Precession Explained, Schwartz, Cal Poly Physics
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- Опубликовано: 20 май 2016
- This video is part of a free online mechanics course with all videos and a textbook.
Please see: sharedcurriculum.peteschwartz....
Torque provides a change in angular momentum, an angular impulse - Added to the present angular momentum. However, we don't need to use angular notation. Precession can be explained with linear momentum of the mass on the wheel rim, which may be more accessible to the physics student. For similar discussion, please see these videos:
Counter Steering a bike: • Countersteering, Pete ...
Gyroscopic Stabilization: • Seakeeper Gyroscopic S... 
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This is the best explanation I've ever seen. I watched several videos and checked my textbook, but this is the best. There should be more views
Thanks for the excellent presentation. I like that you explain clearly the mathematics describing the observed behavior, in addition to the demonstration with the wheel.
mechanical engeniering 4th year student, and thats the best one i saw so far about this fenomen
You made this abstract topic rather obvious. Salute.
This is the best video of precession on RUclips. Thank You Professor. Very lucid and clear explanation. Thank You again.
Have to agree, best explanation I have seen so far!
Best explanation I’ve seen yet, helped me a lot as a fresher Cambridge natural scientist
You look like Kakkar Sir of MadeEasy. A lot of people are turning this wheel on youtube but you made it clear in the best way. Thankyou sir!! Thanks from India
I love the way you presented. This precession phenomenon has been bothering me for last 15 years (I am a chemist, so I could ignore it). However, your explanation is the best. It not only have cleared my idea but also have helped to see a physical phenomenon more logical way)
Extremely good explanation in terms of the actual physics rather than just the mathematics. The connection to translational motion is very helpful.
I have a related manner in which to explain precession.
1. As you say when the wheel is pushed sideways with a lateral force or we should say "torqued", then the distributed lateral accelerations on the circumference of the wheel are represented by the distributed tilt at 4:00 which is really an elliptical function perfectly represented by that scenario. I call this function of lateral pushing as taking place on an " ACTIVE DIAMETER" as it is shifted laterally from its original position on the flat disc, while the dormant diameter is not touched at all.
The DORMANT DIAMETER is not affected by the tilt from to the push, and again using the video at 4:00, the mass particle will start moving out laterally ( up in this case) from the extreme left-hand side of the picture shown at 4:00 and keep on double integrating the acceleration to finish as a location on the extreme right-hand side of the dormant diameter.
On the other far side of the dormant diameter, the active diameter will go in the opposite direction. On our side there is a " hilly hump " type acceleration function over the circumference while on the other side of the dormant diameter there is " a crater hollow" type acceleration" and both are integrated twice along the circumference they exist on in space and time as the wheel rotates. It is more complex than that, but that scenario will do for our purpose.
Rather than an elliptical distributed acceleration lateral function, let us assume that the push caused a constant distributed lateral acceleration circumference then let us integrate the left-hand particle going from west to the east on our side and the diametrically opposite particle right hand going from east to west on the other side of the dormant diameter.
For simplicity let us assume that circumferential "hilly acceleration function" on our near side is a positive constant (+K) while on the other far side the " crater acceleration" the acceleration is ( -K)
Our side.
acceleration is K, hence
velocity is integral of acceleration K.dt = K.t
distance covered is integral of velocity K.t. dt= K.(t^2)/2
Farside.
acceleration is - K, hence
velocity is integral of acceleration - K.dt = - K.t
distance covered is integral of velocity - K.t. dt= - K.(t^2)/2
Then the new diameter will be represented by jointing a vertical point K.(t^2)/2 on the right-hand side and a vertical point - K.(t^2)/2 on the left-hand side.
The time t is the time for the spinning wheel to go through half a spin or one could use the space function through using the distance covered by the circumference.
This is taking place every half a spin as long as the torque is held.
Note this is using an approximation by using the constant +K and - K acceleration function which is not the real case.
