A rope falling off an inclined plane

[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

 

[BvU]

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.

 

[haruspex]

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.

 

[Elvis 123456789]

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.

 

[haruspex]

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 y_i and finishes at height y_f, ΔGPE=∫Δyg.dm=g(∫y_fdm-∫y_idm).
∫y_fdm is the final height of the mass centre, by definition.