Question 9.6: Rope sliding off a table A uniform inextensible rope of mass...

Rope sliding off a table

A uniform inextensible rope of mass M and length a is released from rest hanging over the edge of a smooth horizontal table, as shown in Figure 9.5. Find the speed of the rope when it has the displacement x shown.

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A rope is a continuous distribution of mass, unlike the discrete masses that appear in our theory. We regard the rope as being represented by a light inextensible string of length a with N particles, each of mass M/N, attached to the string at equally spaced intervals along its length. When N is very large, we expect this discrete set of masses to approximate the behaviour of the rope.
Since each particle of the rope has the same speed v(=\dot{x}) , the total kinetic energy of the rope is simply

T=\frac{1}{2} M v^{2}.

The only contribution to the potential energy comes from uniform gravity. If we take the reference state for V to be the initial configuration (Figure 9.5 (left)), then the potential energy in the displaced configuration (right) is the same as if a length x of the rope lying on the table were cut off and this piece were then suspended from the hanging end. In the continuous limit (that is, as N → ∞), this piece of rope has mass M x/a and its centre of mass is lowered a distance b + (x/2) by this operation.
The potential energy of the rope in the displaced configuration is therefore

V=-\left(\frac{M x}{a}\right) g\left(b+\frac{1}{2} x\right).

We must now show that the constraint forces do no work. The reactions exerted by the smooth table on the particles of the rope are always perpendicular to the velocities of these particles; these reactions therefore do no work. Also, the tension forces exerted by each segment of the inextensible string (connecting adjacent particles of the rope) do no work in total. Hence, the constraint forces do no work in total.
Energy conservation therefore applies in the form

\frac{1}{2} M v^{2}-\left(\frac{M x}{a}\right) g\left(b+\frac{1}{2} x\right)=E.

The initial condition v = 0 when x = 0 implies that E = 0. The energy equation for the rope is therefore

v^{2}=\frac{g}{a} x(x+2 b).

This gives the speed of the rope when it has displacement x. This formula holds while there is still some rope left on the the table top.

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