Accelaration of the chain as a function of x

[Tanya Sharma]

Homework Statement

A uniform flexible chain of length L ,with weight per unit length λ , passes over a small frictionless peg..It is released from a rest position with a length of chain x hanging from one side and a length L-x from the other side .Find the accelaration a as a function of x.

Homework Equations

The Attempt at a Solution

I am not sure how to approach the problem.I feel there can be three ways to approach .

1. Work energy method
2 Newtons law F=Ma considering dx element .
3. Centre of Mass

i don't know how to proceed?? Please help...

 

[Sourabh N]

Since the question talks about acceleration, Newton's force law would be the right track to follow. You will need the center of mass of either part of the chain. Good luck!

 

[Tanya Sharma]

Can u elaborate??

 

[Tanya Sharma]

Centre of mass will be half the distance on either part..then?

 

[Sourabh N]

Can u elaborate??

Since the question talks about acceleration, Newton's force law would be the right track to follow.

Since Newton's law has acceleration explicitly (the a of F = ma) and conservation of energy usually has velocities, Newton's laws will lead you to the result more conveniently (There is some intuition also involved here, which I cannot really explain).

You will need the center of mass of either part of the chain. Good luck!

The two segments of the chain, one on each side of the peg, can be treated as blocks of mass λs₁ and λs₂ (where s₁ and s₂ are the lengths of each segment) hanging from a massless string. This reduces your problem to a standard 'two blocks on a pulley' scenario, which I'm sure you have come across before.

As the chain slides, s₁ and s₂ change, thus changing the acceleration.

 

[Tanya Sharma]

In the standard 'two blocks on a pulley' scenario the masses on the two sides are constant, here the masses are constantly changing ??

 

[Sourabh N]

Yes. The scenarios are not the same, but similar.

Once you do the calculation as I outline and put values for s₁ and s₂ as x and (l-x), your acceleration will be a function of x.

In a normal two blocks on a pulley, the mass of blocks on either side is a constant, hence the acceleration is constant.

 

[Tanya Sharma]

Sorry ...i m not able to get it...Can i get more insight ??

 

[ehild]

In the standard 'two blocks on a pulley' scenario the masses on the two sides are constant, here the masses are constantly changing ??

They are constantly changing. The two pieces interact at the peg, with some tension. You can imagine that, for an instant, you have two rods, length x and L-x, and both with linear density λ, connected with a piece of string wrapped around the pulley. This string provides the tension. Can you find the acceleration of such system?

ehild

 

[Tanya Sharma]

Yes i can find the accelaration in this case ... so that means the accelaration is continuously changing since the masses are changing ...increasing continuously...

 

[Tanya Sharma]

What is the force acting on this system ? Both the parts are being pulled down by gravity...How is tension acting ?? What is the motion of Centre of mass ??

Can we approach the problem in some other way?

 

[ehild]

Yes i can find the accelaration in this case ... so that means the accelaration is continuously changing since the masses are changing ...increasing continuously...

What is the result you got? What does it mean?

ehild

 

[Tanya Sharma]

a={(2x-L)/L}g is the answer . since L continuously decreases 'a' increases .

 

[ehild]

L is the length of the chain, it is constant. x increases, so a increases, but how long? And what happens if x=L? Can the acceleration increase any further?

What happens if x=L/2 at the beginning? ehild

 

[Tanya Sharma]

If x=L then chain undergoes free fall i.e accelaration is 'g' .If x=L/2 at the beginning then system remains at rest . am i correct ?

 

[ehild]

Right. In case of x=L/2 at the beginning, but it has some velocity, x will change, and it will accelerate with time.

ehild