Chain problem to do with Newton's 2nd law.

[AJ2357]

Hello,

I have a problem to do with a chain.

You have a table with infinite drop and a chain of length l. The chain is let off at the end of the table. Assuming a frictionless table, calculate an expression for the speed of the chain.

Could anyone help with this please?

Best wishes,


Andrew

 

[Simon Bridge]

Welcome to PF;
Is this a homework problem?

Model the chain as a series of small ideal masses joined by a short, ideal, string.
Use the same approach as you did for just two masses.

 

[AJ2357]

Hi it is a problem given to me by a colleague. I mentioned differential equations and he gave me this problem. He is a Doctor of Physics so I am struggling slightly! I am from a pure Mathematics base. Could I model the acceleration as d^2 x /d t^2?

 

[Simon Bridge]

Context of calculus means that you should treat the chain as continuous rater than a set of distinct links.
However, it may help you understand the problem if you divide the chain up into a series of discrete masses and take the limit that the number of discrete masses becomes very large.

The second time derivative of position is acceleration so that's a good start.
You also need Newton's laws and a free-body diagram.

Have you done the problem for two masses connected by a string - one mass is hanging off the end of the table?

This one is just the same except that the mass hanging off the end is now a function of time.

For a chain length L and total mass M on a frictionless table, (I don't like lower case for L when I'm typing), what is the instantaneous acceleration for the chain when length y<L is hanging off the edge?

If the chain is initially at rest with length y0 hanging off the edge, then:
y(t=0)=y0, v(t=0)=0 - are the initial values.

 

[WannabeNewton]

Just find the potential energy of the chain as a function of height below the table and use conservation of energy.

 

[AJ2357]

Ok I've tried again and got this.

F= Ma where a=d^2 x / d t^2

Also F=mg where m is the mass of the chain over the edge of the table

so m =M(x/L) where x is the length of chain over the table at a given point in time.

Equating these two expressions we get:

M(d^2 x / d t^2) = M (x/L) g

(d^2 x / d t^2) = (x/L) g

Then integrate.

Would this be correct?

 

[ZapperZ]

Thread has been moved to the appropriate forum. Please note the notice that are clearly stated in this Stickied thread:

https://www.physicsforums.com/showthread.php?t=89899

This thread would normally be deleted where it was posted since it did not follow the template for the HW forum. However, since the OP has shown an attempt, it was moved as is.

Zz.

 

[AJ2357]

Sorry about that. Too keen to get started! Thanks for your help.

 

[D H]

Just find the potential energy of the chain as a function of height below the table and use conservation of energy.

Energy isn't conserved in most of these "rope on a table" type problems.

 

[D H]

You have a table with infinite drop and a chain of length l. The chain is let off at the end of the table. Assuming a frictionless table, calculate an expression for the speed of the chain.

How is the chain arranged initially? Is it

1. Laid out in a straight line on the table, perpendicular to the edge of the table, with a tiny bit hanging over the end, or
2. Nicely coiled up just at the edge of the table, so all the falling chain has to do is unravel that coil?

You will get two different answers for these different setups.

 

[Simon Bridge]

Ok I've tried again and got this.

(d^2 x / d t^2) = (x/L) g

Well done.
You should solve the DE as an initial value problem.

BTW: we've been doing the "laid out in a line".
What would be different if the chain were arranged right on the edge of the table?

 

[AJ2357]

Great thanks very much for your help guys.