Phyiscs Friday 104

Consider a massive rope, with length L and mass per unit length of λ. Let this rope be held so that it hangs vertically, with a thin sheet of paper just below the bottom end of the rope. This sheet of paper can support without breaking a maximum weight equal to twice the weight of the rope. If one releases the top of the rope, will the paper stay unbroken until the rope is entirely atop it? If not, what will the height of the top of the rope (relative to the paper) be when the paper breaks?

The mass of the rope is λL, so it has total weight λLg, and so the maximum force the paper can withstand is thus 2λLg.

Now, let h(t) be the height of the top of the rope, and let F be the magnitude of the upward force exerted by the paper on the rope (and thus the magnitude of the downward force exerted by the rope on the paper). Choosing the upward direction to be positive, the net force on the rope is FλLg.

Next, we note that only the portion of the rope still above the paper is moving. Using dot notation for time derivatives, this portion at a given instant has as a velocity and λh as mass. Thus, the momentum of the rope is . Note that since , p<0 (negatives here indicate the downward direction).
The time derivative of this momentum is
, and by Newton’s second law, this is the force; setting them equal,
and solving for F tells us that

But the portion of rope still above the paper is in free fall; when the top has height h, this portion has fallen through a distance Lh. Thus, it has velocity and acceleration . If we plug these into our formula for F, we get
which goes from zero at h=L to 3λLg when h→0+; thus the paper will break for some h>0; setting F=2λLg, and solving for h,

and we have our height when the paper breaks.

Lastly, noting that since the dropping portion of the rope is in free-fall, until the paper breaks, we have height as a function of time given simply by . Thus, we can find how long after the rope is released the paper breaks:


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