Finally, I think that the reason for assuming a much faster rotation than precession for the top, is to simplify the calculations and consider the top as a gyroscope. The angular momentum has to be conservate: i. As cedric said, the gravity, works for the axis of the spinng mass to fall horizontally on the plane: if this happens also the angular momentum as to torque!
Then u can consider that the magnitude of the angular momentum is proportional to the spinning speed: so as the spinning velocity gets higher it gets, for lack of a better word, "easier" for the top to resist the gravity..
If u try to spin a top on an inclined plane you will need to spin it faster to obtain the same "resistance to gravity"! From your linked article :. Spin a top on a flat surface, and you will see its top end slowly revolve about the vertical direction, a process called precession. As the spin of the top slows, you will see this precession get faster and faster. It then begins to bob up and down as it precesses, and finally falls over.
The drawing shows a circle instead of a spiral due to leaving out variables like friction and gravity. The quick answer is that, for the top to fall over due to gravity, each fragment of the top that is moving around the spin axis has to change its individual direction of movement. They are already changing direction around the spin axis, due the rigidity of the top keeping them moving in a circle. And as it slows down, the effect of gravity has more effect, and it falls over.
This is a nice example which shows understanding does not come automatically after completing a calculation. But calculation still serves the perhaps the most important guide. I think these discussions have already elucidated the issue. Unfortunately, they proceeded using Euler angles. I have reformulated their discussions here. Hope that helps. Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group.
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The spinning resists the torque imposed by the gravity-reaction pair. If the material point is not fixed to the rotating platform, but forced to perform a rectilinear uniform motion with respect to the platform, the dynamics can be conveniently described from the non-inertial, rotating reference frame.
The action-reaction counterpart of these real forces, acting on the platform, are the reactive central force, R C , and the reactive Coriolis force, R COR. The sum of these two forces can be referred as the reactive force R.
Note that the magnitude of these forces are not the same as for the rotating precessing reference frame. Table 1 offers an overview of the reference frame dependent interpretation of the forces acting on, and the angular momentum of the square wheel system.
In this table, the forces acting on the elements of the fluid or chain, as well as those acting on the support, are separately marked. Note that the net force including the forces present in the pivotal suspension point O and net torque acting on the support has to vanish, since the support itself is considered to be massless.
Conceptually, the inertial frame allows the simplest, intuitive explanation of the gyroscopic precession. A straightforward treatment can be conducted by computing the contribution of torques acting on the support.
Since the support is assumed to be massless the net torque acting on it has to vanish. Let us consider a bent square wheel , with the radius of curvature, H , equaling the length of the axis attached to it see Figure 2. We study the configuration with the angular momentum of the top pointing outwards. The centripetal forces keeping the chain or fluid on the curved trajectory have larger magnitude in the lower, and smaller magnitude in the upper fluid or chain segment. Accordingly the reactive central force acting outwards on the lower support segment will also have a higher magnitude than the one acting, also outwards, on the upper segment see Figure 3.
Therefore, two external torques will act on the support: one, N G , originating from the weight of the chain or fluid, while the other one, N R C , is raised by two reactive central forces, having different magnitudes and acting on the upper and lower square wheel segments. Figure 3. Note the lower magnitude of the forces acting in the upper segment. Notation of forces see section 2. The pair of reactive central forces will exert a torque on O which can counterbalance the torque of the weight.
This is the essence of the gyroscopic effect phrased in the inertial reference frame. Note that the square wheel model also provides an intuitive understanding on the sequence of events happening when a spinning top hanged on its axis extremity is released.
During the fall a pair of forces will arise in the vertical segments which will lead to a transient angular acceleration around the vertical and precession is initiated. Simple analytic treatment of the problem from the inertial reference frame is allowed by the flat square wheel model Figure 4C.
Equilibrium conditions are computed considering the forces and torque acting on the support. The result of the calculation is the precession angular frequency. Figure 4. A,B Centripetal forces, F CP , acting on the fluid or chain in the middle of the horizontal segments marked by black continuous lines, reactive central forces, R C , acting on the corresponding support segments are marked by dotted green lines.
