Visit ilectureonline.com for more math and science lectures! In this video I will explain how torque is applied to a spinning gyroscope. Next video in this series can be seen at: • Physics 13.6 The Gyro...
i understand why the gyroscope doing circular motion, but i still dont understand why its didnt fall down, can someone make in fbd of gyroscope and draw what force that make the gyroscope stay in the air, and also the equation
@@bocilreplay5349 torque = rxF = rxmg. torque > F. Therefore, the gyroscope rotates in the direction of the torque instead of falling down due to the effect of force (F=mg).
You, dear sir, are the absolute best teacher I could ever hope to find. I've gone through this topic on so many platforms, lecture slides, textbooks, and even MIT OCW... They helped in either getting the math right or in visualizing it happening. This one enabled both to occur simultaneously in one smooth flow. Thanks a billionzillion.
Nice explanation. I noticed 2 mistakes. Formula for ω (precision) is in his form correct only if R=r! ( Length of the radius of the wheel equals handle.) 1. L = I*ω(rotation) I = R2*m Put this two equations together L = R2*m*ω(r) He used the letter r which indicates the length of the handle! So the real formula is: ω (precision) = R(handle length)* g(gravitational constant) / ( r2(length square of the wheel radius length) * ω(rotation)) 2. In reality you do not divide vectors but absolute vectors or norms. Norm or absolute value of a vector is written and defined as follows: ‖ vector ‖ = length of the vector or in this case the numerical value of the L. Mathematically correct it should be written ‖ dL ‖ / ‖ Lo ‖ ( L is always with arrows, because it is a vector.) Nr. 1 is real mistake, 2 you can call it splitting hair.
With all due respect shouldn't the moment of inertia be half of mr^2.?? A disc is being considered here.... So half is essential. So I=1/2mr^2 Note that: For ring the formula in the video is correct... I=mr^2
Thank you so, so much. This made procession finally click for me. I especially appreciate how you compared a spinning gyroscope to a non-spinning one & that you went step-by-step through the algebra. (My textbook always skips over the algebra and it makes it all much less intuitive!)
The r's shouldnt cancell each other because the numerator r is the distance of the disc from the pivot , and the R in the demnominator is the radius of the disc , which are different quantities
Please answer my previous question on what is "rigidity in space" of a gyro. Rigidity in Space Rigidity in space refers to the principle that a gyroscope remains in a fixed position in the plane in which it is spinning. An example of rigidity in space is that of a bicycle wheel. As the bicycle wheels increase speed, they become more stable in their plane of rotation. This is why a bicycle is unstable and maneuverable at low speeds and stable and less maneuverable at higher speeds. By mounting this wheel, or gyroscope, on a set of gimbal rings, the gyro is able to rotate freely in any direction. Thus, if the gimbal rings are tilted, twisted, or otherwise moved, the gyro remains in the plane in which it was originally spinning.
Ah yes. Gyroscopes stay in position because of the conservation of momentum. It is the conservation of momentum that gives the gyroscope the motion of precession.
@@MichelvanBiezen According to your video the axis rotates due to precession. PLEASE ANSWER THE QUESTION that I originally submitted. Where/why "Rigidity in space" derives from so that it causes the axis to not rotate?? It is not the same as precession per all the web sites I have visited.
The induced Torque by the action of gravity can be mathematically expressed as = L x W, where L is the induced angular momentum and W is the initial spin angular velocity of the wheel. In general, this induced Torque counteracts all the external Torques on the system to keep the precession on the horizontal plane.
Thank you so much for your amazing videos! I have been following your channel since more than half a year now and watching them really made me understand many concepts (back then, the pulley-block systems) and helped me a lot in my college entrance examinations! Now I'm in college and out teacher started gyroscopes a few days ago but I could barely follow what he did. Your explanation makes it so easy and clear! :"D
Doesn't your final equation for angular frequency of precession only hold when the radius of precession is equal to the radius of the wheel? Since the angular momentum on the spinning wheel, L = mR^2(omega_spin), where R is the radius of the wheel. This is different to the radius of precession, r, that you've defined. Forgive me if I'm misunderstood. Thank you for the video though!
