Fall 2007 1 Mech

A rope of mass M and length L is suspended in the earth’s gravitational field, g, with the bottom end of the rope touching a surface. The rope is then released from rest and falls limply on the surface (i.e., without the elements bouncing upwards). Find the force F(t) on the surface as a function of time, 0 < t < $\infty$, and make a sketch of it. At what time does F(t) reach its maximum? What is the value of this maximum force?

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Answer

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All the elements of the rope are in free fall. It takes time T = 2L/g for the last element to reach the surface. Hence, at t > T we have

(1)
\begin{equation} F(t) = Mg , t > T \end{equation}

Consider now 0 < t < T. Here F is a sum of two terms, F = F1 + F2. The first one is the the weight $F_1 = \mu l g$ of the part that has already fallen. Here $l(t) = gt^2/2$ is the length of the fallen piece and $µ = M/L$ is the mass per unit length. The second term is the transfer of momentum F2 = dP/dt from the element of length dl = vdt that comes to rest during the time (t, t+dt).

The velocity of this segment is v = gt, and so $F_2 = \mu v(vdt)/dt = \mu g^2 t^2$.
Accordingly,

(2)
\begin{align} F(t)= \frac{1}{2}\mu g^2 t^2 + \mu g^2 t^2 = \frac{3}{2} \frac{M}{L} g^2 t^2 = 3 M g \frac{t^2}{T^2}, \; \; 0 < t < T \end{align}

We see that F reaches a maximum F = 3Mg at t = T, then experiences a sudden drop to the three times smaller value, after which it remains constant.

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