Solve Kinematics Problem: Find x(t) for Chain on Frictionless Table

[naggy]

A chain, total length h meters long, is sitting on a table. C meters hang down from the table. The chain is let go from rest at time t=0 . The table is frictionless. Find the equation of motion, x(t)

Hint: Second order differential equations that can be reduced to cosh if x is defined posative downward.

I need a jump start I guess. No masses are given or anything...

I can write a DE if there is air resistance.

a + kv = g ?

 

[naggy]

I should add the part of the chain that is on the table is straight, not curled up.

MY ATTEMPT

The part on the table is of length h-c and the part hanging down from the table is c. I'm just can´t think of what to use to find the differential equation. Maybe there is air drag?

If I would write a force equation then it would be mg = ma because the table is frictionless?

 

[Doc Al]

I should add the part of the chain that is on the table is straight, not curled up.

That does make a difference--the entire chain moves at a single speed.

The part on the table is of length h-c and the part hanging down from the table is c. I'm just can´t think of what to use to find the differential equation. Maybe there is air drag?

Forget air drag (that just adds complexity).

If I would write a force equation then it would be mg = ma because the table is frictionless?

Yes, something along those lines: Apply Newton's 2nd law.

 

[naggy]

Applying Newtons second law is sort of the problem. I can´t see how I'm supposed to treat the chain as one mass.

Can I think if it as two blocks attached together with a string? There's no friction on the table so the force on the chain on the table( tension in the string) will just be equal to m_cg, where m_c is the mass of the part of the chain hanging from the table.

What other forces are there?

I then have m_cg - T =m_ca

but m_c changes as more part of the chain drags from the table right?

 

[Doc Al]

What other forces are there?

All you care about are forces parallel to the direction of motion, so that's the only force you need. (The weight of the chain on the table is balanced by the normal force.)

I then have m_cg - T =m_ca

Forget about tension--that's an internal force and thus cancels out. (You want the net force on the chain.)

but m_c changes as more part of the chain drags from the table right?

Of course. Write it as a function of x, where x is the amount of chain hanging off the table.

 

[naggy]

what about mx/h where m/h is mass per unit length?

[tex]\frac{mxg}{h}[/tex] = ma

that leads to x'' -xg/h = 0 a second order DE

 

[Doc Al]

what about mx/h where m/h is mass per unit length?

[tex]\frac{mxg}{h}[/tex] = [tex]\frac{m}{h}[/tex][tex]a[/tex]

Good, but correct that right hand side.

 

[naggy]

Good, but correct that right hand side.

already did I think it's just ma not ma/h

 

[Doc Al]

Good. Now all you need to do is solve the DE.

 

[naggy]

Good. Now all you need to do is solve the DE.

There's a sequel to this problem. Now there is friction on the table

Could I write the frictional force as umgx/(h-c)?

u=frictional coefficient

 

[naggy]

No wait. It must be

[tex]\frac{umg(h-c-x)}{(h-c)}[/tex]

 

[naggy]

No wait. It must be

[tex]\frac{umg(h-x)}{h}[/tex]

this should be correct

 

[Doc Al]

That latest version looks good to me.