Hope that this helps. If the students actually carry this drawing of the function repeatedly from the new diameter they will see the precision happening before their very eyes just with paper and paper.
Just start at one diameter and draw a parabola till it meets the circumference and repeat form the new point on the circumference on the other side of the new diameter.
THAT IS PRECESSION.
Mate, this is gold!
I've been wrestling with bloody gyroscope torque, angular momentum and the feckin vectors causing precession, for over a week (I'm writing some training notes for engineering students, and I thought I knew what I was talking about!).
You nailed it in this video...thank you, thank you, thank you
I absolutely love his energy, and the editing was really helpful
Without doubt, the best explanation !! Complete and clear !! Thanks a lot !
Thank you this is a best explanation that I never see ..
I had so many realizations during this video that I can't recover from the happieness.
honestly the best explanations I have seen thus far. Really makes me understand it! Thank you.
Excellent explanation. Best i've found on youtube. Thanks
Excellent explaination sir, i came across various videos on precession but nobody explained it better than u, now my doubts are clear... Thank u so much sir...
Beautiful intuitive explanation! Well done!
After watching several videos on gyros confused, I think I am now wrapping my head around that the gravity is acting horizontally onto the top and bottom of wheel, but since top and bottom of wheel are already moving, it tends to change the direction by precessing the wheel. Thanks for this video!
Crystal clear. Thank you!
Best explanation I’ve seen yet
Great video Pete.
BEST EXPLANATION EVER
By far the best explanation for a layman
Very impressive Mr. Schwartz.
Good video, thanks! Hope to see your channel grow!
Thank you for the explanation!
best explanation yet!!
Best explanation ,I ever had seen
Best explanation by far
thank you sir for clearing concepts of precession
A great teacher!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
So good and complete
Impressive lecture. Thanks a lot.
Thank you sir,
You finally made me understand
Love form India 🇮🇳
Good to hear. Love back to you India. I'm going to be visiting India in April.
Thank you! Just saved my sanity after being lost in lecture
Well done. Thanks
Wow, best explanation i have seen
Sir you explained it very well. Thank you so much 👍
Best explanation I’ve found :0
excellent explanation
This is simple and best
Best explaination 👍
Excellent explanation👌👌
thank u sir.... great explanation
better than Walter Lewin explanation.....thanksssss
Thank you
Best explanation
Woow. Such an awesome explanation
Thanks. I explain this in section 7.5 of my textbook: sharedcurriculum.wikispaces.com/Textbook+-+Mechanics+in+Parallel
Very nicely explained!
Well done this is the best explanation. I have a question : what is counter acting the downward force of gravity? I understand the Torques are operating in the “ horizontally plane” but gravity and tension force on the string is operating in the “vertical plane”. What is “holding it up”???
Dude! that's such a great question... no one has ever asked me that before. If we look at forces, we know that the vector sum of the forces = ma... so in the vertical direction, the forces must sum to 0 because there's no vertical acceleration. So the tension in the string (or the normal force of my finger) must be equal and opposite to the force of gravity. For rotation, it's more interesting. Gravity provides a torque, and there's nothing counteracting it. So we think, that just like forces, the vector sum of the torques = moment of inertia * angular acceleration and we should see angular acceleration... and we do when the wheel isn't spinning. However, this relationship isn't true, and in fact the forces relationship isn't really correct. The vector sum of the forces = the rate of change in momentum, and the vector sum of the torques = change in angular momentum. There's not much a distinction for linear forces, but for torques, there's a huge difference. The torque of gravity changes the angular momentum of the wheel. And we do see this because the wheel changes orientation. I made another video about this concept: ruclips.net/video/zmEKWI489I8/видео.html, Let me know what you think... cheers!
I don’t understand why you ‘go off at an angle when already pushed. Which initial direction (reference) and what angle is added ?
Also, how does friction with your hand impulse affect its direction towards you ?
