# Finding the tension of the rope with a mass

I have seen an interesting problem as follows:

Both ends of a rope with the mass of $$\rho$$ kg/m and length of $$L$$ are attached to a horizontal surface (see photo attached). At $$t=0$$, one end of the rope detaches from the surface and starts falling vertically. When this end falls the height $$X$$, what is the tension on the other end of the rope?

I also have seen two different solutions from a professor which result in two different answers!

Approach 1 is based on falling part of the rope is freely falling (acceleration $$= g$$).

Approach 2 is based on energy conservation $$KE + PE =$$ Constant (no loss).

Now I would like to ask

1- Is the falling part of the rope really freely falling?

2- Is energy conserved when the rope is falling (for $$X \lt L$$)?

• Note to those voting to close: the post isn't asking how to solve the problem. Jul 6 at 21:21
• Chain Drop Answer 2 Jul 7 at 2:57

A point not fully explained in the cited references is that when the free end of the rope is falling at speed $$v$$, the tension in curved bit of the rope is $$T={\mu v^2}/{4}$$ on both sides of the fold. Consequently the falling bit of the rope has a force of $$T={\mu v^2}/{4}$$ pulling it down in addition to gravity. To see that this is so recall that in the absence of gravity a chain moving with speed $$u$$ can maintain an arbitrary planar shape because the centripetal acceleration of its links is automatically provided by the tension according to $$\frac{\mu u^2}{r}= \frac{T}{r}, \quad (\star\star)$$ so the tension takes the value $$T= \mu u^2$$ independent of the radius of curvature $$r$$. For our falling rope, and in the reference frame that is descending with the fold at $$u=v/2$$, the rope/chain is moving through the fold at $$u=v/2$$. If we can ignore the effect of gravity and the non-inertial reference frame, the tension throughout the fold must be $$T= \mu (v/2)^2=\mu v^2/4$$. Ignoring these effects is a safe approximation for our sharp fold because when $$r$$ is small the forces in $$(\star\star)$$ completely dominate all other forces.