The gyroscopic force in general is due to the Centrifugial force in the radial direction.This force can be decompesed into components in spherical coordinates as in the case of earth experiencing precessional motion around its rotational axis. The component of this force on the meridianal axis is harmonic and gives rise to the torque that causes the precession of the earth around its rotational axis.
Drew this out and followed along and defined all the terms I didn't know. Cleared up 99% of what I needed, Thank You! Now to use the math to define my gimbal. Couldn't be happier!!
Bravo: this video explains precisely the tail dragger airplane left turning tendency: the propeller spins clockwise from the cockpit, the angular momentum straight out of propeller center pointing forward. When the engine starts, the tail dragger tail lifts up and exerts a downward force similar to G force on the spinning propeller. The gyro precession creates a left 90 degree torque which drags the nose leftward. This video does a much better job explaining gryo precession than the FAA aeronautical knowledge handbook ✈️😍🤓👍
So there are 2 (two) systems at work here? The spinning disk alone and the spinning disk with the armature. The disk alone has both the torque and the angular momentum in the same direction (frame 8:05-8:15). And that is what holds the gyro in place? And it's the torque of the armature that processes the whole system?
The spinning disk along with the force of gravity provides the torque to make the whole system spin around (precess). The precessional motion produces the angular momentum which must be conserved. That causes the gyro to rise up (to make the moment arm smaller) which forces the precession to increase in angular velocity.
6:06 shouldnt the moment of inertia of disc be 1/2MR² [here R is the radius of disc and not the moment arm 'r']. And you shouldnt have cancelled out this 'R' with 'r' in the very next step of simplification ??
He has considered it to be a ring and not disc. One can imagine it as a wheel with spokes. Mass of spokes very small. And yes you are right about the cancelling ' R'
Firstly thanks for your video. Naming conventions I have used below- i - x axis j - y axis so this white board is x-y plane. k - perpendicular to x-y plane ( i.e. inward or outward) Here we begin- Clearly the change in angular momentum in the i-k plane is in the direction of the torque due to gravity. But how is the angular momentum conserved vertically? When the complete axle will rotate in the i-k plane, then the angular momentum will be in j direction also. Which requires another torque in the y direction. This torque can only be generated using an external force at the beginning in the i-k plane perpendicular to R. (i.e. k axis) This simply means that initially a push will be required to rotate it. It also satisfies the conservation of energy principle as per my understanding. Please share your thoughts in comments.
Sir, can I say the reason that the gyro is not falling down because there is not torque to change the angular momentum in vertical direction, and since there is no torque is acting in vertical direction, the vertical angular momentum is conserved, and that is why the gyro is not moving in vertical direction?
Thanks for wonderful video and great teaching. Although most of the puzzling aspects are answered but one thing which I am still unable to understand is that why is it still not falling? Mathematically the direction of torque is considered to be perpendicular to plane of distance and force but practically if we imagine the torque caused by weight will cause the body to rotate in clockwise direction. In order to keep the rotating body upright, an equal and opposite torque is required without getting into any kind of mathematics to avoid fall in vertical direction. Isnt it? So in summary imagining a cancelling force to counter weight in vertical direction is getting difficult. Or may be I have missed the real point.
It would have fallen down if there was no initial rotation given to the Gyroscope! And that fall would have necessarily caused angular momentum in the direction of Torque! Thus agreeing with the fact that torque equals change in angular momentum and also is in the direction of such change. But in case when you give a rotation to the Gyroscope and release the system on its own, there's already an angular momentum in the direction shown. Now the torque will produce change in angular momentum toward inside of the board just the same as it did when no rotation was given, so net resultant angular momentum will be shifted in position toward inside of the board, causing the so called precession! Summary - if there is torque, angular momentum has to change. Imagine it this way. First comes Torque, then comes dL in direction of the torque. This dL will decide the motion of the object!
