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?




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

\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 $F_2 = \mu v(vdt)/dt = \mu g^2 t^2$.

\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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