frac{1}{2}m ...


2

Actually you do hear a constant whoosh (although often drowned out by engine noise) when you are directly below the helicopter. The tips of the blades cause a wave to propagate outward at the speed of sound. This wave does not have the same strength in all directions. If you are a distance away laterally, then you hear these waves in succession produced ...


2

This is simple, take the linear and angular velocity of the center of mass (point B) and combined them with the inertial properties Linear Velocity of B : $\vec{v}_B = (0,d \omega_1,0)$ Angular Velocity of B : $\omega_B = (0,0,\omega_2-\omega_1)$ Mass of disk $m$ Mass Moment of Inertia of Disk $I_{zz} = \frac{m}{2} r^2$, where $r$ is the radius of the ...


1

The movement of air around the blades of the rotor produces a "white noise" due to turbulence. However, the "white noise" is actually "colored" and a certain band of frequencies predominates. As the blade approaches you there is a Doppler shift toward a higher frequency, and as it turns and goes away from you the Doppler shift reverses. It is this ...



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