Physics 68 Lagrangian Mechanics (11 of 25) The Compound Atwood Machine (1 of 3)

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In this video I will find the equation of kinematics of a compound atwood machine using two Lagrangian equations (Part 1 of 3).
Next video in this series can be seen at:
• Physics 68 Lagrangian ...

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Sir please finish this series and D alemberts and Hamiltonians Please sir :) exams are approaching :)

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[@beoptimistic5853]

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[@amira8749]

at 3:14, iam a little confused about the sign for the velocites. why did you say that m2 moves downward in the +ve of x but in -ve of y? don't x and y have the same sign for the direction which is +ve downwards and -ve upwards? so if m2 is moving upwards, as you said, so the velocity is going to be V2=-y(dot)-x(dot). if anyone can explain to me that would be much appreicated

[@samiulhaque3338]

Yeah , he made a mistake I think

[@Phyzzics_ch]

Because in the sistem of x coordinates (A pulley), the whole mass associated with B pulley is going downwards. Then In the sistem of y coordinates (B pulley) m2 is going upwards for y but still downwards for x and m3 is going down for y an x.

[@brendanward2991]

2:30 - You've got the wrong signs on the three velocity expressions. Velocity is by definition the time derivative of position. So, if x_1 = x, then v_1 = x-dot. Ditto for the other two. Of course, when you square these terms in the Lagrangian, the negative signs disappear and your error is automatically corrected for. It's still an error, however.

[@heysoymarvin]

True!

[@GMYProducciones]

I don't think it was a mistake, I believe it has to do with generalized coordinates, it just happens that in the reference plane he selected the positive x direction goes downwards. Which is the opposite of traditional examples dealing with potential energy. The fact that you can use any coordinate system is where the potential for the Lagrangian mechanics lies. It was not that the mistake 'corrected' itself it was that the Lagrangian works for both coordinate systems, the one shown in the video and the one you are thinking about.

[@jjunluo]

Good point! The error is thus in the mismatch between the diagram and the equations. And that mismatch is a little bug in the video.

[@RazorIance]

at 8:55 I believe you meant the chain rule, not the product rule, I was confused for a while why the product rule would apply in this case. M is a constant.

[@vater3349]

Dont know if you will ever read this, but its actually the product rule. You can ofcourse say its the chain rule, which is totally valid, product rule works the same way here: (x' - y')^2 = (x' - y')*(x' - y') And then differentiate it. The same way how f(x)^2 can be differentiated or how you get the formula for differentiation of polynomials. d/dx x^2 = d/dx x*x = (d/dx x) * x + x * (d/dx x) = 2 (d/dx x) * x = 2x

[@beoptimistic5853]

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[@animal579]

for aynone else that got confused. when taking d/dt (dL/dX_dot), we get the y_dotdot, because it is a time derivative, so the coordinate does not matter.

[@beoptimistic5853]

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[@Peter_1986]

This looks very similar to "Constrained Motion". I remember doing something very similar in those kinds of problems, like I decided which lengths that were variables and then just threw everything else together into a constant and differentiated everything with respect to time to get velocity and acceleration.

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[@parikshitkulkarni3551]

We can even use the fact that total work done by the tension force in the strings is zero, so summation of tension times displacement is zero. Double differentiating we get relation between the accelerations of all masses. Next we can write the equations of motions for all the masses and we get 3 more equations. 4equations with 4 variables can be solved. It looks lengthy but is very short.

[@rohankulkarni8384]

V3 will be derivative of x3 which is -(x+y). Sir why have you taken it x+y. Thanks

[@apoorvasahu]

He took negative PE because L = (T-V) and while doing dL, dT becomes 0 and only -dV is remaining. So instead of writing the whole Lagrangian, he skipped that step and did dL/dx. If we find L and do it, we should get the same result

[@sudenureker9233]

no, look at previous question

[@pbillings808]

Probably should have emphasized there are two generalized coordinates and therefore two "Lagrange Mechanics" mechanics equations ( shown at the top left of the board). Otherwise, well done!

[@MichelvanBiezen]

Thanks for your input.

[@beoptimistic5853]

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[@sudenureker9233]

We choose below +x and +y. So why the potential energy is Negative sign? (pe=mg(position) and positions are + where did minus sign come from?

[@albendsouza]

It wont be difficult if we consider effective mass of pulley system B which is Meff=4m3m2/(m3+m2), then you could do a=(Meff-m1)/(Meff+m1)g

[@beoptimistic5853]

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[@RaviSingh-rq6qe]

Great Work but M3

[@beoptimistic5853]

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[@gosnfrenki]

I dont understand why velocity two is not: +xdot+ydot, because mass m2 is going downwards if you write -y+x it looks to me that mass is going downwards and upwards in a same time... i simply dont understand that...m1m2 so m2 is going to +y for me! Isnt it??

[@rajatnair8580]

X is decided wrt to the main system support and Y is decided wrt to pulley B, when m1 goes up m2 and m3 both go down ie the whole B pulley system goes down so wrt to X both are going down hence X is +ve for m2 and m3.

[@clydewebster9295]

If I extend out the brackets in the KE equation and then take the partial derivatives I get a different answer, and I can't tell if it's correct... I didn't quite follow how you applied the product rule I the differenciation.

[@beoptimistic5853]

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[@dianak.2368]

Why PE is negative? If you defined your positive direction downward and PE=0 on the level of the ceiling - everything below PE=0 should be positive. At least in the very first video of the course it was the case...

[@cdralda]

I'm also interested in the answer. Could you please answer.

[@liwenchang3260]

all objects below(x>0) must have negative potential since they are closer to the earth.

[@DanyAlejandro]

I think it's cause gravity is pulling the objects away from the height zero point (not towards it as usual) so the more the object moves in the direction of gravity, the less potential energy it has... But Li Wen Chang said what the professor said.

[@milosmaric6412]

You have sign Minus over there because our z axis is in same direction with force.Work done on body by gravitational force for dz traveled in the direction of axis he choose is dA=mgdz and potential energy will be negative of work done dU= -mgdz .You can see that increasing the Z will decrease your potential energy.U (z +dz)= U(z)+dU

[@Sans-d8z]

Thanks sir❤🙏

[@MichelvanBiezen]

Glad you like our videos.

[@prakharasaiya5002]

hey u didn't calculate Lagrangian how did u took partial derivatives??

[@oscarchivas900]

I had the same doubt at first, but if you look closely, L = KE + PE. When you take the partial of L (even though he did not calculate it) wrt x, all the terms in the KE are wrt x_dot so their partial derivatives wrt x would be 0, thats why he only considered PE. Same thing with the y. And when he does the partial derivative of L wrt to x_dot, he only considers KE because all of its terms are wrt x_dot, unlike PE. And same thing for the y. I know its kind of confusing but that is the answer to your question. *wrt means 'with respect to'

[@sreeharie7002]

Why do you always get your signs reversed.

[@lutfiyahatika1284]

FIGHTING MAN

[@yahiazakarialadhem9411]

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[@beoptimistic5853]

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[@cleisonarmandomanriqueagui9176]

I did it in simple terms with x and y without la and lb ... obeying to my common sense

[@Mike-vl3fn]

What happens if we assume all masses are equal?

[@sudenureker9233]

Why PE has minus sign?

[@AMIRULHAQE]

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