Oscillations of a pendulum with a freely sliding base
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- เผยแพร่เมื่อ 15 ส.ค. 2023
- Using Lagrangian mechanics to solve for the time evolution of a pendulum with a freely sliding base.
About me: I studied Physics at the University of Cambridge, then stayed on to get a PhD in Astronomy. During my PhD, I also spent four years teaching Physics undergraduates at the university. Now, I'm working as a private tutor, teaching Physics & Maths up to A Level standard.
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One might also observe that the maximum value of theta dot is proportional to 1+m2/m1 and inversely proportional to length l. Thus, the simplifying assumption that the “third order term” (theta times the square of theta dot) is negligible would be false if m2 is much larger than m1 or if length l is very small.
I believe this problem is in lev landaus book it's funny all the other classical mechanics texts like morin and goldstein are 600 pages long but lev gets right into he's book is 200 or less
I've been loving my analytical mechanics class so far, but videos like these make it even better. Thank you for the wonderful explanation.
Thanks for your kind words, glad I could help!
Thank you.
Well done!
Thank you so much!!!
Your analysis about what the solutions tell us about the system is invaluable . Thank You.
Thanks, always important to consider the meaning of the solution!
Cảm ơn thầy rất nhiều ạ . I am from Vietnam ❤
Thanks for watching!
Hey I have a question. I have a simmilar system with the difference that my cart has an external input which means I would have a nonharmonic differential equation for my xddot. Do I have to solve this differential equation or do you suppose there is an easier way to go about it. Thank you in advance.
Or would it also be valid to cancel the coupled second derivates in each equation by substituting Equ 1 into 2 and vice versa?
While calculating the kinetic energy shouldn't we square both x2 dot + y2 dot?
Yes, we do exactly that from 5:06 onwards! Square and then add, rather than add and then square.
Thanks! I always tend to sum components and square the result afterwards, that is messier 😅
It actually needs to be the other way around because the velocity vector v is (xdot, ydot), and the kinetic energy is proportional to |v|². You always calculate the squared magnitude of a vector by adding the squares of the components, rather than squaring the sum of the components.
Hello Teacher, I have a question about this analysis. You choose a reference system with Oy pointing up so the gravity g in the system will have a negative value?. Hmnn 17:36 . Therefore omega squared will have a negative value. .. I hope to receive feedback from the teacher.
........Sorry for using Google translate. I hope you understand what I am saying. Thanks a lot
By definition, g is the magnitude of the gravitational field and is therefore positive, regardless of which coordinate system you use.
I hope to receive feedback from the teacher......
I have a test with similar content but choose the Oy axis to point down so there are some changes in the sign in the expression. Detail :
3:55 : y₂ = l. cos𝜃
4:25 _ y dot = - l. sin 𝜃
7:00 _ v = m₂gl.cos𝜃
7:35. L = T- V = .... - m₂gl . cos𝜃
9:20 : 𝜎L/ 𝜎𝜃 = .... + m₂gl.sin𝜃
11:15 : ..... -m₂gl. sin𝜃
13:00 : ..... ẍ. cos𝜃 + l.𝜃 double dot - g.sin𝜃=0
15:00 . ...... ẍ + l.𝜃 double dot - g𝜃 = 0
17:40 . Ẅ = -{g/l .[(m₁ + m₂)/ m₁ ]}
i know it's a mistake but i don't know how to find the solution because omega squared is always positive. Hope to receive your feedback. Thank you so much
Your expression for V needs a minus sign in front, the same as in the video. When 𝜃 increases, the pendulum bob is moving up, gaining GPE and therefore V should increase. However, cos𝜃 is a decreasing function, so the minus sign is necessary regardless of your choice of coordinate system!
Hi I am student from Serbian, and your videos are helping me a lot with Lagrangian problems. Would you mind doing a similar problem but with two pendulums and sliding base. I am struggling to find solutions for movement of all three bodies. FYI both lengths of a ropes are same.
I'm glad the videos are helping! Thanks for the suggestion - not sure when I will get around to this but I've added it to my to-do list.
If this video did help you, then you should be able to do the next problem yourself. If you need Dr Ben to do the next problem for you as well, then you haven't really learned from this problem.
I have also, simulate this using MATLAB in my channel