# Tag Info

## New answers tagged rotational-dynamics

1

The velocity jacobian is $\vec{\omega}_B = J\, \dot{q}$ with $q=(\phi,\theta,\psi)$. This is used to transform between the generalized forces/torques $Q$ and the vector torques $\vec{M}_B=(\tau_x^B,\tau_y^B,\tau_z^B)$ $$Q = J^\intercal \vec{M}_B$$ The power through the joint is $$Q \cdot \dot{q} = Q^\intercal \dot{q} = \left(J^\intercal ... 0 If the person is moving in a straight line and then takes a right turn then the answer simply is because of inertia. The person initially moving in a st line would continue to move in that direcn and hence when he takes a right turn it would appear that he turns left but actually he is simply trying to maintain his earlier motion.. If he is moving,in a ... -1 I do not think so. When you make a left turn, you have the tendency move to the left in order to limit the centrifugal force \frac{mv^2}{r}. On the other hand, equilibrium on a bicycle is maintained by making a (small) left turn when you start falling to the left side and a (small) right turn when you start falling to the right side. The negative feedback ... 0 As in the comments, on a flat surface, one needs to raise a square's center of mass for it to roll: its center's distance from the ground varies cyclically between r and \sqrt{2}\,r as it rolls where r is the square's halfwidth. So there is a potential barrier to the motion: one needs to input energy to raise the wheel's potential energy - and that of ... 0 A slab can slide, but it will not move unless the static friction force (which is proportional to the weight Mg of the slab by the static friction coefficient \mu_s) is exceeded by the force F pulling on the slab parallel to the surface on which it is supposed to slide. So, there will be no movement unless$$ F>\mu_s Mg Alternatively, if you want ... 2 The basic physics in laymen's terms Okay, so the basic idea is: an object in motion tends to stay moving at the speed that it's moving. When we apply this to rotational dynamics we have an interesting effect: an objects speed goes linearly with the radius it is from the center it rotates around. So if something is rotating with a period T, it must go a ... 0 If the body is a point mass, and even if frames are stationary, still the angular velocity and accelerations may or may not come out to be the same. The question is a little too generic. Consider for example, the case of a uniform circular motion, if the angular velocity is measured about an axis which does NOT pass through the centre, it will be a ... 1 This issue is a bit confusing because there are two types of angular momentum. There's spin, where a rigid body rotates about an axis through its center of mass, and there's orbital, where the center of mass of a rigid body rotates about an axis. For example, the Earth spins about its axis and rotates around the Sun. The total angular momentum can always be ... 0 First of all ,, how could u be sure that the same force F will produce rotation about different axes simultaneously ? When u Apply some force on the body , calculating from ur body fixed frame , it will have certain torque about the origin ,, If u change the system, then u will get different values of the torque , but here torque has all the 3 components ... 4 If the bearings were to be considered frictionless, then the maximum speed of the fan will not decrease, though it will take the fan longer to reach the maximum speed. Because as the moment of inertia of the impeller increases its angular acc. will decrease (for the same torque applied), therefore it will take the fan longer to reach its maximum speed. The ... 0 Yes, your equation is correct and you can calculate the torque at the engine by measuring the torque at the wheels. However the number you get will be less than the true engine torque because of friction in the drive train. The figure you calculate will be the engine torque minus the transmission losses. Rolling roads estimate the transmission losses by ... -4 Yes, the rotation speed will decrease. The reason is that the fan gets a constant power (energy per second) from the system which translates to a constant torque that rotate the fan against air resistance (and such). Greater mass density of the fan means greater value of the relevant Tensor of Inertia component of the fan. This means it takes greater force ... 3 This will depend on exactly what kind of motor you have. If your fan is a brushed DC motor, then the fan speed will be slightly lower, since the new impeller is heavier than the old, so there will be slightly greater bearing friction. The added friction will serve as a power loss, and the motor will have to run slightly slower. If the motor is a brushless ... 6 Your intuition was correct - the shaft will rotate in one direction and the housing/stator will rotate in the other. If you look up "moment of inertia" you will find that it is the rotational equivalent of mass. For almost any reasonable motor the moment of inertia of the shaft/rotor windings will be smaller than the moment of inertia of the housing/stator. ... 2 The astronauts working on the Hubble space telescope had to bring special low torque wrenches to counteract the effect of the torque of the motor spinning them around, due to conservation of angular momentum, although this meant far more use of muscular power to hold them in place. And also to avoid damaging the equipment they worked on, such as screws ... 2 Since a system must obey the law of momentum conservation, the center of mass of a system (which can be made of one or many bodies) must have constant velocity if no external force is applied. Hence, a body can rotate around its center of mass, or it can rotate around any other point, but only if under the influence of an external force. Therefore one can ... 