Physics 68 Lagrangian Mechanics (9 of 25) Example: The Atwood Machine (with Mass)

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In this video I will find the equation of kinematics of an atwood machine (where the pulley has mass M) using the Lagrangian equation.
Next video in this series can be seen at:
• Physics 68 Lagrangian ...

[@jacobrose4805]

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[@beoptimistic5853]

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[@bhoopendragupta4782]

Great explanation, never thought lagrangian could be this easy.

[@madspihl]

Oh the clarity of this. Fantastic, thanks!

[@beoptimistic5853]

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[@anshgupta7930]

Why you haven't consider P.E. of "Pulley" since it is hanging ?

[@borfenglin6150]

I suppose that you could also add P.E. of the pulley into the equation of potential energy, but the result of the differential equation will also be the same, since the position of the pulley remains constant throughout the process.

[@likitasathvik4162]

Because the force acted upon it is non conssrvative unlike gravity in case of the masses. Hence energy cant be stored. But if can say that the pulley is torsional then you can consider P.E. of the pulley

[@beoptimistic5853]

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[@parikshitkulkarni3551]

The above two comments are correct, and you could also say that we are neglecting the dimensions of the pulley

[@astyutechick7980]

4:04 I think you forgot to copy the minus sign. Love your videos btw. They've been very helpful!! Edit: Never mind, you correct it right away at 4:58

[@beoptimistic5853]

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[@044_kumarabhinav2]

best teacher

[@beoptimistic5853]

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[@luisjanuario1]

It´s L=(1/2m1+1/2m2+1/4M)Ẋ^2-(-m1gX-m2gl+m2gX), you forgot one minus on m1gX Forget it, he corrects it at 4:58.

[@MichelvanBiezen]

That is great that you caught that. That means you are understanding that concept. 🙂

[@biswajitghosh6443]

Thank you so much sir . I learned a lot from you ❤❤.

[@MichelvanBiezen]

You are welcome. Glad you found our videos. 🙂

[@hfgjjhkjjbvkkb3442]

Hi. Dont you have to add the potential energy of the pulley when calculating PE ? I already know it wont add any important modification because partial derivatives will take it to zero; Just wondering, and just that. By the way, thanks for your videos !

[@MichelvanBiezen]

Since the position of the pulley doesn't change, we do not have to take it into consideration when calculating the PE.

[@ghostzart]

The answer seems obvious in hindsight but I was ten seconds away from typing out the exact same question.

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[@jozefinagramatikova8278]

Thank you for saving my life again!

[@beoptimistic5853]

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[@manishmks8049]

Why don't you consider mass of pulley in calculation of potential energy.. Dear sir you have mistakenly taken the PE=0 , for the roof it should be a line passing through centre of pulley

[@amalguptan6716]

MANISH MKS you can choose whatever you want

[@aaronjs99]

You may take that into consideration. However, it turns out to be that the PE of the pulley differentiated with respect to x is 0 as the PE of the pulley is fixed in this case. So, it makes no big difference.

[@jayrashamiya2810]

If the mass of the disk is uniformly distributed over its surface, then the potential of the disk at every point would be constant, and it is not wrong to take reference to be zero at the roof. However, We would have to add a constant in the potential term, but it won't affect our further calculations if we chose to not to do so. One more point, it was quite obvious in the beginning that mass is uniformly distributed, since M.I is taken to be 1/2 MR^2 ... Although, I do not claim that there does not exist any other type of mass distribution having the same M.I .

[@beoptimistic5853]

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[@roopjeetbpatro3065]

Hello there, have you considered the pulley to be at negligible distance from the ceiling, due to which there is no term of the potential energy of the pulley ?

[@MichelvanBiezen]

The reference point (or the zero potential energy point) for potential energy can be chosen to be any point.

[@roopjeetbpatro3065]

Michel van Biezen Thanks for the timely reply, so is it more apt to consider the pulley as the reference ?

[@MichelvanBiezen]

Typically the reference point is chosen to make it easier to solve the problem, or to reduce the calculations.

[@happyhedgehog6450]

You're a life saver thanks a lot!!!

[@beoptimistic5853]

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[@Steven_Salah-Eddine]

Hi, I believe you forgot to negate the 1/2*m2*(v2)^2 term in your Kinetic Energy equation. Therefore KE for the second mass is 1/2*m2*(v2)^2 = 1/2*m2*(-xdot)^2, since v2 = -xdot

[@Mysoi123]

actually kinetic energy is 1/2 m |v|^2 because kinetic energy cannot be negative by definition, and the absolute value of velocity is just speed!

