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New answers tagged rotational-dynamics

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I don't have that text, but I can find the table of contents on the internet. Somewhere in that text (most likely chapter 5 on non-inertial reference systems), there should be a derivation that for any vector quantity $\boldsymbol q$, the time derivative of that vector in an inertial frame and a rotating frame that share the same origin are related by $$... 3 The statement is really about the transformation between inertial co-ordinates and co-ordinates fixed to the body. This is expressed by:$$D_t = d_t + \omega(t)\times\tag{1}$$where D_t is the "total" derivative, i.e. the time derivative in the inertial frame, d_t is the time derivative in the frame fixed to the body. Since there there are no torques ... 0 Indeed. It is due to the law of conservation of angular momentum. The angular momentum of the rotating element within the motor will exactly cancel that of the rest of the motor, thereby giving zero net angular momentum, as with the initial conditions. 0 Allowing the bar to pivot about point G, you will see that R1 (if it was the only force) would rotate the bar clockwise and R2 (by itself) would rotate the bar counterclockwise. The direction of rotation is determine by the direction of the force (up or down) and where it acts in relation to the pivot (right or left). 0 The casing will spin in the opposite direction. That is the principle of reaction wheels. 2 The angular momentum of the earth && mouse wheel system does not change. When the earth pulls on the mouse, the mouse pulls on the earth, so no net moment is seen any arbitrary point in the universe. 21 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 ... -1 Oh, absolutely. With a Superman-like body you can also 1) Lift an entire mountain in two hands without the rock disintegrating under the local pressure, and without the mountain falling apart due to preexisting weaknesses in the rock, 2) Travel faster than light 3) Travel faster than light and travel backwards in time 4) Hear sounds so faint that they ... -1 I believe you could overcome gravity (in the atmosphere) in the same way as birds do - you just need enough muscles to do that. 0 The short answer is, yes. The slightly longer, but slightly more accurate answer is yes, but not noticeably. As you say, increasing humidity by evaporating existing water will cause the mass of the vapor to move to a greater distance from the earth's center, and this will increase the moment of inertia, reducing the rotation rate. Consider that the ... 2 I suspect this is an example of the spinning-egg problem, in which a prolate spheroid (such as an egg) spun on a table about one of its "short" axes will tend to "stand up" so that it's spinning about its long axis. A few explanations have been proposed for this phenomenon, most notably: H. K. Moffatt & Y. Shimomura, "Spinning eggs — a paradox ... 0 Your entire analysis is entirely correct and complete (and in my opinion the best way to go about it). There's no more information that need to be derived about this issue. Other analyses are just a different representations of the same facts. 1 The rotation of the Earth's dipolar magnetic field produces an electric field in space. Because the electric field is zero in the rotating frame, it is equal to$$ \mathbf E=-(\omega\times \mathbf r)\times \mathbf B $$in a fixed frame, where \omega is the angular velocity of the Earth, \mathbf r the radial distance and \mathbf B the magnetic field. ... 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 ... 2 The direction of the motion at any time t is the direction of the velocity vector \textbf{v}(t) as derived by solving the equations of motion; likewise \omega(t) gives you back the direction of rotation according to the right hand rule. friction is the force that causes rotation is not entirely correct. Any force with non-zero torque generates ... 1 What data do you have for linear motion? Your equations are correct if you have the acceleration as a function of time and the orientation is constant. The angular accelerometer can give you the angles as a function of time with integration. Unfortunately, drift can be a problem. The received wisdom is to use an accelerometer (linear or angle), integrate ... 2 You correctly identified there are two angles of interest, labeled \theta and \phi. I actually want to pick two different angles for my analysis - see this diagram: First thing to note is that if the cylinder rolls without sliding, the length of the green arc and the red arc must be the same. Length of green arc: (\phi + \theta) a Length of red ... 1 The hammer can be thought as a mean to deliver enough energy to the nail to deform the underlying material and let it penetrate deeper. Ideally, all the energy the hammer gets from your arm as kinetic energy is transferred to the material and results in its (hopefully) permanent deformation, but as always, that's not the case in practice. What should be ... 0 A hammer delivers an impulse (i.e., a change in momentum, calculated as the integral of force over time, https://en.wikipedia.org/wiki/Impulse_%28physics%29) to its target. In the case of a hard, rigid hammer striking a hard, rigid surface, the impulse will be a relatively large force delivered for a relatively short duration. A mallet (including the ... 2 Lots of aspects to this question. Mass per se does not increase drag - volume of the bob of the pendulum might. The increase in mass (inertia) makes the stored energy of the bob larger (thus - longer time for motion to decay); but the projected area (in the direction of the motion of the bob) will presumably also increase, which result in greater drag. ... 0 I don't know what you mean by "drag" exactly, but if you've got friction forces at play, then the increased mass can certainly increase both inertia and restoring force, making the friction forces less important by comparison (though not decreasing them per se). However in general if friction is not an issue, the gravitational force that tries to bring the ... 3 Your first quote is correct for an idealised model. There is no rolling friction then. Both wheel and surface are considered completely rigid. Ideal model - no rolling friction Non-ideal/more realistic model - rolling friction comes into the picture These pictures are from this link that gives a very good graphic view on this. Going away from an ideal ... 0 Firstly, friction is the resistance to lateral motion between two surfaces and so is required for there to be no motion at the point of contact. There may be confusion between rolling friction and rolling resistance. The former being the friction between the rolling object and the rolling surface required for rolling motion to occur. The latter being the ... 1 Never seen that before, so I just tried it. Cool. I believe that the membrane between the yolk and the white is elastic, so when you first, gently, give the egg a little angular momentum, you are only spinning the white. As the yolk catches up the effective moment of inertia drops, and conservation of momentum therefor implies a higher angular velocity. 2 If you have a good pendulum clock, you may be able to extract a minute amount of energy from it - but usually any additional friction will cause the clock to stop ticking. This is of course a function of the amount of energy you try to extract. Here is the experiment to do: Tape a strong disk magnet to the back of the pendulum. Make a coil of magnet wire ... 0 A magnet on the pendulum could generate energy through coils. It sure can, but the load would quickly bring the pendulum to a halt. Pendulum clocks are a very careful balance of swinging mass, falling mass, springs, gears and escapement. It doesn't take much to upset the balance, and then the clock doesn't run. The clock will run all day because the ... 0 This answer is valid under the assumption that the wire also rotates with the disk. @cag's answer reveals two things : (1) that the electric field is independent of the distribution of charge in the disk. We know that this distribution will vary for materials with different conductivities. (2) that the field is independent of the shape of the conducter ... 1 You are right. In an optimal system, there isn't any friction. But in real life there is, because there aren't any balls or surfaces that are so perfect that they touch just in one point. Also the surface of the ball is never perfectly flat, so there will always be friction between the surface of the ball and the surrounding air. So for calculating the ... 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 ...

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