2 @ChrisDrost's answer is correct, but we can actually remove the assumption that the friction is constant by considering conservation of angular momentum instead. If we put our origin at a point along the ground, then there is no net torque on the sphere: The frictional force always points directly towards (or away from) the origin, and the normal force ... 1 So this is a phenomenon which is known in billiards as "backspin": you hit a ball off-center and it simultaneously has a motion "forwards" but a spin that imparts a force on the ground to send it "backwards". Trick shots where you induce extreme amounts of backspin by hitting the ball almost vertically downwards are known sometimes as "massé shots", if you ... 1 Try to think of this problem using a polar coordinate system. x is essentially the radius r or \rho, measured from pivotal. w is simply the angular velocity. So the position vector of the object is x\hat{\vec r}+\theta\hat{\vec \theta} So the velocity vector is \dot x\hat{\vec r}+w\hat{\vec \theta} The hatted vectors are unit. So the ... -1 Yes it will rotate due to magnets pushing and so providing motive forces, however not for too long at all this is because if you take the friction in account for the object that is generated between the pivot and the magnets. Furthermore, the air-resistance might not be too much but it will definitely extrapolate in results over few tens or minutes and will ... 0 When the balls collide, they can transfer linear momentum. But the smooth surface means they transfer almost zero angular momentum. In the case where the cue ball takes a glancing shot, it transfers little momentum. There might be some spin on the ball, but the large amount of linear momentum means that it behaves almost exactly as you see in the ... 0 I also doubted it to begin with. Then I considered the case of A colliding with B directly on the A-B line. Then A stops cold and all its momentum is transferred to B. Now deviate the collision angle to the left by a very small amount, like 1 degree. What happens? A will almost stop. It will be left with a small velocity to the right, at right angles to ... 0 This actually makes sense if you look at the vector addition. If you add \vec{v}_{2a} and \vec{v}_{2b}, you'll see they add up to \vec{v}_{1a} for all elastic collisions with pool balls of equal mass. (Your vectors in your graphic are not to scale, so you can't do it graphically there.) This makes sense, because when the cue hits the other ball, it is ... 0 In 2D, torque is \tau = (r \sin \phi) F = r ( F \sin \phi). These are equivalent statements. You either consider the perpendicular distance to the line of action of F, or the perpendicular component of F along r. Either way the result is the same. It really doesn't matter how you interpret this expression, as both ways are valid. In 3D, torque is ... 0 A way to understand this is as follows . According to a definition, moment arm is the perpendicular distance from the axis to the line of action of force and the lever arm is the length of the line that connects the axis to the point of action of force. Hence, when the force is perpendicular to the lever arm, the lever arm and the moment arm coincide. For ... 1 A spinning top is a gyroscope; it doesn't need air. The gyroscopes on the Hubble telescope work just fine up there. 11 anyway, how likely is it the ice ages could be explained by the earth 'realigning' so that polar regions would migrate over the surface of the earth? How about zero? The geological evidence of the Ice Ages clearly says that, between the ice episodes, the ice did not move. It's just that the polar caps shrank. For instance, the extent of the last ice ... 2 It will spin about its axis. In general terms, if you attach a flywheel to a big motor in space and turn it on, the flywheel will spin in one direction with some angular velocity, and the motor and whatever is attached to the motor will spin in the opposite direction with a different angular velocity. This is the basic principle behind a reaction wheel. ... 3 They don't want to get in front of the rolling wheels -- at least, not intrinsically. When you are turning, your front wheels stop going forwards so much, and start going sideways. As Newton taught us, a great way to think about such things is that an object in motion will tend to keep moving however it is moving. There are sideways forces on the front ... 2 First, braking causes the car to pitch forward on its suspension and decrease the static friction available on the rear tires and increase that on the front. The rear tires begin to slide because the available friction is less than the friction to prevent sliding. This means they are now in the realm of kinetic friction. The kinetic friction on the rear ... 1 There isn't a simple formula for fan noise, but the physics can be worked out from the fundamental equations of fluid mechanics and acoustics. It isn't a simple problem however. The noise created by fans is complex and from several fundamental sources, and its amplitude depends on frequency. Here is an example of a fan noise frequency spectrum for a cooling ... 0 If I understand the situation you intended to draw correctly, it would be this: except that while your diagram suggests vertical movement, your words suggest horizontal. But I am sticking with the vertical arrangement of your original diagram. The relationship between \theta_2 and \theta_3 follows from this diagram: but while (with all the ... 1 Depends on the nature of the collision. If there is a mechanism that takes energy from the system, i.e. a deformation, than energy is lost. You could think of your example as the center of mass of your rod as a point mass that starts rotating on a massless string once is passes the pivot. As usual, energy and momentum are conserved. You could do the ... 2 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.

22

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 ...

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.

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