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.
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
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
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'
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!!
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 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.
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.
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!
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 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.
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 ✈️😍🤓👍
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
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
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
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
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?
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.
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?
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 :)
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?
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)?
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.
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.
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 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?
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.
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.
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, 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?
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!
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 ?
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?.
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......
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)
The math describes what you see analogously, and science has derived the causal explanation from the math -but math was only EVER an analogy. Follow only the momentum that you put into the system. A spinning rigid non-flexing disc has momentum that is already accounted for and does not change. Tilt it all you want and rotational velocity does not change from tilting. So, follow only the momentum you put into it, and you will find that your input momentum has to transition a fulcrum axis. A fulcrum axis running through a tilting plane of rotation does not move, while a seesaw effect is happening with mass passing through it VERY RAPIDLY. Now reference the equal-and-opposite-reaction mnemonic. Once realized, then you will start to understand the difference between math and reality. The best part is that the first fulcrum axis reacts as the anomalous tilting action to create a second fulcrum axis, and this is the most amazingly beautiful part about it, but I'm not going to hand it to you. You have to think about it and stop regurgitating math derived crap. Angular momentum does not exist.
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.
For the rotation of the disk the moment of inertia = (1/2) mR^2 where R is the radius of the disk, and for the disk spinning (precessing) around the vertical pole it is assumed to be a point object with I = mR^2 where R is the radius of the precession motion.
@@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.
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.
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).
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.
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).
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. ;)
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.
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.
You have to use the initial momentum of the disk, Lo=I*w , not the momentum created by the weight! This can be also found in my Serway textbook, where Ω=(mgd)/(I*w), (where d is the length of the axis) and not the result presented in the video!
Also vectors are divided by vectors in the video, which is an undefined operation. The issues as far as I can see was confusing the vector norm of L with the vector itself.
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?
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.
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The best explanation on gyroscopic precession
I've ever encountered on RUclips! Bravo!!!
He doesn't explain why it's 90 degrees offset though, which is the real question
Porca madonna sei anche tu del poli?
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).
I've been looking for videos on this platform to actually understand this and no one even came close! Thank you
Glad you found us.
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.
We appreciate the comment.
Thank you! I probably spent 2 hours looking at different videos and you provide the clearest and most intuitive explanation.
Thank you. Glad you found our videos. 🙂
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
^
yeah!! you are absolutely right
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
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 ??
Agree
I was supposed to say that
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'
A ring has I=mr^2 but a solid disc has I = mr^2/2
when u need something the most and u get it, fells a lot more than awesome, thank u sir, i ll pray for ur success, thank u once again sir
You are most welcome
Probably the best explanation of gyroscope physics in the internet. Thank you
Great to be able to re-visit these basic undergraduate physics concepts at leisure and pleasure of 40 years ago…. !
Yes, it is also fun to go back and understand concepts we didn't get a chance to understand as students.
small correction: The r in the moment of intertia of the ring (mr^2) L is not by definition the r of the gravity torque.
I != m * r^2 because we're talking about a rotating disc of its own radius R, not related to r, the length of the gyroscope arm
So it wasn't just me.. Is this a mistake?
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!!
Great! Glad you found our videos! 🙂
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!)
Glad you liked it.
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.
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.
this video is THE BEST explanation so far. Can't thank you enough, prof.!!!
Glad it was helpful!
This feels like a glitch in mathematic carried into real life...
Thanks for the great video. But at 6:04 Inertia of disk should be 1/2 m r (disk) ^2, not m r (precession)^2.
نعم صحيح صححه عندك حتى r لا يجب أن تلغي بعضها فrتعبر عن طول القضيب والثانية تعبر عن نصف قطر القرص
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!
For the precessional motion, the spinning disk can be taken as a "point" object.
Hi Sir,
In the video, the angular momentum, L is I*w = mr2w, where as, this r is radius of the disk, isn't? It is not the radius vector of the gryo.
Yes, you are correct, it is the radius of the disk. (I should have used two different symbols).
Thank you.
yes... You are Right!
@sir how both r was cancelled while calculating anglr precession vel at time 6:36
I was thinking the same thing. Another trip-up can occur in substituting angular momentum(L) with the precession radius vector(R).
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 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.
