Worked examples in classical Lagrangian mechanics

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Dr Mitchell's physics channel

Dr Mitchell's physics channel

3 жыл бұрын

Classical Mechanics and Relativity: Lecture 9
In this lecture I work through in detail several examples of classical mechanics problems, which I solve using the Lagrangian formalism:
2:00 Single pulley system
14:10 Double pulley
29:14 Planar pendulum
35:18 Spherical (3d) pendulum / particle in a bowl
44:42 Particle in a cone
55:34 Bead on a spinning wire
1:06:45 Bead on a spinning ring
1:08:43 Ball in an elevator
1:22:48 Bead on a rotating ring
1:37:09 Trebuchet mechanics!
Theoretical physicist Dr Andrew Mitchell presents an undergraduate lecture course on Classical Mechanics and Relativity at University College Dublin. This is a complete and self-contained course in which everything is derived from scratch.
Full lecture course playlist: • Classical Mechanics an...
Course textbooks:
"Classical Mechanics" by Goldstein, Safko, and Poole
"Classical Mechanics" by Morin
"Relativity" by Rindler
Mathematica demonstrations project credits:
Bead on a ring: Ryan K. Smith
Trebuchet simulation: Erik Mahieu

Пікірлер: 24
@eamon_concannon
@eamon_concannon 2 жыл бұрын
I think that having videos of worked examples like this is a huge help for self learners if they decide in the future to gain more fluency by working textbook problems. Thanks a lot!
@nirmalk97
@nirmalk97 3 жыл бұрын
I'm a physics major undergrad from Sri Lanka. Your Classical Mechanics lectures helped me a lot to understand the concepts and to practice questions for my final exam. Thank You very much sir.
@smoosq9501
@smoosq9501 2 жыл бұрын
underrated channel and underrated lecture videos, thank you for sharing
@wenhanzhou5826
@wenhanzhou5826 2 жыл бұрын
Thank you for making my learning experience more enjoyful!
@lauramariaboterovillegas1108
@lauramariaboterovillegas1108 2 жыл бұрын
thank you very much for this type of content, it is really necessary to someone that has the enogh patience to clarify all the doubts we can have and how better if it's making examples to make you understand better the theory and how to apply it.
@unkownuser1042
@unkownuser1042 2 жыл бұрын
Tanks sir..for providing such lecture series ....helped a lot
@shivanshsrivastava9264
@shivanshsrivastava9264 Жыл бұрын
Awesome examples sir!!!
@nilslorand
@nilslorand 3 жыл бұрын
Good Video, lots of examples, helped me a lot! Also greetings from Germany
@danielkonstantinovsky108
@danielkonstantinovsky108 2 жыл бұрын
Could you make a video covering how to incorporate constraints _not_ through choice of coordinates, but through the Lagrange method where you set the euler-lagrange equation equal to the constraint (I think?) It would be really useful for understanding things like molecular dynamics with rigid chemical bonds, where there isn't a suitable coordinate system that removes all the bond length constraints.
@amishkakru5407
@amishkakru5407 11 ай бұрын
great work!
@shishengwang1310
@shishengwang1310 11 ай бұрын
Great job ! thanks a lot!😄
@NovaWarrior77
@NovaWarrior77 2 жыл бұрын
Thank you so much!
@marcoghislanzoni
@marcoghislanzoni 10 ай бұрын
At 22:43 there seems to be a missing square for the (q2 dot - q1 dot) in the second term of the kinetic energy
@awaiszakeer759
@awaiszakeer759 2 жыл бұрын
Great
@nomachinesinthisroom
@nomachinesinthisroom 5 ай бұрын
Thank you so much for these!! At min 7.51 shouldn't there be a minus from x_2 dot = -x dot? Therefore should the two terms of the kinetic energy have a minus in between? My result has (m_1-m_2) in the denominator.
@MrPopololo22
@MrPopololo22 3 жыл бұрын
At 26:14 you write the partial derivatives with respect to q1 and q2 but I think the signs of the masses are supposed to be reversed because of the -g constant. Am I wrong?
@hershyfishman2929
@hershyfishman2929 3 жыл бұрын
I think you are right. The error stems from 23:00 where V is given as -mgx_i and therefore ends up as +mgx_i in the Lagrangian. I think it should be -mgx_i in the Lagrangian as in Example 1 at 7:58. There also seems to be an unrelated error in both Examples 1 and 2 regarding the sign of L (length). It should -L_i in all the equations. However, the sign of L doesn't affect the eom. Since I'm commenting I just have to say I love these lectures by Dr. Mitchell. They're among the best videos on KZfaq.
@drmitchellsphysicschannel2955
@drmitchellsphysicschannel2955 3 жыл бұрын
You're right! Thanks for correcting the error.
@oak6677
@oak6677 2 жыл бұрын
@@hershyfishman2929 please pin that comment Sir
@oak6677
@oak6677 2 жыл бұрын
@@drmitchellsphysicschannel2955 Sir I'm getting m2=-m3 from the EOM for second component. What is the physical significance of this? Mass can't be negative so does that mean that the system cannot be in equilibrium?
@markbrightwell3536
@markbrightwell3536 3 ай бұрын
In the two pulleys example, it seems like the second pulley (the lower one) becomes an accelerated frame of reference when the mass m1 is accelerating. Do we need to account for that by changing the value of g for the second pulley to g plus the second time derivative of q1?
@curtmcd
@curtmcd 8 ай бұрын
The content of this video, and especially the pace of exposition, were both ideal. As an amateur armchair physicist, I can't believe I watched a physics and math video this long and complicated where I understood every stroke of the pen, if not anticipated it, and felt empowered to use it for other applications. You're an excellent teacher!
@mariogalindoq
@mariogalindoq Жыл бұрын
It is not a good idea to use the same symbol for two different magnitudes. At 4:55:00 and 33:47:00 you are using L for two different concepts. Anyway, I like your videos. Good job.
@lindsaywaterman2010
@lindsaywaterman2010 Жыл бұрын
I think that a mistake was made in defining the potential energy; V is not mg*cos(theta). V = - mgL*cos(theta). If the zero potential was taken to be the origin of the string, then Z = -L*cos(theta; if the potential was taken to be at the lowest possibe vertical position of the string, then V = mgL(1 - cos(theta). The correct equation of motion for theta double dot, is (theta double dot = sin(theta)cos(theta)*phi dot-squared - g/L*sin(theta). IN OTHER WORDS. IN THE EQUATION, THE SIGN INVOLVING GRAVITY SHOULD BE NEGATIVE.
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