All explanations given here (also in the video) are not complete. The reason for this is that the reaction forces at the gyroscope's contact point are always neglected. For example, a centripetal force is required for the precession movement, which is applied by the bearing in the centre. The explanation in the video only shows that a torque causes a tilting movement at the first instant. Euler's gyroscope equations are necessary to describe the entire movement over all times. The reaction force through the bearing can then also be derived from these equations. It can be seen that the reaction force has both a tangential and radial component as well as a normal force vertically upwards, which on average compensates for the weight force. However, the force components of the reaction force are oscillating, resulting in additional nutation. It is quite easy to see that the explanations here are incomplete: The gyroscope apparently suddenly begins a precessional motion and thus gains rotational energy, in contradiction to the conservation of energy. In reality, the gyroscope falls a little and converts potential energy into rotational energy. In addition, the gyroscope no longer rotates around a principal axis of inertia (addition of the angular velocities), which makes everything much more complicated. Unfortunately, you will only find a complete explanation in older literature: archive.org/details/elementarytreatm00crabiala/page/n5/mode/2up
Today i went to a science museum at Valencia, Spain, there was a thing with two spinnable heavy wheels and a weight where the distance between the CM and the weight was less than the distance from CM and the wheels, and i started to accelerate one of the wheels with the weight closer to the spinning wheel, and it happened exactly that it started to rotate , then you could spin the other wheel, change the distance of the weight and see how it precess it. And at the end of the day, i can understand how it works
Is there possibly an error with 1 / L term, because in this case L is angular spin momentum of the disc whose value is (m r^2 omega-spin)/2? But you instead substituted angular orbital momentum, which is (m r^2 omega-orbital)?
Interesting demonstrations! I wonder, if there two gyroscopes opposed to each other, and a motor in the middle as pivot to accelerate artificially the precession movement faster than its natural speed, would it bring a fraction of the precession force up?
@@MichelvanBiezen Let's say there would be only 1 gyroscope mounted like in the video, but instead of a neutral pivotal point, there would be a motor to force a 10x faster precession speed (or make it spin in the direction opposite to natural precession), what would be the effect on the gyroscope's speed and torque's direction? Thank you for answering :)
Can you please explain what happens if you don't allow the precession to occur? Does the spinning disc fall? what are the force and moment components like?
That is a really good question. And I am not sure about the answer. Next time I have a chance to set up the gyroscope, I will give that experiment a try.
I have a doubt sir. I the very last equation that you derived. If the angular velocity of the gyroscope is zero. It will mean the the angular velocity of precession becomes infinite! But how? Actually if the angular velocity is zero the precession won't be there. Second question is the earth axis also had precession. Who provides the torque for this. Pls answer. And great explanation. Thanks.
Good morning professor☀️First day after the spring break. I’m confused why you mentioned tan theta instead of just using sin theta, as I couldn’t see the reason why I should care much about that in the very beginning sin theta equals tan theta approximately.
Depending on what wheel and what spin you are talking about, it does make a difference. The rotation of the disk itself needs to be such that the angular momentum vector points outward. After that you can rotate it either direction.
Thank you for the video upload. I understand Torque = r×F, where r is the distance where the force is exerted on the mass and the F is the mg, where g is the gravitational acceleration due to gravity. But what is the torque, the r component and the F component as they relate to the Earth's precession in r×F?
thank you for the good explenation. Only one thing is not clear for me: at 6:31 there is division by r. But one r is the lever arm of de gravity force and the r from r^2 was from the moment of intertia of the spinning disc. So i thought that r from r^2 was the radius of the spinning disc. Then how can they cancel eachother out when the r's are from two different radius? or what am i seeiing wrong?
Still doesn't explain why it seems to defy gravity, indeed it explains why it processes but I don't see why it processes and fall simultaneously. Can you please explain?
If the speed of the disk would remain constant (no friction) then the gyroscope would precess indefinitely. SInce the disk will slow down, the gyroscope will start moving upward and the precessional velocity will increase. If the gyroscope would move downward, it would violate the principle of conservation of momentum.
I don't want to be pedantic, but I was only with you up until the dL divided by L and the Torque divided by Ang. Momentum part. Dividing by a scalar OK. But how can you divide a vector by another vector? Or do these equations only hold when treating L, T, et al as magnitudes? For example is omega actually the torque scaled down by the magnitude of the initial angular momentum?