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 ✈️😍🤓👍
Thanks for this excellent example of the gyro effect on airplane propellers. 🙂
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
You are so welcome!
Thank you! I have always been searching for the correct explanation
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
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
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
@@MichelvanBiezen yeah, I understand, but that you say, don't be should the angular precession speed?, or I'm wrong?
I didn't quite understand your last question. The moment of inertia chosen for the various parts of this example are correct in the video.
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.
The precessional motion will start on its own, providing the disk is spinning.
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?
@6:08 Why is I when calculating angular momentum (L0 = I * w) mr^2 and not (1/2)mr^2? Are we not calculating angular momentum of the spinning disk?
+NoHow
Notice that r is the distance from the pivot point to the disk, (not the radius of the disk).
@@MichelvanBiezen No. the radius in moment of inertia is the radius of disk.
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?
He made a mistake, you cant divide vectors by vectors. He should take the magnitude of the vectors.
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.
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?
Hi Michel, studying for the physics GRE and this was really helpful! I completely forgot the step that needed the small angle approximation
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?
I an not quite seeing the picture. Gyroscopes tend to be unitary designs (one gyroscope at a time).
@@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 :)
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?
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)?
Because in that case we consider the whole disk as a point object.
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.
If you spin the wheel to the other side nothing would change right ? The angular momentum would change to the same side ?
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.
Very nice explanation. Very good indeed. Thanks a lot sir. You are great.
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 ?
No, conservation of momentum will limit how fast it can precess.
Does he explain why the rotating mass is not falling down around the pivot point? What is the balancing force against gravity?
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?
Really well explained the concept of gyroscope. However, I would suggest you to highlight the mathematical correction in the description section.
That is a good suggestion, thanks.
Sir, here if u use 2 symbols for radius and length you will get for the precision velocity as 2g/rw
2g/r*w or 2g/(r*w)?
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.
That is correct. For small angles sin(theta) is approximately equal to tan(theta). The form used makes it easier to derive the result.
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)
The length of the arm holding the disk is the radius of the orbital motion of the disk (not the rotational motion).
@@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.
In the video, we didn't use the radius of the disk, only the radius of the moment arm.
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?
That is correct
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!
So simple and so excellent.
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?
Correct. That is discussed in the other videos.
Thanks for sharing.
But would you mind to use the new marker ? I cannot see the words you wrote clearly
You sir are a blessing
Hello sir, why is r pointing in the outwards direction? Is it because the gyroscope is exerting a force on the stand?
The direction of R is typically FROM the point of rotation TO the point is space.
Really good explanation
Glad it was helpful!
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!
SI units.
Very precise and rewarding
Glad you enjoyed it. 🙂
your explanation is also pretty amazing! :D
Thanks! 😃 Glad you are enjoying the videos.
Kudos now it is clear for me thanks
Thank you and welcome.
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.
From the torque created by the weight of the spinning wheel. The torque will cause the spinning wheel to gain the precessional motion
@@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 ?
Thank you very much. I could understand all about gyroscope.
Great!
well said sir ,superb
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?.
Thank you, Sir, for an excellent explanation.
You are welcome. Glad you enjoyed it. 🙂
is it the moment of inertia wrong? the moment of inertia of disc is 1/2 mR^2 ,right?
Yes, if you consider the rotation of the disk. But with the precessional motion it is considered a point mass at the end of the rod.
@@MichelvanBiezen ohh, okay,tq
Great explanation
Glad you think so!
At 6:10, how L is equal to mr ^ 2 w !!? It should be Iw because we do not know the radius of the spinning disk and cannot become this way
For the precessional motion the disk is considered a point object since all of the mass is at a distance r from the center of rotation.
What if you had two counter rotating wheels spin?
If earth acts like a spinning top and a spinning top precesses because of mg then what plays the role of mg for the spinning earth and its precession?
The gravitational force between the Earth and the Moon and the Earth and the Sun, and to a lesser extent between the Earth and the other planets.
@@MichelvanBiezen That seems a little vague, given how sensitive gyros are, and these influences are all constantly moving?
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......
The gravitational forces are not chaotic at all. They are very stable and nearly constant.
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
That is an approximation that we use in many such physical situations.
