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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 ...


25

Velocity is relative. There is no special reference frame that would be "at rest". But acceleration is not and was never claimed to be. Reference frames in free fall are special and reference frames that are accelerating relative to the ones in free fall contain inertial forces (circular motion involves acceleration towards the centre; the corresponding ...


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

Earth is about 100x more massive than the moon, and since $F \propto M / r^2 $, the distance from Earth to the astronaut would have to be about $\sqrt{100}$ = 10x further than from the moon to the astronaut. Therefore, the astronaut falls "up" about 90% of the way to the moon. [The earlier answers go a lot more into detail (and are more technically ...


9

Special relativity deals with "inertial" or "non-accelerating" frames. Physics in inertial frames are equivalent independent of their velocity and the velocity of inertial frames are relative. You are free to assume any inertial frame is stationary and all other frames are moving relative to it. Rotating frames are not inertial, they are accelerating ...


8

Suppose you and the table are floating in space. If you push the table it will go in one direction and you will go in the other direction. Your and the table's acceleration will be different so you will end up travelling at different speeds. This is obvious from conservation of momentum. The momentum in the centre of mass frame is initially zero, so after ...


6

Damon writes: Essentially, the second law is the mathematical formulation of the first one, f=ma, f being the unbalanced force acting upon the other body. Actually is the other way round: the first law is a particular case of the second law, where F = 0. If no force is acting on a body its acceleration does not change $F = 0 \rightarrow a = 0$


5

General covariance applies only to freely falling observers -- once you invoke non-gravitational forces, like the inward pressure of the wall, the observer is no longer freely falling.


5

In general relativity, angular motion actually does have some "relativity" to it as well. When you're in close proximity to a spinning object, you'll actually be dragged along with it. This is known as the Lense-Thirring effect, or just "frame-dragging". The most dramatic example is the ergosphere of a spinning black hole, a region where no object can remain ...


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

For your 2 minute egg timer here on Earth it comes out to be 4 minutes 54 seconds on the Moon because: $t_{Moon} = t_{Earth} \sqrt{6}$ Full explanation below. Q: What is the relationship between hourglass flowrate and local gravity? As in the excellent answer to a related question (hourglass flowrate vs. sand grain size) and this published paper, the ...


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 ...


3

If the occupants of the space station were not aware of its design and could not look out a window then there is no way to tell if it is rotating or they are near a earth size planet that causes the gravity. Orbiting around another space station will causes a sensation of gravity, and it seems you are contradicting yourself. If there is any rotational ...


3

Easy way to distinguish between gravity and rotating space station: Throw a ball straight up in the air. If it comes straight down, gravity. If it moves away from you (behind your tangential velocity), it's a rotating space station.


3

It does work - poorly. The sail has to be large enough, and it takes a very strong fan. You would be be better off turning the fan and using it to propel your boat directly as they do on swamp boats in Louisiana. Proof: Mythbusters video. The boat will actually move forwards if you have a strong enough fan and a big enough sail. Imagine that instead of a ...


3

You are wrong at assuming constant friction. Rolling Friction increases when you increase speed of the car (See the formulae at the bottom). Also, aerodynamic drag increases with the square of speed (See the formula at the bottom). So, at higher speed, the car engine needs to counter higher rolling friction and air drag to maintain that speed. While the ...


3

In a perfect vacuum, on a frictionless road, you could just turn off the engine and the car would keep moving, never slowing down. However, in the real world, there are several effects that exert a force on a moving car, slowing it down, such as: rolling drag between the tires and the road surface, fluid drag from the air that the car moves through, and ...


3

Essentially, the second law is the mathematical formulation of the first one, f=ma, f being the unbalanced force acting upon the other body.


3

Further to Damon Blevins's pithy answer, you need to be stating what an inertial frame is, so that you can measure your acceleration. A practical answer: you carry with you an accelerometer, and if this measures "nought" then Damon's formulation is good and Newton's first law is that in the absence of any nett force on it, a body will be either comoving with ...


3

Theoretical Answer Consider that you travelled in your car at $10~{km}/{h}$ for one hour, then at $100~{km}/{h}$ for the next hour. First hour of the Journey: You travelled a distance of 10km, so the work done is $$W=F\times s=10F$$ Second hour of the journey: You travelled a distance of 100km, so the work done is $$W=F\times s=100F$$ The work done is ...


3

Note that Fitzpatrick states towards the beginning, The following solution method exploits the fact that the Coriolis force is much smaller in magnitude that the force of gravity: hence, $\Omega$ can be treated as a small parameter Generally, when statements like that are made, powers (greater than 1) of the term in question are considered to be zero: ...


3

Ahh, Richard Fitzpatrick. Great guy. Ok, If you start with the second set of expressions, use the appropriate double-angle-formula and then assume the "angle" $2\Omega \sin\lambda t$ is small (note that the $t$ is not within the sin function!), you get the first expressions, e.g. $$\cos(\theta+\phi) = \cos\theta\cos\phi - \sin\theta\sin\phi,$$ and then ...


3

This is essentially Pascal's law. I would like to add that it depends on the fluid being in equilibrium, as well as incompressible. Otherwise local pressure differences can build up and the fluid can behave rather differently. I like to look at this problem from a thermodynamics point of view. One way to think of this is that the pressure of a fluid is a ...


2

KInetic energy is never negative, the answer you got was wrong. Of course, you could say methaporically that when the particle is tunneling the KE is negative, but that cannot be measured, and as far as I know, the particle that tunnels is either on one side or the other of the barries, but never inside. In quantum theory the eigenvalues of kinetic energy ...


2

A quick search and it dows not look like this has been asked before. For a high school project I suggest that you might think about atomic physics. The way the negatively charged electrons travel around the positively charged nucleus. I suggest this for two reasons. 1) It is a clear case where Newton's laws do not work. 2) There are simple experiments ...


2

If the kinetic energy of both the ball decreases, how can their velocities be equal?? The front one ( B ),from the beginning, had low KE; if it decreases during the deformation, how can its velocity be equal to the velocity of the rear ball ( A )? In order to get a clear picture, let's consider the extreme case when the velocity of B = 0 Let's ...


2

If an object is accelerating upwards at a rate of $a$ m/s$^2$, then the gravitational force felt by this object is effectively, $$ g_{eff}=g+a $$ where $g\sim9.8$ m/s$^2$ is the canonical gravitational acceleration we all know and love. In your particular case, the common equation of motion for a pendulum, $$ \frac{d^2\theta}{dt^2}=mg\sin\theta $$ replaced ...


2

If the block is resting on a frictionless surface, it won't fall over. It'll set in horizontal motion even with little force. To create torque, one edge of the bottom surface of the block needs to be fixed with high static friction and there must be a non-zero angle between the force and the line connecting force application point and fixed point (you can ...



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