In reality you do not divide vectors but absolute vectors or norms. Norm or absolute value of a vector is writen and defined as follows: ‖ vector ‖ = length of the vector or in this case the numerical value of the L. Mathematically correct it should be written ‖ dL ‖ / ‖ Lo ‖ ( L is always with arrows, because it is a vector.) But remember, angular momentum of the body behaves and it is defined as a vector. If you change the position, orientation or rotation you can quantify the change of L with the vector contraction between final and initial position.
Hi, great lecture but I am a bit confused, surely if the gyroscope precesses into the board there must be an angular momentum vector pointed vertically up?
one question: in the minute 6:09, that L don't should be equal to the product of the moment of inertia of the disc and the omega?, if that's right, then the moment of inertia should be = (mr^2)/2,?
The moment of inertia of a solid disk spinning about its center of mass is indeed (1/2) mR^2. But when that disk is rotating at the end of a spinning beam (as shown in the video) then it becomes a "point" mass and its moment of inertia is then mR^2
Earth rotates at 15 degrees an hour. Since gyroscopes maintain their spacial orientation, would it be possible to have a machine that would convert gyroscopic torque from that rotation, into energy in order to power itself? The machine would be stealing some of the planet's rotational momentum in the process.
The surface of the Earth is essentially an inertial reference frame and thus it would be extremely difficult to devise a system that could harness energy from the rotation of the Earth (while being on the surface).
1) The weight has no torque on the body or spinning disk.. In fact the reaction from the pivot gives the torque on the spinning disk . Neverthless they have the same direction 2) How is the moment balance about the pivot satisfied for the spinning body for equilibrium? The weight gives a moment about the pivot, the pivot is hinged, so it cant give a reaction torque or bending moment there..but still the moment seems balanced..Im confused??
@@MichelvanBiezen Thank you...Can u suggest me some books explaining it? Is it based related on virtual work principle , potential energy minimization etc?
We have some videos on virtual work and the principle of least action. Most books are good, but often difficult to follow (unless you already understand the material).
The secret is in the conservation of angular momentum. The weight of the rotating mass produces a torque which causes an angular acceleration around the z-axis. In order to conserve the angular momentum, the radius of the spinning disk rotating about the z-axis needs to decrease to compensate for the increasing angular velocity around the z-axis, thus forcing the spinning disk upwards.
Sir, can I say the reason that the gyro is not falling down because there is not torque to change the angular momentum in vertical direction, and since there is no torque is acting in vertical direction, the vertical angular momentum is conserved, and that is why the gyro is not moving in vertical direction?
The gyroscope precesses inward because of the inward torque generated by its downward weight but does it not depend on the spin direction? How do you explain it precesses outward if the spin direction is reversed? Thanks.
Sir, thank you for all your videos. One thing I am unable to understand. The precision motion has angular momentum in the vertical direction. But initially I found the presence of no torque to cause that vertical angular momentum. Where am I getting wrong ? How to explain that angular momentum ? Please help.
The weight of the spinning wheel will push down on the wheel. Since the angular momentum of the spinning wheel (pointing outward horizontally) will experience a torque causing the wheel the spinning wheel into a precessional motion.
@@MichelvanBiezen thank you sir for your quick reply. But still I am missing something. Sir, imagine that the spinning wheel is already in its prcissional motion, then it already has an angular momentum in vertical direction. According to the law of conservation of angular momentum it should try to maintain that angular momentum in vertical direction since there is no net torque on the wheel in vertical direction. Then why do we need to explain the precision as horizontal dL changing initial horizontal L of the wheel. My question is where did the spinning wheel get the angular momentum of the precission motion ? dL due to gravity is in the horizontal direction. It cannot cause the vertical anguler momentum of of the wheel due to precision. Please help.
@@MichelvanBiezen thanks again. I think I am getting the point. The problem with me was that I was thinking that the direction of the torque due to the weight of the spinning wheel is horizontal and the angular momentum of the precisional motion is vertical. So, that torque can not cause the angular momentum due to precisional motion. But I was wrong. Torque causes angular acceleration that changes angular velocity. In uniform circular motion the centrepital acceleration does not bring the particle toward the centre but changes the direction of tangential velocity and keep it moving along the circular path. Similarly in gyro torque due to weight of the spinning wheel does not rotate the wheel in a vertical plane but changes the direction of the angular momentum it already has due to spin. This causes the precision. Just in case of uniform circular motion centrepetal acceleration does not bring the moving to object towards the centre along the radius but changes the direction of the momentum it ready has and , hence, make it follow the circular trajectory. Sir, please tell me whether I have rightly got the point. Thank you very much sir.