[@lincolngovender2137]

Thank you Sir. You are a legend

[@beoptimistic5853]

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[@anesmerazi603]

sir , shouldn't be the velocity of the mass m2 : X dot 2 rather than X dot ?

[@hecz250]

I think the reason we don't have to include subscript is because every point on the string moves with the same linear velocity (assuming the string does not stretch or slip around the pulley).

[@beoptimistic5853]

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[@physicsacademybyzakirullah8825]

Nice Job...Sir

[@beoptimistic5853]

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[@mohdshariquealam1296]

Can you please explain how you got P. E for m2 =(l-x).... I'm kind of stuck here🤔

[@jamesonpayne3490]

"l" is just an arbitrary constant representing the total distance from of the ceiling between m1 and m2, since as m1 gets farther m2 gets closer so it's conserved. So really you can think of it as m2 = (x1+x2)-x1

[@beoptimistic5853]

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[@animals42life8]

I want to know the traditional way of solving atwood's machine which includes the mass of the pulley. I haven't tried it yet. Thanks.

[@gastonrwegasira8154]

Good explanation

[@Sans-d8z]

Thanks sir❤🙏

[@MichelvanBiezen]

You are welcome.

[@burakoner1596]

does not PULLEY has potential energy ?

[@tkftns]

why is the inertia of the fixed disk = 1/2 M(R^2).. shouldnt it be M(R^2) ? Thank you so much for everything you do sir , i really benefit alot!

[@MichelvanBiezen]

Different shapes have different moments of inertia. In the case of a flat solid disk it is: (1/2)mR^2

[@beoptimistic5853]

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[@natepolidoro4565]

what is electroline?

[@MichelvanBiezen]

It is actually "Ilectureonline", which is the name of the channel.

[@beoptimistic5853]

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[@arpitshrivastava3036]

happy teachers day

[@MichelvanBiezen]

Thank you

[@beoptimistic5853]

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[@trlanmmmdova8588]

Thank you🤍🤍

[@bouhababrahim2292]

Tq sir

[@GamerTheTurtle]

this feels like cheating compared to using newtonian from high school. Or maybe I'm just a lot better at physics problems then I was back then...

[@EdwardSnowden125]

I dont understand why they didnt teach this to me as a part of the syllabus💀 This honestly is just as good as Newtonian mechanics

[@emeraldeyegaming7054]

why didn't we took PE of the pulley?

[@MichelvanBiezen]

The height of the pulley doesn't change.

[@gregofuente1119]

Are X1 and X2 independent?

[@MichelvanBiezen]

In this case the change in one will equal the negative change of the other.

[@gregofuente1119]

@@MichelvanBiezen Understood. Thanks!

[@beoptimistic5853]

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[@TamandaniNgwira-hb4vc]

😊😊

[@Ayayron_e3]

how come you didn't change signs in the parenthesis at 4:10 and you mention to be careful with the signs? shoudn't it be................ "-(m1gx+...-...)" since the negative is on the outside.

[@MichelvanBiezen]

+aaron brooks Yes, I caught my mistake later in the video.

[@farhanomar7839]

Why moment of inertia is 1/2 MR^2 and not just MR^2?

[@nitrozox212]

Because a pulley is a disc and the MOI for a disc is mr^2/2

[@prakharasaiya5002]

I wanna know why u took PE of m1 to be negative??

[@physicsacademybyzakirullah8825]

The P.E for every mass below zero level should be considered as negative

[@madhuriganguly1445]

@@physicsacademybyzakirullah8825 could you explain this further please?

[@physicsacademybyzakirullah8825]

@@madhuriganguly1445 yes plz why not. Please simply consider real number lines in which at midel we have zero number and at right side we have positive numbers and equal in magnitudes but of opposite in signs numbers are in the left side that are negative numbers. Exactly in that scenc if we consider a surface at zero P.E then the values of P.E above the surface will be positive and its values will be negative for the surface below the zero level. And that's it😊😊😊😊😊😊

[@anasblh6094]

i still don't understand what (l) represente ! que représenter le (l)!

[@MichelvanBiezen]

I is the symbol for "moment of inertia"

[@anasblh6094]

m talking about petit L in the exemple x=L-x ...thank you so much

[@quotestants8650]

@@anasblh6094 it represents the distance or length

[@beoptimistic5853]

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