# Tag Info

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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 ...

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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 weels, the direction in which the center of mass will not change directly, but the angle in which the rest of the body points will ...

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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 ...

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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 ...

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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 ...

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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 ...

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Your answer seems to be good. I solved it in the same way. However there are two points in which i have no clarity. The first thing i do not have clear is the conservation of energy in the problem. We use the energy conservation, but how do we guarantee that the normal force is conservative? And, still, in the conservation equations we do not consider its ...

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If the person starts with an initial velocity equivalent to the tangent velocity at the edge of the station, he will not slide across the surface on the inside regardless of friction (friction is meaningless in this case, as it cannot act if the person has 0 velocity in respect to the floor beneath him). Imagine we are looking at this space station as a ...

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Friction between the person's shoes and the floor/outer wall of the station would only exist if the person were attempting to start walking around the circumference of the station and thus participate in non-uniform circular motion. Think about how his bowling ball would behave if he just put it down at his side.

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Where are you getting these books? Here it makes absolutely no sense to mention indefinite integrals and I have never seen such confusing statements. You need definite integral, usually in multiple dimensions (ideally 3D, as we live in 3D space, but you can simplify in some cases). The integration goes over the entire volume of the object, you have ...

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I think the book authors might have the Huygens-Steiner theorem in mind when saying that. The theorem says that if the body has moment of inertia $I_{cm}$ with respect to axis crossing its center of mass, then moving the reference axis to another parallel axis gives a new moment of inertia $I'$, related to original one by $$I'=I_{cm}+md^2,$$ where $d$ is ...

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Unless I have the rest of the book, it is difficult to understand what he meant. But many books have equations that could make not much sense or whose explanations are very unclear (no author is perfect). The only thing you really need to know is what you already seem to know. In an actual calculation you use a definite integral, and the constant $C$ doesn't ...

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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 ...

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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, ...

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Now, torque τ⃗ is given by the above relation. But in which direction does it act? Torque, like angular momentum is a pseudovector and not a vector. It is a conventional way of showing the direction (anti/clock -wise) ot the rotation. It is devised in such a way that you can apply the right-hand rule

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That's not the definition of torque -- in fact rolling or not, the perpendicular force causes torque about the point of contact with the ground. The reason the rolling disc does interesting things is that it has a bunch of angular momentum which leads to precession of the axes. It's a brain-mangler to work out (tho' there are plenty of intro Mechanics ...

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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 ...

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I would say an axis of rotation, at center mass, would produce rotation maximization and symmetricality.

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You can always imagine torque as being represented by two opposing forces at either side of the object. In this case imagine the torque being generated by two forces of magnitude F at the top and bottom of the disk, the top one pointing forwards and the bottom one, backwards. In order to generate torque $\tau$, these two forces must be of magnitude ...

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The question (even after the edit) is not very clear. So I will make some general statements about forces, objects and rotation. In order to cause a change in the angular momentum of an object (which is one interpretation of "rotate", although it can mean "stop rotating" too), you need to apply a torque. A torque is a force applied along a line of response ...

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See https://www.andrew.cmu.edu/course/24-352/Handouts/logdecrement.pdf and http://en.wikipedia.org/wiki/Logarithmic_decrement For a damped oscillation you measure the peaks, and record the decreasing amplitudes. From that you calculate the log decrement $$\delta = \frac{1}{n} \ln \left( \frac{x_0}{x_n} \right)$$ where $x_0$ is the initial amplitude and ...

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The way I see it movement of a rolling ball in a bowl shaped like half a sphere may be quite tricky to model exactly and the equations may not be straightforward The one thing that is certain is that damping causes energy to be lost from the motion and you can measure the ammount of energy whenever the ball gets to the highest point of the oscillation. So ...

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Dust sticking to things is a complex process but can be broken down into several stages and analyzed. First though lets define our dust. Dust Size The aerodynamics of dust are most easily approximated by pretending all of the particles are spheres with a density equal to water ($1000 \frac{kg}{m^3}$). Each particle is assigned an aerodynamic diameter that ...

