A chain of length d lies on a table, $d/4$ of it hangs loose on the side of it. Friction coefficient is 0.2. It is released and begins to slide. What is the final velocity with which it falls of the table?
My attempt:
$Wf = \bigtriangleup Ug = – \int_0^{0.75d} \! \mu_k g \frac{m}{d}(\frac{d}{4}+x) \, \mathrm{d}x = -0.4687\mu_k mgd$
Which is the change and loss in energy during the slide. My plan was to equate the initial gravitational potential energy to the final kinetic energy plus the energy lost from friction (calculated above), however there is no way to define the friction for the rope since it has two parts, with apparently distinct potential energies. Is there another way to do it?
Best Answer
The conceptual problem seems to be
Don't worry about the "distinct potential energies". You can just compute the gain in potential energy for each part separately. For the bit already hanging down, as the rope slides by a certain distance, the center of mass moves by that same distance. For the bit of rope that starts off horizontal, when the last part goes over the edge its center of mass has moved half as far.
That leaves the calculation of the work done by friction. If you consider your chain like a train with lots of cars, then for each car in the train you can consider the work done due to friction. In other words - for an infinitesimal element $d\ell$ that starts out at a distance $x$ from the edge, the force of friction is $\mu \frac{d\ell}{L} m g$ and the work done is $F \cdot x$.
Summing the work for all these elements will give you the total work done. That is a simple integral.
In the spirit of "homework like questions", I will leave you with these hints.