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)
Shivam Sharma thanks man. ive studied some of the math a little more and the answer is clearer now. (mclaurin approx. and geometric arguments)
The math describes what you see analogously, and science has derived the causal explanation from the math -but math was only EVER an analogy. Follow only the momentum that you put into the system. A spinning rigid non-flexing disc has momentum that is already accounted for and does not change. Tilt it all you want and rotational velocity does not change from tilting. So, follow only the momentum you put into it, and you will find that your input momentum has to transition a fulcrum axis. A fulcrum axis running through a tilting plane of rotation does not move, while a seesaw effect is happening with mass passing through it VERY RAPIDLY. Now reference the equal-and-opposite-reaction mnemonic. Once realized, then you will start to understand the difference between math and reality. The best part is that the first fulcrum axis reacts as the anomalous tilting action to create a second fulcrum axis, and this is the most amazingly beautiful part about it, but I'm not going to hand it to you. You have to think about it and stop regurgitating math derived crap. Angular momentum does not exist.
Unless you tilt it at 1:30 or 7:30 position!
@@gensyed I'm always asleep then!
are we talking about ws or wp?
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.
You have assumed that the radius of the disc is similar to the perpendicular distance (r). I sit right?
That is not required nor assumed here. (if I understand your question correctly)
Thank-you for the explanation.
r= radius of gyration is not equal to moment arm this at time 6:34
For a point mass, the radius of gyration is equal to the moment of inertia.
@@MichelvanBiezen Iw=mr1^2w but moment mgr2 r1 not equal r2 thank you sir
For the rotation of the disk the moment of inertia = (1/2) mR^2 where R is the radius of the disk, and for the disk spinning (precessing) around the vertical pole it is assumed to be a point object with I = mR^2 where R is the radius of the precession motion.
ジャイロ効果は、首ふり運動による遠心力でも理解できますね！
Angular momentum shouldn't be (Icm (of disc) + mr^2)w ??
Not if it acts like a point object rotating around a moment arm.
@@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.
You are a good man.
It is a life long journey to be become a good man or woman. I haven't reached the destination yet.
Michel van Biezen I feel your love and patience in these intelligently designed lectures.
Thank you. I do enjoy teaching, it has always been a passion to try and make any topic understandable. (that is also a life long quest). 🙂
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.
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??
You wouldn't be the first person to be confused. It is a complicated situation and you need to go through it one step at a time.
@@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).
So when a body is rotating (at any speed and direction), all force acting on it is converted to torque?
I typically don't like to use the word "all" except in very few situations. If the rotation is too slow, the object will just fall due to gravity.
Why can we consider a dL?
How exactly does he "divide" by a vector?
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.
are there any hidden approximations or assumptions in this?
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).
Good explanation!
I love his bow tie
very Amazing teaching...
Thanks for these videos!
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?
If the disk spins in the opposite direction, it wouldn't work as you see in the video.
it wouldn't work at all? like nothing would've happened?
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. ;)
What happened to the moment of inertia of the disk??!
How should u handle case when w tend to 0, i see the L tend to 0 and OMEGA= infinity but from practice we know that it will fall down.
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.
If Spinning Gyroscope rotates in the opposite directions, does the direction of the torque is still the same?
If the disk spins in the opposite direction then the torque vector will point in the opposite direction.
@@MichelvanBiezen Thank you sir, but the direction of vector R and Gravity force still the same.
Great explaination and easy to follow. well done!
prefect explanation, thank you
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?
Also, if it doesn't fall, that means there must be a reactionary force acting back up, why doesn't that cause a torque in the opposite direction?
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 too dont understand why at 6:08 it is (L0 = I * w) mr^2 and not (1/2)mr^2 someone pls help
In this case the disk is a point object revolving in a circle of radius r (not the radius of the disk).
But I thought L stands for the angular momentum due to spinning and not precession
There are two angular momenta, one due to the spinning and one due to the precession.
You have to use the initial momentum of the disk, Lo=I*w , not the momentum created by the weight!
This can be also found in my Serway textbook, where Ω=(mgd)/(I*w), (where d is the length of the axis) and not the result presented in the video!
Also vectors are divided by vectors in the video, which is an undefined operation. The issues as far as I can see was confusing the vector norm of L with the vector itself.
What are the units of measurement for precession? I get 1/rad s?
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?
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.
came here for nmr , stayed for the wonderful explanation