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This is a problem I thought up. If I'm correct as far as I've gone, the answer is rather surprising.
A thin but 'massive' rope lies coiled on a floor. Treat the coil as a point.
One end passes vertically up and over a smooth cylindrical drum radius r.
The centre of the drum is height h above the floor.
The other side of the drum, the rope descends to a second coil height d below the first coil.
In steady state, the rope runs at constant speed from the first coil to the second.
1. What is the shape of the descending arc of rope?
2. How far horizontally from the centre of the drum is the second coil?
Do I have something wrong?
A thin but 'massive' rope lies coiled on a floor. Treat the coil as a point.
One end passes vertically up and over a smooth cylindrical drum radius r.
The centre of the drum is height h above the floor.
The other side of the drum, the rope descends to a second coil height d below the first coil.
In steady state, the rope runs at constant speed from the first coil to the second.
1. What is the shape of the descending arc of rope?
2. How far horizontally from the centre of the drum is the second coil?
Using (s, ψ) co-ordinates, s is the distance along the rope from the point where it loses contact with the drum, and ψ is the angle to the vertical made by the arc.
T = T(s) is the tension.
ρ = mass per unit length.
v = speed of rope
Along the rope there is no acceleration: ##(1): \frac{dT}{ds} = - \rho g \cos(\psi)##
Perpendicular to the rope there is centripetal acceleration: ##(2): v^2\rho\frac{d\psi}{ds} = T\frac{d\psi}{ds} + \rho g \sin(\psi)##
Differentiating (2): ##v^2\rho\frac{d^2\psi}{ds^2} = \frac{dT}{ds}\frac{d\psi}{ds} +T\frac{d^2\psi}{ds^2} + \rho g \cos(\psi)\frac{d\psi}{ds}##
Applying (1): ##v^2\rho\frac{d^2\psi}{ds^2} = T\frac{d^2\psi}{ds^2}##
Since T is clearly not constant: ##\frac{d^2\psi}{ds^2} = 0##
Whence the trajectory is a circular arc(!).
T = T(s) is the tension.
ρ = mass per unit length.
v = speed of rope
Along the rope there is no acceleration: ##(1): \frac{dT}{ds} = - \rho g \cos(\psi)##
Perpendicular to the rope there is centripetal acceleration: ##(2): v^2\rho\frac{d\psi}{ds} = T\frac{d\psi}{ds} + \rho g \sin(\psi)##
Differentiating (2): ##v^2\rho\frac{d^2\psi}{ds^2} = \frac{dT}{ds}\frac{d\psi}{ds} +T\frac{d^2\psi}{ds^2} + \rho g \cos(\psi)\frac{d\psi}{ds}##
Applying (1): ##v^2\rho\frac{d^2\psi}{ds^2} = T\frac{d^2\psi}{ds^2}##
Since T is clearly not constant: ##\frac{d^2\psi}{ds^2} = 0##
Whence the trajectory is a circular arc(!).