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

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8 жыл бұрын

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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 ...

Пікірлер: 85

@jacobrose4805 8 жыл бұрын

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@beoptimistic5853 3 жыл бұрын

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@bhoopendragupta4782 3 жыл бұрын

Great explanation, never thought lagrangian could be this easy.

@madspihl 4 жыл бұрын

Oh the clarity of this. Fantastic, thanks!

@beoptimistic5853 3 жыл бұрын

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@astyutechick7980 4 жыл бұрын

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 3 жыл бұрын

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@animals42life8 7 жыл бұрын

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 6 жыл бұрын

Good explanation

@roopjeetbpatro3065 7 жыл бұрын

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 7 жыл бұрын

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

@roopjeetbpatro3065 7 жыл бұрын

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

@MichelvanBiezen 7 жыл бұрын

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

@anshgupta7930 6 жыл бұрын

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

@borfenglin6150 5 жыл бұрын

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 4 жыл бұрын

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 3 жыл бұрын

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@parikshitkulkarni3551 Жыл бұрын

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

@hfgjjhkjjbvkkb3442 8 жыл бұрын

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 8 жыл бұрын

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

@ghostzart 7 жыл бұрын

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

@happyhedgehog6450 6 жыл бұрын

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

@beoptimistic5853 3 жыл бұрын

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@biswajitghosh6443 Жыл бұрын

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

@MichelvanBiezen Жыл бұрын

You are welcome. Glad you found our videos. 🙂

@lincolngovender2137 6 жыл бұрын

Thank you Sir. You are a legend

@beoptimistic5853 3 жыл бұрын

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@044_kumarabhinav2 6 жыл бұрын

best teacher

@beoptimistic5853 3 жыл бұрын

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@anesmerazi603 7 жыл бұрын

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

@hecz250 6 жыл бұрын

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 3 жыл бұрын

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@burakoner1596 3 жыл бұрын

does not PULLEY has potential energy ?

@jozefinagramatikova8278 5 жыл бұрын

Thank you for saving my life again!

@beoptimistic5853 3 жыл бұрын

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@mohdshariquealam1296 6 жыл бұрын

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

@jamesonpayne3490 5 жыл бұрын

"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 3 жыл бұрын

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@physicsacademybyzakirullah8825 4 жыл бұрын

Nice Job...Sir

@beoptimistic5853 3 жыл бұрын

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@manishmks8049 6 жыл бұрын

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 6 жыл бұрын

MANISH MKS you can choose whatever you want

@aaronjs99 6 жыл бұрын

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 6 жыл бұрын

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 3 жыл бұрын

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@tkftns 4 жыл бұрын

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 4 жыл бұрын

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

@beoptimistic5853 3 жыл бұрын

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@luisjanuario1 Жыл бұрын

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 Жыл бұрын

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

@Steven_Salah-Eddine Жыл бұрын

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 Жыл бұрын

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!

@emeraldeyegaming7054 8 ай бұрын

why didn't we took PE of the pulley?

@MichelvanBiezen 8 ай бұрын

The height of the pulley doesn't change.

@prakharasaiya5002 7 жыл бұрын

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

@physicsacademybyzakirullah8825 4 жыл бұрын

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

@madhuriganguly1445 4 жыл бұрын

@@physicsacademybyzakirullah8825 could you explain this further please?

@physicsacademybyzakirullah8825 4 жыл бұрын

@@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😊😊😊😊😊😊

@user-nl1sy7sr4n 10 ай бұрын

Thanks sir❤🙏

@MichelvanBiezen 10 ай бұрын

You are welcome.

@bouhababrahim2292 3 жыл бұрын

Tq sir

@natepolidoro4565 5 жыл бұрын

what is electroline?

@MichelvanBiezen 5 жыл бұрын

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

@beoptimistic5853 3 жыл бұрын

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@gregofuente1119 5 жыл бұрын

Are X1 and X2 independent?

@MichelvanBiezen 5 жыл бұрын

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

@gregofuente1119 5 жыл бұрын

@@MichelvanBiezen Understood. Thanks!

@beoptimistic5853 3 жыл бұрын

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@trlanmmmdova8588 3 жыл бұрын

Thank you🤍🤍

@JohnnyYenn 7 жыл бұрын

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@FadingMemoriesHS 7 жыл бұрын

Dont donate. ITS A SWINDLE

@Ayayron_e3 8 жыл бұрын

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 8 жыл бұрын

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

@farhanomar7839 6 жыл бұрын

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

@nitrozox212 6 жыл бұрын

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