Hi, Dave. Thanks for your question. I'm not sure I understand what you're describing, so if I miss the mark, please ask again with more detail.
For the second question, friction is always present, but friction is not necessary to describe the process, so you can approximate this as a frictionless system.
Why it goes off at an angle... it's maybe better explained in this other video I made. Please see it and let me know if it helps: ruclips.net/video/KNDpsBlxUIY/видео.html
Again, thanks for your question. - Pete
Great explanation ! Small correction: angular momentum and torque are pseudo-vectors en.m.wikipedia.org/wiki/Pseudovector
1) What would happen in the last part if the wheel is rotated in the opposite direction? Will it rotate in the other direction if it is held from one side, if so how would the you draw the vectors for angular momentum.
Sultan Alhammadi , direction of precession would be opossite..
Perfect
wow..really nice..
Sir great demo. on gyro.
Love the explanation. This is very hard concept to understand. But his explaination was mind blowing. But still I have question. Can gyroscopic precession be explained with just translational motion without referring to angular momentum and torque just using force and linear momentum?
Yes, it can... but Don't I do that in the video? Did you watch the whole thing? Do you see where I explain it using just force and linear momentum?
Thx sir
Best!
The work where is described the solution for the precession of Mercury’s perihelion, but like the precession of a gyroscope (derived from the Lagrangian mechanics and the General Theory of Relativity) is the following (from 2014):
article.sapub.org/10.5923.j.astronomy.20140302.01.html
U R Great 👍 👌
Why is the angular momentum perpendicular and following the right hand rule. I understand that it is a cross product, and therefore perpendicular. But why wouldn't it follow the left hand rule?
The global physics community uses the RHR. You could use left hand too if you used it consistently for all the other directions such as the direction of angular momentum... but we use the RHR.
@@pvs242ful Ah, therefore it is a pseudovector if we can also define the vector to be the opposite of the RHR. It is not a "real" vector such as a force or normal vector. Thank you!
so, since the torque is equal to the change in angular momentum, if the wheel was motorized would it still spin ? there would be no change, because the angular velocity would be constant. or does the friction in the bearings provide enough force to cause precession?
If there is no friction in the system, the wheel still precesses. The torque is perpendicular to the angular momentum (and the angular velocity), so the torque doesn't increase or decrease the angular momentum, it just changes its direction. This change of direction is precession.
But you mentioned the change in angular momentum. is that created by the torque?? also, can you explain how the precession works if the gyro is supported by both sides?
If the axel is supported on both ends, there is no torque and the wheel spins without precessing. It may be helpful to see another video I made more recently about gyroscopic stabilization: ruclips.net/video/KNDpsBlxUIY/видео.html
So, in theory, why would a artificial horizon gyro precess if it was held at both side of the axle? is the only reason for the pendulous vanes for constantly adjusting for the earth curve while traveling around it? i seem to remember the rate gyros having precession and thier rotating discs have mechanical bearings on both sides, so i guess im confused as to what could cause precession there. ill check out the video. thanks!
Physicists apply torque.
Chefs apply toques.
ooh danke excellent
Hey, I finished my textbook. I explain precession in section 7.5: sharedcurriculum.wikispaces.com/Textbook+-+Mechanics+in+Parallel
I agree. Best and fastest mathematical and physical explanation yet watched on RUclips. Keep it up. Fuck MIT, fuck all of them!!! Just keep it up man an that humour too:)
Hi thanks, I'm glad it works for you. But I still like MIT too. But thanks for the comment.
Everybody tries to make the subject clear, but you can tell, at the end, nobody feels they get it.
