Formulating lagrangian for a system

[Gloyn]

Homework Statement

http://fotozrzut.pl/zdjecia/ad3bbdd9f6.jpg

Mases are as stated on the picture, I is the moment of inertia of the pulley, angle marked is ω and a is a radius.

Homework Equations

L=T-V

The Attempt at a Solution

I try to formulate a lagrangian for this system, but I don't know what coordinates shall I use. How to relate the angle and position of the mass m?

 

[Dick]

The distance the mass descends is equal to the amount of string that comes off your pulley when it rotates by angle ω, right? There's only one real coordinate in your lagrangian, ω.

 

[Gloyn]

So is it ok to write:

T=(dω/dt)^2(a^2)(m+M)/2+I(dω/dt)^2/2

V=-mgaω-Mga(cosω)

?

 

[andrien]

seems o.k. to me.

 

[Gloyn]

Ok then, so after writing Euler-Lagrange's equations I get:

(d^2ω/dt^2)(ma^2+Ma^2+I)=agm-Msinωag

And I need to know the angular velocity. If i integrate wrt. time i get:

dω/dt=agm/(ma^2+Ma^2+I)∫dt-gam/(ma^2+Ma^2+I)∫sinωdt

But it seems strange that angular velocity would depend explicitly on time. Did I mess up somewhere? First term in V depends on ω and not on trig. function of ω, so when I differnentiate the lagrangian wrt. ω it disappears, and so after integration there is 't' in angular velocity.

 

[Dick]

Ok then, so after writing Euler-Lagrange's equations I get:

(d^2ω/dt^2)(ma^2+Ma^2+I)=agm-Msinωag

And I need to know the angular velocity. If i integrate wrt. time i get:

dω/dt=agm((ma^2+Ma^2+I)∫dt-gam/(ma^2+Ma^2+I)∫sinωdt

But it seems strange that angular velocity would depend explicitly on time. Did I mess up somewhere? First term in V depends on ω and not on trig. function of ω, so when I differnentiate the lagrangian wrt. ω it disappears, and so after integration there is 't' in angular velocity.

Why would that seem strange? If you put M=0 and analyze that then you just have a constant torque on your pulley. Angular velocity should increase linearly in time, right?

 

[Gloyn]

You're right, it was just my autosugestion as question is formulated:

find angular velocity in terms of a, m, M, I and theta. So how to do that? The answer that I found is:

\sqrt{\frac{2ga(m\theta - (M + M)\cos\theta)}{I + Ma^2 + ma^2}

and ther's no t explicitly.

 

[Dick]

You're right, it was just my autosugestion as question is formulated:

find angular velocity in terms of a, m, M, I and theta. So how to do that? The answer that I found is:

\sqrt{\frac{2ga(m\theta - (M + M)\cos\theta)}{I + Ma^2 + ma^2}

and ther's no t explicitly.

You should probably figure how tex tags work https://www.physicsforums.com/showthread.php?t=8997 Here's what you wrote:

$$\sqrt{\frac{2ga(m\theta - (M + M)\cos\theta)}{I + Ma^2 + ma^2}}$$

Aside from typos (why M+M?) it looks like they used conservation of total energy E=T+V setting the energy E to 0. I don't know why they did that. There should really be an arbitrary constant in there.

 

[Gloyn]

Ok, so maybe I'll post the full question and the solution:

http://fotozrzut.pl/zdjecia/d9fba96b75.jpg

Solution:

http://www.thestudentroom.co.uk/showthread.php?t=856635&page=2&p=17921549#post17921549

And yes, as you see guy obtained it by conservation of energy, I'm just wondering how to get the result with lagrangians.

 

[Dick]

Lagrangians aren't the way to go. You won't able to solve the equation of motion in any simple way.

 

[Gloyn]

But if this equation is correct:

$$\frac{dω}{dt}=\frac{agm}{(ma^2+Ma^2+I)}\int dt -\frac{gam}{(ma^2+Ma^2+I)}\int sinωdt$$

Shouldn't it be able to transform it to required form? I mean, if those equations express the same quantity, why do they differ so much?

 

[Dick]

But if this equation is correct:

$$\frac{dω}{dt}=\frac{agm}{(ma^2+Ma^2+I)}\int dt -\frac{gam}{(ma^2+Ma^2+I)}\int sinωdt$$

Shouldn't it be able to transform it to required form? I mean, if those equations express the same quantity, why do they differ so much?

No, you can't get a useful solution that way. What are going to do with the integral sinωdt? sinωdω would be easy but that's not what you have. Just use conservation of energy.

 

[TSny]

But if this equation is correct:

$$\frac{dω}{dt}=\frac{agm}{(ma^2+Ma^2+I)}\int dt -\frac{gam}{(ma^2+Ma^2+I)}\int sinωdt$$

Shouldn't it be able to transform it to required form? I mean, if those equations express the same quantity, why do they differ so much?

Go back to the equation of motion (Lagrange's equation)

$$(ma^2+Ma^2+I) \frac{d^2ω}{dt^2}=mga -Mgasinω$$

and multiply through by $dω/dt$ before integrating with respect to $t$. This will lead to the energy equation.

 

[Gloyn]

Damn, it works :). Thank you very much, I didn't think of doing that trick.