Solving Rope Falling Off Table Homework

[torq123]

Homework Statement

A (smooth) rope of length L and mass m is placed above a hole in a table. One end of the rope falls through the hole, pulling steadily on the remainder of the rope. Find the velocity of the rope as a function of the distance to the end of the rope, x. Ignore friction of the rope as it unwinds. Then find the acceleration of the falling rope and the mechanical energy lost from the rope as the end of the rope leaves the table. Note that the rope length is less than the height of the table.

Homework Equations

F = mA= (mg/L)x

The Attempt at a Solution

My thoughts are that since a(x)=x\frac{g}{l}, x(t) = e^{t\sqrt{g/l}}, so v(t) is just the derivative of that, and v(x) =x\sqrt{g/l}

I cannot figure out what I have done wrong up to this point. The problem is that, as you can see, that equation simply leads to KE(gained) = PE(lost).

The correct solution starts with:

mg = m\frac{dv}{dt} + vm\frac{dm}{dt} which yields v^{2} = m\frac{2gx}{3}.

which corresponds to a loss \frac{mgL}{6} of mechanical energy.

I just don't get it...

 

[TSny]

Hello, torq123.

The wording of the problem is not all that clear. Based on your equation F = mA = (mg/L)x, you seem to be letting m stand for the mass of the entire rope and assuming that at any instant, all points of the rope have the same acceleration.

But the intended interpretation is that the rope is coiled up and sitting above the hole. Essentially, only the part of the rope that has passed through the hole (or is about to pass through the hole) is in motion. So, more and more of the rope goes into motion as time passes. See the attached figure.

Note that in the differential equation of the solution that you gave, m stands for just the mass of rope that has passed through the hole. But it looks like there is a typographical error in the differential equation, as you can see that the last term does not have units of force like the other two terms. Likewise your expression for $v^2$ does not have the right units.