Question

18 A flexible rope of mass m and length L slides without friction over the edge of a table. Let x be the length of the rope that is hanging over the edge at a given moment in time (a) Show that r satisfies the equation of motion/dt2 -gr/L. Hint: Use F-dp/ dt, which allows you to handle the two parts of the rope separately even though mass is moving out of one part and into the other (b) Give a physical explanation for the fact that a larger value of r on the right-hand side of the equation leads to a greater value of the acceleration on the left side (c) When we take the second derivative of the function r(t) we are supposed to get essentially the same function back again, except for a constant out in front. The function e* has the property that it is unchanged by differentiation, so it is reasonable to look for solutions to this problem that are of the form = bect, where b and c are constants. Show that this does indeed provide a solution for two specific values of c (and for any value of b) (d) Show that the sum of any two solutions to the equation of motion is also a solutioin (e) Find the solution for the case where the rope starts at rest at t 0 with some nonzero value of x

Answer 1

If at any time t the length of the rope hanging from the table is X then the force acting on the rope is gravitational force acting on the portion of rope hanging from the table as given by

$F=mg\frac{x}{L}$

This is equal to the mass of the rope times acceleration of the rope at that time which is given by

$F=m\frac{d^{2}x}{dt^{2}}$

So the equation of motion is given by

$F=m\frac{d^{2}x}{dt^{2}}=mg\frac{x}{L}$

d2

This is the equation of motion.

b) if a larger part of the rope hangs from the table so the amount gravitational force on that part will increase that will also increase the acceleration of the rope.

c) let $x=be^{ct}$ is a solution of our problem. So we can use this solution to the equation of motion as shown below

$\frac{d^{2}}{dt^{2}}(be^{ct})=bc^{2}e^{ct}=c^{2}x=g\frac{x}{L}$

$c^{2}=\frac{g}{L}$

$c=\pm(\frac{g}{L})^{\frac{1}{2}}$

So we see that we have two values of c one is positive and other is negative for any value of b.

d) the general solution is the sum of two different solutions for two values of c as given below

$x= c_{1}be^{(\frac{g}{L})^{\frac{1}{2}}t}+c_{2}be^{-(\frac{g}{L})^{\frac{1}{2}}t}$

Where $C_{1}$ and $C_{2}$ are constant.

This is also a solution of equation of motion as it satisfies equation of motion as shown below

$\frac{d^{2}x}{dt^{2}}=\frac{d^{2}}{dt^{2}}(c_{1}be^{(\frac{g}{L})^{\frac{1}{2}}t}+c_{2}be^{-(\frac{g}{L})^{\frac{1}{2}}t})=\frac{g}{L}(c_{1}be^{(\frac{g}{L})^{\frac{1}{2}}t}+c_{2}be^{-(\frac{g}{L})^{\frac{1}{2}}t})=\frac{g}{L}x$

So we see that sum of any two solutions is also a solution of equation of motion.

e) so for this we find the velocity of the rope is given by

$v=\frac{dx}{dt}=\frac{d}{dt}(c_{1}be^{(\frac{g}{L})^{\frac{1}{2}}t}+c_{2}be^{-(\frac{g}{L})^{\frac{1}{2}}t})=(\frac{g}{L})^{\frac{1}{2}}(c_{1}be^{(\frac{g}{L})^{\frac{1}{2}}t}+c_{2}be^{-(\frac{g}{L})^{\frac{1}{2}}t})$

From the question at t = 0 velocity was zero because the rope was at rest.

Using this condition we get

$c_{1}+c_{2}=0$

$c_{1}=-c_{2}=C$

is a arbitrary constant.

So the solution is given below

$x=Cb(e^{(\frac{g}{L})^{\frac{1}{2}}t}-e^{-(\frac{g}{L})^{\frac{1}{2}}t})$.