C Sketch for the calculations of the torques raising in the symmetric points marked by r up and r low see Equations 2, 3. Here we mention that an alternative interpretation of the forces in the inertial frame, relying on the d'Alembert principle, may also be instructive. In this approach, we compute the forces and torque acting on the upper and lower square wheel segments from the reference frames moving together with the circulating chain or fluid for the case when the nutation angle is right, i.
The precession angular frequency can immediately be obtained as before. The flat square wheel model enables an analytic solution leading to identical results to the one presented in section 2. These individual comoving reference frames have different distances from, and orientations with respect to the precession axis. Centrifugal forces, as well as the corresponding reactive central forces have to be computed for all of these reference frames individually.
This complicates the calculation of the net deviation torque. In this case, though we have to settle with an approximation. It is more intuitive to apply for the bent square wheel model [ 21 , 57 ] Figure 2C. This circular setup facilitates the calculation of the individual centrifugal forces. In this case, when summing up the elementary centrifugal forces arising in the upper and lower segments, we consider them as being parallel to each other and the axis of the bent square wheel Figure 5.
Figure 5. The vertical fluid or chain segments will not contribute to the deviation torque. It is especially instructive to determine the torque in a reference frame precessing together with the center of mass of a flat square wheel see Figure 6.
As a consequence, besides the centrifugal forces, Coriolis forces will also arise. As we will see, only Coriolis forces in the upper and lower segments will contribute to the deviation torque. Figure 6. A Lower flow speed, when the centrifugal force dominates over the Coriolis force, yielding an outward pointing net inertial force.
B Higher flow speed, when the Coriolis force dominates over the centrifugal force, and an inward pointing net inertial force will result. As centrifugal forces do not depend on speed their torques cancel. Therefore, only the Coriolis forces will contribute to the deviation torque. Their magnitude on the upper and lower segments is the same, but they point in opposite directions.
Note that the net reactive force raising on these segments can point radially inwards or outwards as well, depending on the relative magnitude of the Coriolis and centrifugal forces.
The deviation torque exerted on the suspension point O will therefore be. We obtain, without any approximation, the equilibrium condition expressed by Equation 4.
Since the calculations are tedious, we just remark two important conclusions. First, the point of application of the resultant of the centrifugal forces will not be anymore the center of the square wheel. Supplementary Figure 3 shows the schematic structure and functioning of a device that demonstrates the lifting torque of the forces arising in the horizontal segments. The precession of bodies having, besides an angular momentum, an additional magnetic moment, bear special importance: the interaction of the magnetic moment with an external magnetic field will also contribute to the torque.
Therefore, the precession will be altered compared to the purely gravitational case. In the following, we examine a generalized square wheel model involved in both gravitational and electromagnetic interactions. This structure is placed in uniform magnetic B and gravitational g fields see Figure 7.
Figure 7. The calculations rely on the formal similarity between the expression of the Coriolis and the Lorentz force [ 59 , 60 ]. In this paper we quantified the square wheel model of spinning tops.
Although there were several attempts to find an intuitive, force-based explanation of the phenomenon of precession these approaches are not simultaneously simple, quantitative and intuitive [ 7 , 8 , 47 , 48 , 51 , 52 ].
Circular tops do not allow easy calculations, dumbbells are oversimplified, while the first, forgotten paper of the square wheel model did not provide any quantification. The proposed model consists of a heavy chain or fluid which frictionlessly circulates in a closed loop determined by a square-shaped tube, and therefore an angular momentum will arise.
Our approach is more intuitive than the classical explanation relying on the Newton-Euler equation on the alteration of the angular momentum.
We discussed an extended model as well, where the flow carries an electric current and an external magnetic field is also present besides the gravitational one. These thoughts contribute to the better and more intuitive understanding of the difficult, but intriguing problem of the precession of heavy tops, which is the ground for understanding various phenomena ranging from mechanical gyroscopes to magnetic resonance. Note that variants of the square wheel model can be applied to intuitively explain the dynamics of free tops as well see Supplementary Sections II.