Sorry, Friend. If you want, check out the end of chapter 7 of my textbook: sharedcurriculum.peteschwartz.net/textbook---mechanics-in-parallel/
based
私もこれに関し投稿しています。
Nope! You are giving life to math and using mnemonic devices as causes, like the right hand rule and torque vectors. This is the explanation:
ruclips.net/video/edrp-ag2TfY/видео.html
This is a causal mechanism. It has nothing to do with a spin axis. The cause is found in the fulcrum of a tilting plane of rotation. A chain traveling in a square path creates precession from the straight sections of travel, so a spin axis just helps math to get hold of it.
Maybe you'd appreciate this presentation better. It boils down to basic physics a little more completely: ruclips.net/video/KNDpsBlxUIY/видео.html. I do like that you built that chain-gyro. My explanation in this video is consistent with your experiment. Thanks.
Yours is close, but you still use angular momentum which has nothing to do with it. And, the mass doesn't just start momentum in an axial direction. Helicopter rotor blades do not react gryoscopically at exactly 90 degrees. It's close, which is related to what you are saying, but more complicated than a single paragraph. My degree was physics, and they laughed at me for this:
Consider a plane of rotation (the plane, not the rotating mass). While tilting that plane, you create a fulcrum line running through the plane -that line being a place where mass must reverse directions with each pass through. View a rotating disk edge on within that plane, and as the mass passes through the fulcrum created from the plane's tilting, you can see the path of the mass creating centrifugal force. You don't even need a spin axis for that, just mass passing through a fulcrum. I did it with 4 water hoses (although messy and complicated). It even works using triangular paths, it's just clearer to the point using 4 paths.
John, it seems you reject angular momentum as anything other than a way to memorize how to get an answer. Is this correct? I recognize that the right hand rule for instance is arbitrary, and you could just as well use a left hand rule consistently. Yes some quantities are kind of derived in a way that may not be fundamental. However, you use "fulcrum line" and "centrifugal force", also not fundamental physics. Maybe you don't like the use of angular momentum because not all things exhibiting gyroscopic behavior are rotating solid bodies, such as your chain drive or water in a hose? However, these devices also have angular momentum as does a point mass as I discuss in this short video: ruclips.net/video/4CyvrS86uYs/видео.html
Some applied concepts are more accepted than others as fundamental. In particular, "centrifugal force" is not accepted in standard physics because a force is a push or pull, such as the tension in the chain that accelerates the chain around the turns in your gyro. I won't go so far as to say that your paradigm is incorrect, because it may consistently explain for you the phenomenon. However, I don't think you should be surprised that the physics community doesn't accept your explanation. Again, I appreciate the experiments that you've done and I may not be understanding your objection to my use or anyone's use of angular momentum. If so I apologize for missing your point.
Centrifugal force, or whatever you call it, can kill you, so to me it is real in some sense. Everyone is using angular momentum as if it's causal, but it is a concept derived AFTER the math. Angular momentum only exists in the math. The cause of the gyroscopic effect is found when looking at a spinning mass edge on. You can see motion that causes what is known as centrifugal force that happens at the fulcrum when mass passes through it. In my video, the chain jumps into the air when passing through the fulcrum, and it will do that from linear motion, not angular. This is the cause of the effect. Modern physics is full of these invented concepts that contain the cause, but it's not addressing the true cause. It is merely redefining it. The gyroscopic math takes all inertial data and attaches it to a spin axis. It is addressing the real data, but it displaces the cause to the spin axis, when it's actually the fulcrum running through the spin plane that is the cause. There is momentum passing through a fulcrum, and viewed edge on, it looks like centrifugal force. Classically trained physicists are the worst at grasping this concept, and most just stop talking to me. Non-physicists get it immediately.
If you jump off a cliff and hit the ground, is it the LINEAR force that kills you? I think we'd agree that it is the normal force when you hit the ground that kills as this normal force accelerates you linearly. Similarly, if you are spun in a centrifuge, there's no centrifugal force. There's a normal force that accelerates you inward. No force is pushing you outward. In any case, you are welcome to express yourself as you like. I think it's unlikely that the physics community will change.
Very impressive Mr. Schwartz.