The work that is necessary to pull a hanging chain

[Rikudo]

Homework Statement

A chain with length L and mass m is placed on a table with frictionless surface.If one-fifth of the chain is hanging at the end of the table, how much work that is necessary to pull the chain so that all parts of that chain is on the table.

Relevant Equations

Work-Energy theorem

This is the solution from my textbook, and I have some questions about the method

The mass of hanging chain : $$m_h =\frac m 5$$
the center of mass of the hanging chain : $$h_1 = - \frac{1} {2} \cdot \frac L 5 = - \frac L {10}$$
(the minus sign here means that it is under the table surface)

The Work that is necessary to pull the chain : $$W = \Delta E_p$$ $$W = m_h \cdot g (h_2 - h_1) $$ $$ W = \frac m 5 \cdot g (0 +\frac{L} {10}) $$ $$ W = \frac{mgL} {50}$$

Here is some parts that I don't understand:
1. In $\Delta E_p$, why do we use the center of the hanging chain mass instead of -$\frac L 5$ (the length of the chain that is initially hanging) as the initial height? Doesn't this means that the other $\frac L {10}$ is still hanging after the chain is pulled?

2. The initial formula of the Work-Energy theorem is $W= \Delta E_p + \Delta E_k$ . In the solution, there is no $\Delta E_k$ ,which means there is no change in final and initial velocity. When the person who is pulling the chain starts to stop giving force, at this instant, isn't the chain is supposed to have velocity?

 

[PeroK]

Here is some parts that I don't understand:
1. In $\Delta E_p$, why do we use the center of the hanging chain mass instead of -$\frac L 5$ (the length of the chain that is initially hanging) as the initial height? Doesn't this means that the other $\frac L {10}$ is still hanging after the chain is pulled?

2. The initial formula of the Work-Energy theorem is $W= \Delta E_p + \Delta E_k$ . In the solution, there is no $\Delta E_k$ ,which means there is no change in final and initial velocity. When the person who is pulling the chain starts to stop giving force, at this instant, isn't the chain is supposed to have velocity?

1) Each part of the chain requires a different amount of work to pull it up onto the table. Using the centre of mass is a quick way of calculating the average work per unit mass.

2) If the chain is pulled slowly enough, then the final KE can be made as small as possible, Small enough to be negligible. We are really calculating the minimum (or threshold) work required here. Any final KE of the chain would be additional to that. This is a common theme in these problems.

 

[Rikudo]

1) Each part of the chain requires a different amount of work to pull it up onto the table. Using the centre of mass is a quick way of calculating the average work per unit mass.

Let's say we already pulled $\frac L {10}$ of the chain onto the table, which means that we do the amount of works : $W = \frac{mgL} {50}$. How about the other $\frac L {10}$ that is still hanging? Shouldn't we also need to pull the chain again until it is on the table (which means we need to give more Work)? And also, this may sounds so dumb, but could you please explain why the chain requires different amount of work?

 

[PeroK]

Let's say we already pulled $\frac L {10}$ of the chain onto the table, which means that we do the amount of works : $W = \frac{mgL} {50}$. How about the other $\frac L {10}$ that is still hanging? Shouldn't we also need to pull the chain again until it is on the table (which means we need to give more Work)? And also, this may sounds so dumb, but could you please explain why the chain requires different amount of work?

Parts of the chain are closer to the tabletop than others. $\frac L {10}$ is the average distance each part of the chain must be raised. If you use $\frac L 5$, then that assumes that all the $\frac m 5$ that's overhanging is $\frac L 5$ from the top of the table. That's definitely too much work.

 

[Rikudo]

Ah...I understand now. Thanks!

 

[Delta2]

The work that is needed to raise an infinitesimal mass $dm$ that is located at distance x below the table is $dW=dmgx$. The total work is $$W=\int dW=\int_0^{\frac{L}{5}}gx dm$$
On the other hand the definition for the center of gravity of the hanging chain is $$h=\frac{\int_0^{\frac{L}{5}}gx dm}{\frac{M}{5}g}=\frac{W}{\frac{M}{5}g}\Rightarrow W=\frac{M}{5}gh$$

 

[haruspex]

If the chain is pulled slowly enough, then the final KE can be made as small as possible

Alternatively, pull it at whatever speed (but not so much that it becomes airborne as it reaches the tabletop) and stop pulling when it has enough KE to finish by itself.