I and J. PH came to the idea of the square wheel, made the initial version of the calculations, prepared the figures, the referred patents, the numeric simulations, and equally contributed to the writing of the paper.
ZL refined the calculations, performed the calculations in the Supplementary Material , and equally contributed to the writing of the paper. The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest. We are indebted to A. Lakos, Gy. Orosz, P. It drops below the horizontal as it slows down.
The other end bobs up and down rapidly due to nutation. The string swings out from the vertical to provide a centripetal force on the gyro since the gyro rotates slowly in a circular path about the vertical axis. That is simply amazing. It looks like a magic trick with a hidden spider thread holding up the other end. The torque about the center of mass is directed into the screen in the photo. The angular momentum of the disk lies along the spin axis and points to the end supported by the string.
The change in the angular momentum points into the screen, since that is the direction of the torque, meaning that the tip of the angular momentum vector moves into the screen and the opposite end moves out of the screen towards you. It still looks like magic, especially since the film is viewed in slow motion and the disk seems to be rotating at low speed.
Here is another explanation. The red particle is moving up at velocity v1 and the blue particle at the top is moving into the screen at velocity v1. What happens next? The disk could rotate out of the screen or it could rotate downwards, which is the intuitive expectation.
And vice-versa if the disk rotates downwards. The change in v is directed left to right in each case, as shown in the diagrams below, so there must be a left to right force F on the particle to change its velocity. The torque is due to the string pulling up on the axle and it acts in a direction into the screen.
A horizontal force applied to increase the precession frequency would cause the disk to rise. Here is a gyroscope from www. If the wheel is not spinning then the counter-balance at the opposite end is not heavy enough to hold the wheel up.
If the wheel spins, the gyro generates an upward torque to balance the counter-weight, and the gyro precesses slowly. The upward torque is proportional to the precession frequency. If the support is pushed horizontally to speed up the precession, the gyro rises vertically since the upward torque increases. In , Professor Eric Laithwaite showed that he could lift an 18 kg disk above his head, using one hand, when the disk was spinning on the end of a 0.
In Feb , Derek Muller repeated the experiment. The disk acts a gyroscope and precesses in a horizontal plane rather than falling vertically. If the disk was not spinning, it would be impossible to support the disk in that manner. Note that Derek pushes the axle horizontally at the start to speed up the precession so that the disk rises more easily. If you spin an object on a table, then friction will slow it down until it comes to a stop. A rattleback not only slows down to a stop but it then reverses direction.
There are many theoretical papers on the subject and many videos of rattlebacks on YouTube, but there are no simple explanations. All existing explanations are very mathematical and quite obscure. Nevertheless, there is a simple explanation, and it involves the effect of friction. You can make your own rattleback by cutting the handle off a plastic spoon and attaching two small masses to each end of the spoon.
The secret of a rattleback is that the distribution of mass is not lined up with the geometric axis. So, the two small masses must be located as shown in this 3Mb QuickTime slow motion video taken at fps. I used two small pieces a Blu-Tack, a re-usable adhesive. With reference to the diagram, the spin axis angular momentum is the black arrow and it responds to the applied couple green arrow by changing in that direction red arrow to give a new spin axis blue , and this change in direction is called "precession" orange.
These "arrows" are called "vectors", but never mind that. To figure out the arrow relates to the direction of spin just point along the arrow and then wiggle your finger clockwise. There's an important result from all of this - you get slower precession if the spin is fast or if the couple is small. This is because the smaller the couple, or the larger the spin angular momentum then the change in spin direction red arrow is smaller. The moon, for example, goes around the Earth at constant speed, but its' direction of travel is changing continuously due to the force of the Earth's gravity.
See other gyro links here. Explanation B less technical : What stops a spinning top from falling over? All we need to show is that the force of gravity is insufficient to cause the top to fall.
The only force acting to push the top over is gravity. First think what happens when the top is not spinning.
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