Ball On A Cylinder Using The Lagrange Multiplier Method
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- Опубликовано: 6 июл 2024
- ** NOTE: θ1 is measured relative to r2 and NOT relative to the vertical ** - VIEW: bit.ly/3VhnRP7
Deriving the equation of motion for a ball rolling down the side of a cylinder using the Method of Lagrange Multipliers and determining the point at which the ball leaves the surface.
Download notes for THIS video HERE: bit.ly/48XrZ9Y
Download notes for my other videos: bit.ly/37OH9lX
CHAPTERS:
0:00 Introduction
0:10 Set Up
2:10 Questions
3:08 Constraint Equation
3:45 Kinetic Energy
4:39 Digression
8:15 Potential Energy
8:36 The Extended Lagrangian
9:48 Lagrange's Equation
12:07 What is the Lagrange Multiplier?
17:54 Equation of Motion
20:55 When Does the Ball Lose Contact?
21:35 Integrating the Equation of Motion
24:42 Initial Conditions
26:25 Finding the Point
27:17 Review
29:06 Outtro
Hey! Thanks a lot for your videos, they're always a pleasure to watch.
For the 4:12 and onward bit, I'm not sure I'm convinced. The definition of theta.1 is a bit ambiguous. If it is -- as your drawing suggests -- the angles that a given radius of the small ball does relative to a parallel to the _y_ axis, then I'd argue that both effects described in your digression are already taken into consideration with theta.1.dot: it actually captures the absolute rotation of the small ball, relative to the referential given.
If we wanted to isolate for the rotation of the small ball relative to *r.2*, the radius of the big ball along the line that goes through both centers, then we would write theta.0.dot = theta.1.dot - theta.2.dot. And that theta.0.dot would be the equivalent rotation of the ball if it were to roll on flat grounds.
For instance in the system Earth-Moon, theta.0.dot is actually 0, because of the synchronous orbit of the Moon. In that way, theta.1.dot = theta.2.dot, which is to say that the Moon rotates around itself in the same time that it rotates around the Earth (the rotation relative to a radius of the Earth along a line that goes through both centers of the Earth and Moon). And if we were to say that the angular momentum of the Moon was 0.5*I*(theta.1.dot + theta.2.dot )^2, we would actually overestimate it by a factor of four (because the Moon is not rotating around itself twice in 28 days). I understand that system isn't a no slip condition but it shouldn't matter.
Okay so what am I missing?
Yes, θ1 should be measured relative to r2 and not relative to the vertical. This is how I applied it in the model, however, I mislabeled it on the drawing. Thanks for catching that. I will add a note to the video description in the hope that others will see it.
Thanks for your brilliant explanation about why spinning both of circles of radius r should be same as one fixed and the other is spinning and revolving, which explains the revolving circle should spinning twice so as to go back to its original position, and this leads to spinning 3 times when one circle’s radius is twice than the other circle and angular velocities should be added before squared. What an enlightening lecture this is!
Great question and video
Thank you!
Very well done, many thanks!
Thank you too!
Would you have answered this correctly? ruclips.net/video/FUHkTs-Ipfg/видео.html
Hey
Well come back
Great video
I have seen All your series and have run many problems succefully
I would like to see a video for a rotation shaft with tu load equal at the center and a cantilever load for the transmission to simulate a shaft for a pulley
Thanks in advance
Will add it to my To Do list, but this might require multiple videos to cover it.
Amazing! Can you make videos about orbits?
Thanks. It's a little outside the scope of this material which is generally about vibrations. I might have to start a new playlist with some interesting dynamics problems.
I loved your class. What software/equipment do you use?
The app is called "Paper" by WeTransfer. Running on an iPad Pro 13 inch and using an Apple Pencil. The voice is mine.
excellent video. It seems that the angle of slipping is highly dependent on the moment I
Correct. The higher the value of I, the greater the angle before slipping.
cool. wouldn't the initial condition of theta2(0)=0 correspond to an (unstable) equilibrium where the ball just sits on top of the cylinder?
Yes technically you are correct, however, we assume that we displace the ball ever so slightly from the top. In fact, θ2(0) is not equal to zero, but rather can be though of as θ2(0) = 0.0000000...1 which can be rounded to zero - so it's close enough.
Great video, I'm currently taking dynamic systems control and this explained the concept of why theta 2 matters for kinetic energy much clearer than the textbook!
I was wondering, however, if the ball was rotating INSIDE the cylinder so that the overall radius of rotation is r2-r1 instead of r2+r1 would the kinetic energy then be theta2 dot -theta1 dot instead of plus?
You could do it your way if you change the definition of θ1 to be in the opposite direction from the way I defined it.
Alternatively, if you leave keep the definitions of θ1 and θ2 the same, then the translational kinetic energy would have (r2-r1) instead of (r1+r2), the rotational kinetic energy would remain unchanged, and, the constraint equation would now have a -ve sign, r1·θ1 = -r2·θ2
About T : ok for 1/2 I theta1dot^2 but for theta2^2 , I si not the same. Right ?
Since the cylinder is not rotating, its I is irrelevant. The ball is rotating as a result of BOTH θ1 and θ2 so it is the same I for both.
@@Freeball99 yes you are right. I don’t though about the I of the big ball, but the I of the small around the center of the big.
Does this mean that a tennis ball mass m on a string length L rotating with angular frequency w has a kinetic energy
T=1/2 mL2w2+1/2 Iw2 ?
Yes. However, in this case you'll find that the contribution of the rotatory kinetic energy to the total is insignificant if the radius of the ball, r
@@Freeball99Thank you for the reply. I calculated it using different methods and I believe I understand now but I can assure you that I’m the one who would miss the extra term.
Why the same Inertia for rotation theta1 and thera2?
The ball is the only object that is moving. There is no other moment of inertia involved. Both θ1 and θ2 cause the ball to rotate.
how will it change for a cylinder on a cylinder?
If the rolling object is a cylinder or a disk, then in equation 15 you would use I = 1/2 mr^2. Similarly, if the rolling object is a hoop or a hollow cylinder, then use I = mr^2.
@@Freeball99 I see thank you!