How do the masses and pulleys affect the motion of a double Atwood machine?

[bon]

Homework Statement

My question is about a double atwood machine like this:

http://farside.ph.utexas.edu/teaching/336k/Newton/img1915.png

Now imagine that m1 = 5m, m2=2m, m3=3m

My question is: why does the upper pulley rotate despite the fact that the masses on either side are equal?

And secondly - why do we take x' to be measured from the second pulley, rather than from the top again?

Thanks!

Homework Equations

The Attempt at a Solution

 

[bon]

anyone..?

 

[bon]

I actually understand why the top pulley must rotate now (to keep the COM in the same position)..
It's just the labelling of the coordinate's that I don't get..

 

[ehild]

Well, what do you think? Try to draw the free-body diagram for all masses and for the moving pulley.

ehild

 

[bon]

Well basically..I've solved the problem using the Lagrangian method by using two generalised coordinates x, and x' (both measured from the top pulley)..

The problem is I'm now having trouble solving using NII in a way that agrees with my result...

taking the 3m mass for example, I write 3m-T = 3m * accel of 3m mass
and i write T-2m = 2m * accel of 2m mass.. but when i equate T, I get an equation in terms of x and x' double dot that doesn't agree with the Lagrangian method...

is this because i shouldn't have measured the coords from the top pulley for some reason?

 

[ehild]

That is one choice to give the position of m2 and m3 with respect to the moving pulley. You can do it on other way if you like.

ehild

 

[bon]

I wanted to give the position of m2 and m3 with respect to the stationary pulley... wouldn't this be easier since what I'm told to find is the accelerations of the three masses...

I just can't seem to get Newton's law to give the same result as the Lagrangian...what would be the EOM's according to Newton law?

 

[ehild]

Show what you did.

ehild

 

[bon]

m1 = 5m, m2 = 2m, m3 = 3m

All positions are measured from top pulley. let position of 5m mass be x, position of bottom pulley is thus l-x. Position of 2m mass is x'. Position of 3m mass is l'+2(l-x)-x'

Write out the lagrangian = T - U

find Dl/Dx dot and DL/Dx' dot.. get 2 EOMS

17 d^2 x/dt^2 + 6 d^2'/dt^2 = -g

and

6 d^2 x/dt^2 + 5 d^2'/dt^2 = -g

Solve to get d^2 x/dt^2 = 1/49 g and d^2'/dt^2 = -11/49 g

Then when i do the analysis with Newton Laws i get a different result..

Is this right so far?

 

[bon]

ehild? :)

 

[ehild]

Your result obtained with the Lagrangian is correct, now show, please, how do you apply the other method.

ehild

 

[bon]

Great. Thanks ehild.

Um this is where the problems start...

so I am assuming the tension in the "lower" string is T'

so I have 3mg - T' = 3m (-3 d^2x/dt^2 - d^2 x'/dt^2 ) (-3 because i also have to add relative accel of top pulley)

Also T' - 2mg = 2m ( d^2x'/dt^2 - d^2x/dt^2) (again factoring in the relative accel of top pulley)

but these end up giving -d^2 x'/dt^2 - 11 d^2 x/dt^2 = g

which isn't consistent with the previous solution...

im just unsure about applying NII here really..can't see how to do it for the top pulley in any case..

 

[ehild]

Start from the beginning. Write out the equation of motion for all masses. I do not see the tension of the upper string. And what is x' now? Again the distance from the top pulley?

ehild

 

[bon]

x' is still the distance from top pulley, yes.

What i wrote there was the EOM for the two bottom masses...

I don't know what to write for the top mass..

5m - T = m d2x/dt2 ? Then T - 5m = d2x'/dt2?

 

[ehild]

x' is not the distance of m2 from the top pulley?

Use the constrains (constant lengths of both ropes) to get relations between the accelerations.
Find the relation between T and T'.

ehild