# Tag Info

54

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 ...

48

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 ...

47

I'm going to assume the bottom end of the rod is fixed, so the rod rotates around it. I think this is what you have in mind - shout if it isn't. So at some point during its fall the rod looks like: The mass of the rod is $m$ and the mass oif the weight on the end is $M$, and I've drawn in the forces due to gravity. To write down the equation of motion ...

31

It is because the moment of inertia is not a conserved quantity. The statement that an isolated body can't change its position is more precisely the statement that an isolated body cannot change the position of its centre of mass. The position of the centre of mass, ${\bf R}$, is given by: $${\bf R} = \frac{1}{M}\sum m_i {\bf r}_i$$ where $M$ is the ...

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 ...

23

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 ...

20

In this case, gravity is still an external force. In a zero-g environment, the mouse would also begin to move around the inside of the wheel, opposite the rotation it causes in the wheel, which would keep the angular momentum at zero. This would happen because the only way for the mouse to exert a force on the wheel and rotate it is for it to push itself in ...

17

Calculating the power emitted as gravitational waves is relatively straightforward, and you'll find it described in any advanced work on GR. I found a nice description in Gravitational Waves: Sources, Detectors and Searches. To summarise an awful lot of algebra, the power emitted as gravitational waves by a rotating object is approximately: $$P = ... 14 You are right. To open the door during the same time interval (for pushing at a and b), you should induce the same angular acceleration. Since rotation in both cases is about the same axis, this means you need the same torque, this gives$$ \frac{F_a}{F_b} = \frac{r_b}{r_a} $$where r is the distance from the point of contact to the axis. However the ... 11 This apparent paradox is actually not a paradox at all. Infact, it is because of Newton's second law, that we can say that the object will rotate exactly about it's centre of mass. This is true for any body on which net external force is 0, but net torque is present. In Newtonian mechanics, centre of mass serves to simplify calculations, for exactly the ... 11 John correctly stated that this is possible because re-configuring our bodies allows us to change our moment of inertia, but not our mass. As the question was about an intuitive explanation, consider adding a series of floating weights to get an analogous situation for translational motion: The astronaut stretches their arms above the head, grabs a weight, ... 11 The roll doesn't have to be in a fixed place for a torque to be generated. Torque will be generated regardless of a fixed axis or not and your toilet paper will unroll eventually. Torque is generated because as @CuriousOne said above, the force is acting somewhere other than through the centre of mass thus creating an unbalanced force on one side which ... 11 Not sure if I can add much to Kyle's comment, but I'll try. Looking closely, he starts with no angular momentum about the vertical axis - the take-off is "straight". Then he moves one arm behind himself and then extends it sideways - generating a torque about the vertical axis. By tucking his other arm in tightly, his body can now rotate. At the end of the ... 11 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 ... 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 ... 10 One of my favorite scientific papers of all times (mainly because it's rather bizarre) explains the basics of what's going on here. That paper is Kane & Scher, "A dynamical explanation of the falling cat phenomenon," International Journal of Solids and Structures 5.7 (1969): 663-666. To get even more mathematical, there's Montgomery, "Gauge theory of the ... 10 The physics 101 answer is no: it takes more force, but it is compensated by the smaller displacement so the energy stays the same. If we start with a static door, and we end up with a door rotating at some speed, the energy into the door is the work done by the force and it must be the same independently from the point where the force was applied But let's ... 8 The reason that you get slip at even the smallest forces results not from the fact that the tire is slipping against the ground, but that the tire is elastic. There is no way to completely eliminate slip with an elastic tire. Let's see why this is. To measure the slip, lets put twenty little green splotches of die evenly spaced on the circumference of the ... 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 ... 7 I assume that you're imagining two sticks whose bases stay stuck still to the ground as they tip over. In this case, what we should do to calculate the tipping rate (I wouldn't exactly say "falling"-- I think it's misleading) is consider the torque applied to each stick as it tips. What the answer really comes down to is the struggle between torque (the ... 7 Ernie is close to the correct answer, but the fundamental thing that needs to be considered is how the internal energy of the body flows. I researched this in a very interesting book I''m still reading, Principles of Animal Locomotion . Chapter 7 addresses running and section 7.5 discusses Internal Kinetic Energy . Limb accelerations can store kinetic energy ... 6 Speculating here... I suspect that for light weights the answer is yes - with the right technique. Your center of mass moves up and down which requires energy being absorbed and expended by your legs. Moving your arms with small weights should allow you to even out the motion, lowering the peak stress on your legs so they tire more slowly. In a sense you ... 6 Electrons in a conducting disk in order to maintain equilibrium will have to have a centripetal force on them equal to the local change in potential energy with respect to a change in radius, that is$$ m_e\omega^2 r = -e{d\phi\over dr} $$After integrating, we get a potential difference between the center and a point R out$$ \Delta\phi = -{m_e\omega^2 ...

6

Your intuition is correct. For the ball to change its angular momentum (to go from "backspin" to "forward spin"), there needs to be a net torque acting. There are two forces on the ball: gravity, and the normal force of the slope. Both these forces act through the center of mass - so neither force adds torque. Without torque, there is no change in angular ...

6

No ,the centre of mass would not accelerate,the object will rotate about the centre of mass. If the object is on a table(or anywhere here on the earth) and if you apply the forces as shown in the fig it may not be rotated about the centre of mass because of the frictional forces acting on the object,but in space where there is no friction the object will ...

6

The rotational energy of a body is given by: $$E = \tfrac{1}{2}I\omega^2$$ where $I$ is the moment of inertia and $\omega$ is the angular velocity. For a uniform sphere the moment of inertia is related to the mass of the sphere, $m$, and the radius of the sphere, $r$, by: $$I = \frac{2}{5}mr^2$$ You already have the mass, and you can Google for the ...

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

The Earth's rotation rate and the location of the rotation axis change over time. These collectively are called the Earth orientation parameters. On very short time scales, a day or less, the changes in the Earth orientation parameters result predominantly because of the ocean tides. On the scale of decades to a century or so, the dominant driver is exchange ...

Only top voted, non community-wiki answers of a minimum length are eligible