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Imagine yourself as the center post of the gyro and you lean 15 degrees to the right you have a bucket of water that you spin over your head.(this represents the spin of the gyro) As it spins you will see the angle of the bucket spinning and then have a friend estimate the angle If you left the bucket at the same angle as you lean you might fall over so ...

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It's as simple as adding the two vectors, the vector that determines the orbital rate relative to your reference frame and defined origin, and the vector that defines the spin angular velocity relative to the spin axis of your gyroscope. The two vectors are tip to tail connected and their sum is just the vector connecting the origin to the tip of the spin ...

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Choose a point P on the periphery of your gyroscope and paint it with some color. Then, as the gyroscope rotates take photos and if possible, record the time of each picture. Let's name O the origin around which rotates your gyroscope. Now, using the pictures and the times recorded, calculate the change in time of the angle between the line OP and some fix ...

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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 ...

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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 ...

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You assumption that energy changes is incorrect. In both cases the rotational energy is conserved. Rotational energy is given by $E_{rot}=\frac{1}{2}I\omega^2$. As no work is done on the system the energy of the system must be conserved. We then see that as the man walks out $I$ increases and $\omega$ decreases and visa versa as he walks in just as you ...

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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, ...

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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 ...

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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 ...

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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 ...

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The angular momentum is a concept analogous with the linear momentum p = mv, in which m is the mass of the body and v its velocity. Now, see where the angular momentum comes from. Consider for simplicity a body moving on a circle around some axis, and let ω be the angular velocity, i.e. the angle by which the object rotates, in a unit time. The ...

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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 ...

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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 ...

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Maybe you can see it this way: The modulus of a vector multiplication is like this: $$|\mathbf{L}|=|\mathbf{r}\times\mathbf{p}|=rp\sin{\hat{rp}}$$ where you can see the main feature of angular momentum: position and linear momentum of the matter considered need to be both proportional to $L$ and inversely related to each other. That is how you guarantee ...

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Have a look at the video here I hope that when you see the man spinning around and moving the weights (changing $r$) you can see that $r$ is important. Remember $r$ is the distance from each part of a rotating object to the axis of rotation, (which is not exactly the same as the position)

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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 ...

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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 ...

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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 ...

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Objects in orbit tend to lose their spin on their own axis. However they do not completely lose their rotation and end up rotating with a period that is the same as the orbital period, so that they face always the same side towards the other body. The best know example is the Moon that shows always the same side to Earh. The phenomenon is called tidal ...

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So, in the first case the frictional force have the same direction with the acceleration of the center of mass but it's not in the latter one. Can someone explain the difference between those 2. In your first case, you say "to make the wheel roll faster", but you don't say how this is done. Is is pushed? Do you apply a torque to the axle? If you ...

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Yes, you are right! Only when a force is applied purely through the center of mass it results in the body gaining a linear action with no rotational components.When any force is applied at a distance from the center of mass, it results in the body gaining the linear acceleration mentioned above plus an angular acceleration which depends on the moment arm ...

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That's actually a really interesting question. Stating it in a slighly different way: If there is no friction, would the sphere on the inclined plane experience a torque? And the answer is - no it would not. The reason for this is that the external force on the sphere (the normal force of the plane) acts through the center of the sphere. In other words ...

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Without seeing the full context of what he says, he is most likely saying this in the context of classical physics, and it is most likely that he is saying that in this physics, particle dynamics are described by second order differential equations. Thus, if we know the positions and velocities of all the particles at any one time, these data fully define ...

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First of all, this is not a physics problem, it's pure geometry. Secondly, posting homeworks is something, that's not so much welcome here. However, I'll help you out. It's quite simple: If the box started rolling when the center of it's base was touching the top of the ball, then these lengths (drawn red on the figure) will be equal. Their length is $r ... 1 so the total torque relative to the center is zero and the revolving speed stays constant at$\omega_1\$ This is not possible. When the two forces are unequal, there is a net force on the disk which means its center of mass will decelerate. The net angular momentum of the system (disk plus arm) remains constant - because there is no net torque on the ...

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