Solve Falling Chain (3.11) Length & Speed

In summary, the chain hangs down through the hole in the table after losing contact with it, and the speed at which it hangs down is proportional to the tension T applied to it.
  • #1
Radarithm
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2

Homework Statement



A chain with length ℓ is held stretched out on a frictionless horizontal table, with a length y0 hanging down through a hole in the table. The chain is released. As a function of time, find the length that hangs down through the hole (don't bother with t after the chain loses contact with the table). Also, find the speed of the chain right when it loses contact with the table.

Homework Equations



[tex]F_y=m\ddot{y}[/tex]
[tex]y_0\ge y\ge l[/tex]

The Attempt at a Solution



Conceptually, more of the chain must be through the hole so that y-naught is greater than y (initially). To cause an acceleration to the left (or towards the hole; however you picture the scenario), the tension T must be equal to mg (correct?). What I'm having trouble with is setting up an appropriate differential equation; should I just do the old: [tex]\ddot{y}=-g[/tex] because this would (should) hold true if I claim that the tension T is equal to mg. This means that something fundamental is getting past me, I just don't know what. Also this doesn't have to do with the question, but is K&K harder than the book I already have (Morin)?
 
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  • #2
The m in ##\ddot y## is different from the m in ## mg##. Make a drawing.
 
  • #3
http://postimg.org/image/vprzi0y6z/
The FBDiagram on the left is for a piece of the chain that is present on top of the table, whereas the one on the right is for a piece of the chain that is falling through the hole in the middle.
 
  • #4
frictionless, so T isn't needed.
My point is that mg is not constant because that m = y x m/L
 
  • #5
BvU said:
frictionless, so T isn't needed.
My point is that mg is not constant because that m = y x m/L

Would you mind proving how m = y x m/L? Is that a cross product?
 
  • #6
This is one of these totally artificial exercises that confuse bright students. Just imagine what is needed to make the chain take a 90 degree turn with 0 radius at the rim of the hole. Yuch.

Never mind. Questions about your FBDs: On the left I see d M g. What is that ? T is the accelerating tension I suppose.
If there is an FBD for a (the ?) piece on top of the table, why is there no FBD for a (the) piece hanging vertically ?

On the right I see rather nothing. T is gone ?
 
  • #7
No. If the whole chain has length L, mass m, then a piece with length y has mass ## y\ m / L ##
 
  • #8
This is probably incorrect (I'm sure it is) but here I go:
[tex]dm\frac{dv}{dt}=-dm_2g[/tex]
[tex]dmdv=-dm_2gdt[/tex]
edit: Nevermind the diff eqs, that was before I saw your replies.
Can you at least give me a hint as to how I can relate the length to all of this? I can't seem to set up a diff eq.
 
  • #9
In the left FBD, I now know that:
[tex]gdm=\frac{ym}{dL}[/tex]
Forgot to include the tension in the FBD on the right, and it is the one for the vertical piece.
 
  • #10
Ok so because:
[tex]m\frac{d^2y}{dt^2}=\frac{ym}{dL}g-T[/tex]
for the vertical piece of the chain, then:
[tex]T=m\left(\frac{d^2y}{dt^2}-\frac{gy}{dL}\right)[/tex]

Correct?
I don't know how to relate the diff eqs for the vertical and horizontal pieces of chain in order to find L(t).

I do know that the limits of integration (for the integral that I have to calculate) are [itex]y_0[/itex] and [itex]l[/itex]
 
  • #11
Ok so if what I got for T is correct then:
[tex]m\frac{d^2y}{dt^2}-m\frac{gy}{dL}=m\frac{d^2x}{dt^2}[/tex]
[tex]\frac{d^2y}{dt^2}-\frac{gy}{dL}=\frac{d^2x}{dt^2}[/tex]

I guess the only problem I have now is with setting up the diff eq. If I am right, does the [itex]dL[/itex] have to be on the left side with everything else on the right? (there are multiple dt's; I don't suppose that this is problematic, assuming that I can group them up together if they are similar).
 
  • #12
This is going at a posts per few minutes. I can't follow. What is d in dL and what is x ?
You can't write [itex]gdm=\frac{ym}{dL}[/itex] if you take a differential in the lhs, you also need a d in the numerator on the rhs; it can't end up in the denominator.

We know the chain remains stretched. So vertical piece, length y(t) means horizontal piece length L - y(t).

Vertical: F = y(t) m/L g - T
Horizontal: F = T

Someone else will have to take over; I have to go to a training.
 
  • #13
You may need to go back to F = dp/dt = dm/dt v + m dv/dt if you write a differential equation for only one of the sections (because for a section dm/dt is not 0). Either that , or only write the differential equation for the entire chain.
 
  • #14
Well, the accelerations for x and y must be equal -- otherwise the chain would stretch out or it would create extra slack. How about something like this?
$$\frac{d^{2}y}{dt^{2}} = \frac{g \cdot \lambda y}{M}$$ Where [itex]\lambda[/itex] is the linear mass density -- [itex]\frac{M}{L}[/itex], which reduces to: $$\frac{d^{2}y}{dt^{2}} = \frac{gy}{L}$$

This may be totally off though -- I'll think more about it.
 

FAQ: Solve Falling Chain (3.11) Length & Speed

What is the falling chain problem and why is it important?

The falling chain problem is a classic physics problem where a chain is suspended from two points and allowed to fall under the influence of gravity. It is important because it demonstrates how tension, acceleration, and velocity are related in a dynamic system.

How do you solve the falling chain problem?

To solve the falling chain problem, you need to use the principles of Newton's laws of motion. First, you must calculate the acceleration of the chain using the equation F=ma, where F is the force of gravity and m is the mass of the chain. Then, you can use the equation v^2 = u^2 + 2as to calculate the velocity of the chain at any given point.

What is the formula for calculating the length of the falling chain?

The formula for calculating the length of the falling chain is L = (2h/3)*(1+cos(theta)), where L is the length of the chain, h is the initial height of the chain, and theta is the angle between the chain and the horizontal.

How does the speed of the falling chain change over time?

The speed of the falling chain increases as it falls due to the force of gravity. However, as the chain reaches the end of its fall and approaches the ground, its speed will decrease due to the tension in the chain. The speed of the chain can be calculated at any point using the equation v^2 = u^2 + 2as.

What factors can affect the length and speed of the falling chain?

The length and speed of the falling chain can be affected by various factors, such as the initial height of the chain, the mass of the chain, the angle at which it is suspended, and the force of gravity. Other factors, such as air resistance and friction, may also have a small impact on the results.

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