# A rope falling off an inclined plane

• Elvis 123456789
In summary: If you choose the x axis along the slope, the lowest part of the rope is always at the same height, so U(x) only changes by changes in the height of the mass centre of the part of the rope on the incline (which is always x/2 above the lowest part on the slope). That is, ΔU(x)=mUg(x+x0)/2g. So U=U0+mgh where U0 is an arbitrary constant of potential energy.Why ? You take the horizontal plane as a reference, which you are allowed to do.It's just a matter of choosing a reference point for potential energy. In this case,
Elvis 123456789

## Homework Statement

A rope of length L is falling off an incline. Part of the rope is still on top of the horizontal surface. There is no friction between the incline and the rope. The incline is at an angle theta above the horizontal.

a) What is the magnitude of the acceleration of the rope, when the part of the rope on the incline is x (i.e., the part left on the horizontal surface has a length of L-x)

b) What is the speed of the rope at this moment?

c) What is the value of x, when the acceleration reaches maximum?

## Homework Equations

∂L/∂x - d/dt(∂L/∂v) = 0 where v = x-dot
a = dv/dt = (dv/dx)*(dx/dt)

## The Attempt at a Solution

So I worked out this problem but I am unsure of my result, specifically my result for the potential energy in the lagrangian.

L = ½mv2 - U(x) = ½λLv2 + (λx)(g)(½x sin(θ))

where λ=m/L is the linear mass density

the Euler-Lagrange equation then gives

1.) a = (x/L)g sin(θ)

assuming my acceleration for # 1 is correct, I used the identity a = v*(dv/dx) to find the velocity

a*dx = v*dv → ∫(x/L)g sin(θ)dx = ∫vdv

2.) v = √[v20 + 1/L (x2 - x20)g sin(θ) ]

then # 3 would be x = L

Last edited:
Elvis 123456789 said:
specifically my result for the potential energy in the lagrangian
Why ? You take the horizontal plane as a reference, which you are allowed to do.

Elvis 123456789 said:

## Homework Statement

A rope of length L is falling off an incline. Part of the rope is still on top of the horizontal surface. There is no friction between the incline and the rope. The incline is at an angle theta above the horizontal.

a) What is the magnitude of the acceleration of the rope, when the part of the rope on the incline is x (i.e., the part left on the horizontal surface has a length of L-x)

b) What is the speed of the rope at this moment?

c) What is the value of x, when the acceleration reaches maximum?

## Homework Equations

∂L/∂x - d/dt(∂L/∂v) = 0 where v = x-dot
a = dv/dt = (dv/dx)*(dx/dt)

## The Attempt at a Solution

So I worked out this problem but I am unsure of my result, specifically my result for the potential energy in the lagrangian.

L = ½mv2 - U(x) = ½λLv2 + (λx)(g)(½x sin(θ))

where λ=m/L is the linear mass density

the Euler-Lagrange equation then gives

1.) a = (x/L)g sin(θ)

assuming my acceleration for # 1 is correct, I used the identity a = v*(dv/dx) to find the velocity

a*dx = v*dv → ∫(x/L)g sin(θ)dx = ∫vdv

2.) v = √[v20 + 1/L (x2 - x20)g sin(θ) ]

then # 3 would be x = L
All looks right. I get the same for 2 using conservation of work. In going from x0 on the slope to x on the slope, a length x-x0 has descended from the horizontal surface to be the lowest part of the rope. Its mass centre has descended (x+x0)/2.

BvU said:
Why ? You take the horizontal plane as a reference, which you are allowed to do.
The part that sort of bothers me is the "h" part in mgh for U(x). For continuous bodies I am used to having to integrate over the whole part that contributes potential energy instead of just taking the position of the center of mass.

Elvis 123456789 said:
The part that sort of bothers me is the "h" part in mgh for U(x). For continuous bodies I am used to having to integrate over the whole part that contributes potential energy instead of just taking the position of the center of mass.
If an element of mass dm starts at height yi and finishes at height yf, ΔGPE=∫Δyg.dm=g(∫yfdm-∫yidm).
∫yfdm is the final height of the mass centre, by definition.

## 1. What is the cause of a rope falling off an inclined plane?

The main cause of a rope falling off an inclined plane is the force of gravity acting on the rope. As the rope is placed on an incline, it experiences a downward force due to gravity, which can cause it to slide or fall off.

## 2. Can the angle of the incline affect the likelihood of a rope falling off?

Yes, the angle of the incline can greatly affect the likelihood of a rope falling off. The steeper the incline, the greater the downward force due to gravity, making it more likely for the rope to fall off.

## 3. Is there any way to prevent a rope from falling off an inclined plane?

There are a few ways to prevent a rope from falling off an inclined plane. One way is to increase the friction between the rope and the incline by using a rougher surface or adding more weight to the rope. Another way is to decrease the angle of the incline, making the downward force due to gravity less strong.

## 4. What other factors can contribute to a rope falling off an inclined plane besides gravity and the angle of the incline?

Other factors that can contribute to a rope falling off an inclined plane include the weight and tension of the rope, as well as any external forces acting on the rope, such as wind or other objects.

## 5. How does the length of the rope affect its likelihood of falling off an inclined plane?

The length of the rope can also play a role in its likelihood of falling off an inclined plane. A longer rope will have more surface area in contact with the incline, increasing the friction and making it less likely to fall off. However, a longer rope may also be heavier, which can increase the downward force due to gravity and make it more likely to fall off.

• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
31
Views
348
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
26
Views
2K
• Introductory Physics Homework Help
Replies
18
Views
3K
• Introductory Physics Homework Help
Replies
8
Views
868
• Introductory Physics Homework Help
Replies
10
Views
2K
• Introductory Physics Homework Help
Replies
19
Views
2K
• Introductory Physics Homework Help
Replies
15
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
2K