Confirm the differential equation for a falling rope using Newtons law

[sapiental]

Homework Statement

A uniform chain of length L, measured in feet, is held vertically so that the lower end just touches the floor. The chain weighs 2lb/ft. The upper end that is held is released from rest at t = 0 and the chain falls straight down. Ignore air resistance, assume that the + direction is downward, and let x(t) denote the length of the chain on the floor at time t. Use the fact that the net force F in (18) is 2L to show that a differential equation for x(t) is:

(L - x) (d^2x / dt^2) - (dt/dx)^2 = Lg

Homework Equations

the problem mentions equation 18 in our book which is

F = d/dt (mv)

The Attempt at a Solution

I rewrote the original equation as

(L - x) (d^2x / dt^2) - Lg = (dt/dx)^2

by analyzing the system I'm guessing the weight = 2(L-x)

mass = 2(L-x)/g

since F = ma

2(L-x)/g * a = 2L

a = Lg/(L-x)

a = (d^2x / dt^2) or the second derivative of t with respect to x

then

(L-x)(Lg/(L-x) - Lg = (dx/dt)^2

0 = (dx/dt)


When I tried to reproduce the same d.e. for this system i get

(L - x) (d^2x / dt^2) + (dt/dx)^2 = Lg

*the + instead of the - before the (dt/dx)^2 is what throws me off.

thanks in advance for all your help

 

[anathema]

!

Thank you for your post. It seems that you have made a small error in your attempt at solving this problem. The correct differential equation should be:

(L - x) (d^2x / dt^2) + (dx/dt)^2 = Lg

The reason for the positive sign in front of the (dx/dt)^2 term is because the chain is falling downwards, so the velocity of the chain is also in the downward direction. This means that the velocity term in the equation should be positive, as it represents the rate of change of the length of the chain with respect to time.

I hope this helps clarify the issue. Keep up the good work in your studies!

 

[ogb p]

I would first confirm that the differential equation provided in the problem is correct. Using Newton's second law, F = ma, we can write the force equation for the falling rope as:

F = (2L - 2x)g = ma

Where 2L is the weight of the rope and 2x is the weight of the portion of the rope that has fallen. Since the rope is assumed to be falling straight down, we can take the direction of motion as positive and rewrite the equation as:

2L - 2x = ma

We can also use the chain rule to express acceleration in terms of the derivative of the length with respect to time, dx/dt:

a = d^2x/dt^2

Substituting this into the equation above, we get:

2L - 2x = m(d^2x/dt^2)

But we also know that the mass of the rope is given by m = 2(L-x)/g. Substituting this into the equation, we get:

2L - 2x = (2(L-x)/g)(d^2x/dt^2)

Simplifying this equation, we get:

(L-x)(d^2x/dt^2) = Lg

This confirms the differential equation provided in the problem.

Regarding your attempt at solving the problem, it seems like you have made a few mistakes in your calculations. Firstly, the weight of the rope is not equal to 2(L-x), it is equal to 2L. This is because the weight of the rope is given as 2lb/ft, so the total weight of the rope is 2lb/ft multiplied by the length of the rope, which is L feet. Secondly, the mass of the rope is not equal to 2(L-x)/g, it is equal to 2(L-x)/g. Finally, you seem to have made a mistake in your substitution, where you have replaced a with (d^2x/dt^2) instead of (d^2x/dt^2). This is why you get a different sign in your final equation.

In science, it is important to carefully check our calculations and equations to ensure that they are accurate and valid. In this case, it is important to understand the physical principles behind the problem and use them to derive the correct differential equation.