@@MichelvanBiezen one more confusion sir, the gyro is rotating in precision motion along a horizontal circle. So, it has gained some kinetic energy. Where does it come from ?
Sir, As you said about Precision,isn't it inversely proportional to angular momentum and if it is then won't precision be infinite when the disc is not rotating , therefore it would spin at an immense speed won't it ?
The length of the arm holding the disk, say L is not necessarily the radius of the disk, so it doesnt cancel out in your calculation. More over, the moment of inertia of a disk is 0.5*mR^2. Then the result is: /theta_dot = 2gL/(w*R^2)
@@MichelvanBiezen i missed your point. There are two characteristic radius's here. The orbital motion as you mentioned, is the length of the arm holding the disk (horizontally), say L. And the secons is the radius (R) of the disk which determines the moment of Inertia. The video accidently used the same parameter (r) for both length and canceled them out by mistake.
The only issue I have is that L1 is larger than L0. People argue dL is 0, so L1=L0, but then dTheta is also 0 meaning there is no precession. Can anyone clarify this?
The idea is using differential quantities, to describe change. The principle of this phenomena is that the rate of change of the angular velocity is due to external Torques. In order to write down the following equation we need to describe the rate of change, so we,re using differential quantities.
If I don't allow procession to occur (say, with a fixed vertical rod just behind the horizontal rod of length 'r'), i.e. if I put work in to counteract the processional motion, will the torque imposed by "mgr" still be countered by the gyroscopic motion? I.e. will it still 'levitate' if I can't conserve angular momentum? instead, some linear work (reaction of the vertical rod) is used to conserve total energy?.
Sir, in this case, the r (radius) of the disk is different from the r of the rod. Therefore, we cannot cancel out like that. Same thing to the mass? Am i right?
Sir, so from 6:30-6:34, we cannot cancel out the radiuses and the mass. They should be called radius of disk, radius of rod, mass of disk, mass of rod. They are different so they cannot be canceled out. So the correct equation of the speed of precession is (mass of rod X radius of rod X gravity) / (mass of disk X radius of disk ^2 X omega of disk). Not gravity/(radius X omega of disk as written squared box on the board, i am right? Please let me know because this is very important. Thank you sir!
Hi Michel, If I may say, great video series on a challenging topic... informative, clear and comprehensive!My question relates to the equation developed for the precessional angular velocity (large omega). Q: If small omega goes to zero, will Large omega go to inifinity?
Mathematically yes, but in the real world that wouldn't happen because when small omega becomes to small, the gyroscope will not remain in position and fall.
It would have helped to have a practical example of calculating the speed of precession worked out. The rotational speed in radians/sec is clear enough, but exactly what units are R, Omega and even G being measured in? R measured in meters? G measured in meters/sec squared? Omega measured in radians/second? Interesting ideas, but tough for me to understand without more help!
Not sure what you mean by: "that seems a little vague". I think my explanation was clear and straightforward. The Earth's orbit and motion are constantly affected by those gravitational forces.
@@MichelvanBiezen .....you say "constantly affected by those gravitational forces.".... but these "constantly applied forces" are chaotic in nature whereas the precession of a spinning earth is constant in nature - how does a chaotic sum of forces create a constant precession? that is the question, given how extremely sensitive a gyro is to the net sum of applied forces. Also the quality of the applied forces counts: a force applied to the precessional motion makes it increase the precessional angle, a force applied o the spin rate makes it decrease the precessional angle......
is d theta the vertical angle between initial position of disk in z-component, and after it gets raised up ? Or is it simple the angle between the angular displacement, as in flat surface?
Respected sir, I do believe you missed out a small point... I=1/2mr^2 here. That is the moment of inertia for a solid disc about its natural axis! However,your formula would be correct for rings. (As for rings I=mr^2)
@@MichelvanBiezen Rather the best formula that should be is I=rg/{w(k^2)} Where k is the radius of gyration for the rotating object used... r is the gyro (effective length )
Not only does the rotating disk not drop, it will rise as it precesses around in a circle. The conservation of momentum induces a force that pushes the object upwards.
@@MichelvanBiezen Sir, can you describe the mechanism of conservation of angular momentum that induces a force pushing the object upwards. How is this upwards vector derived?
ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-XHGKIzCcVa0.html maybe this helps, it explains it without math but with intuiton or physics. I might be a bit late but i hope it helps.
sir, what would have happened if the angular momentum of the spinning wheel was in the other direction? i.e. what if the angular velocity was clockwise rather than anti-clockwise? which direction would it spin?
I recommend you check out precisely how angular momentum is defined as vector. If you change the rotation you change direction of the vector for 180 degrees and precision direction changes. Angular momentum is defined as vector product!!!! The precession direction can be determined from the vector product but not the first one. ;)
Because the conservation of angular momentum. When the gyroscope drops just a little bit, the torque that creates will make the gyroscope spin around a circle, which creates an angular momentum pointing upwards, and in order to keep that constant, the gyroscope must go up slowly, counterbalancing the mg.
@@MichelvanBiezen sir, please could you make a video explaining revolution vs rotation and it's mechanics, or provide some source as while watching your 2nd part video, you replied someone, that "we consider it as point object in revolution" Thank you sir !!
The best way to think of it is like this: If the center of mass rotates around a central point at a distance R, then I = mR^2 and we call that motion revolution. If the disk rotates around its own center of mass, then I = (1/2) mR^2 and we call that motion rotation.
I really appreciate these videos, thank you! just two things about this one. Around 3:45, all of the approximate equalities seem very hand-wavy to me. I also don't really understand why the "angular momentum triangle" is necessarily a right triangle and not an isosceles triangle, since the magnitudes of the momentum vectors are equal. I think I'm thinking about it the wrong way but a clarification on that would be very helpful. Thanks again.
Jose Molina Undoubtedly the triangle is isosceles. Here when the angle is infinitesimally small then the other two angles tend to be equal to 90 degrees. (i.e. almost parallel)
still nobody explain why wheel get tilted just before gyroscopic procession. the usual answer is potential energy (due to height of center of mass of wheel) changes into circular momentum of wheel around the rope.i.e energy conservation. but how the tilting work and change in circular motion
Take for example the formula on the bottom right corner with the border around it that describes the angular velocity of the gyro arm in terms of the gravitational acceleration, radius and disc rotational velocity. If you want to do this calculation in 3D when you are given random vectors of acceleration (like the g-vector pointing downward or in any other direction) and disc rotation vector (small omega), here's what you can do if you want to know in which direction will the resultant angular velocity of the arm (capital omega) will result in. You basically "break it up into orthogonal components", and then perform the calculation using the formula on the board and only the magnitudes of the component vectors. You start by finding the unit vector of the disc rotation speed vector (small omega), and divide each of its components by the length of that vector to determine the unit vector of the (small omega) vector. Basically you resize it to a magnitude of one. Call it (u_omega). Do the same with (g) to find (u_g). You then find the component of the acceleration vector (g) that is perpendicular to the unit vector (u_omega). So what you do is find the dot product of (u_omega) and (g). This gives you only a scalar magnitude. Multiply it with (u_g) to find the component of (g) that is parallel with (small omega). But you wanted the perpendicular vector not the parallel vector. So you then subtract the parallel component of (g) from the original (g) vector to find the perpendicular (orthogonal) component. Now that you have the orthogonal component of (g) and you have (small omega) you can simply take the magnitudes (lengths) of each vector and plug it into the formula on the board to find the magnitude of the rotation vector of the swing arm (capital omega). You now have the magnitude of (capital omega) but you still need to find its direction. Here's what you then do: Find the unit vector of the orthogonal component of (g), and then make it negative by multiplying it with (-1). This unit vector points straight up in the opposite direction of (g). It extends from the vertical pillar on the board. When you then multiply the result of the scalar formula with this unit vector, you will find the resulting (capital omega) as a vector.
