# Velocity of the end point of the string.

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A string of uniform mass $M$ and total length $l$ is placed on a table. The length $h$ of the string is falling from the edge of the table. Find the velocity of the right end of the string that is on the table just before falling down.

(Wow !!! Thanks for the built-in image)

Let $x$ be the amount that string falls in time $\mathrm {dt}$

Now the mass per unit length $M_0 = M/l$

Mass of the part below table, = $M_0 (h+x)$ and above table = $M_0(l- (h+x))$

Since h part is pulling the above table part,

$$M_0a(l-(h+x)) = M_0g(h+x)$$
$$\implies a(l-(h+x)) = g(h+x)$$
$$a = g{h+x \over l-(h+x)}$$
$${dv\over dt} = g{h+x \over l-(h+x)}$$
$${dv\over dx}{dx \over dt} = g{h+x \over l-(h+x)}$$
$$v{dv\over dx} = g{h+x \over l-(h+x)}$$
$$v{dv} = g{h+x \over l-(h+x)}dx$$
$$\int_0^v v{dv} = \int_0^{l-h}g{h+x \over l-(h+x)}dx$$
$${v^2 \over 2} = \int_0^{l-h}g{h+x \over l-(h+x)}dx$$

Now I will do a u substitution that I never did before this,
Let $u = h + x$

${du\over dx} = {1} \implies du = dx$

$$\int_0^{l-h}{h+x \over l-(h+x)}dx = \int_0^{l-h}{u \over l-(u)}du$$
$$= -\int_0^{l-h}{-u \over l+(-u)}du = {u+l\loge(\vert u-l\vert)\bigg\rvert{0}^{l-h}}$$
$$= {h+x+l\loge(\vert h+x-l\vert)\bigg\rvert{0}^{l-h}} ={h+l-h+l\log_e(\vert h+l-h-l\vert) -(h+0+l\log_e(\vert h+0-l\vert) )}$$
$$= l+l\log_e(\vert0\vert) -h-l\log_e(\vert h-l\vert) ) = {l -h +l(\text{undefined} -\log_e(\vert h-l\vert))}$$

$$v^2/2 = g(l -h +l(\text{undefined} -\log_e(\vert h-l\vert))$$

$$v = \sqrt{2g(l -h +l(\text{undefined} -\log_e(\vert h-l\vert))}$$

I think I did everything properly still I got "undefined" in my solution.

I think my intial integral itself is wrong.

edited Jan 28, 2017
With origin at the corner, $p_0(t)=h(t)(-\hat y)$ and $p_1(t)=(L-h(t))\hat x$.  The amount of mass falling at time t is $\rho_m ||p_0(t)||$, where $\rho_m=M/L$. This is what I think at least, so the h+x terms you use looks wrong to me. Aha just realized you mean h(0)+x(t), where x is displacement from the original setup. Nevermind then :-) wolfram alpha says $\int u/(l-u)du=-l log(u-l) -u + C$, so the definite integral is the difference of two such terms, and the constant needs boundary conditions of some sort, maybe that the speed is $v_0$ at t=0 is enough, just a guess.
That definite integral is undefined for my limits on the integral. Can you check my limits on the integral ?

Have you tried using conservation of energy? Decrease in PE = increase in KE. You only need to look at the initial and final conditions. No need to integrate.

answered Jan 28, 2017 by (28,746 points)
selected Jan 28, 2017 by Motu
$-\Delta PE = \Delta KE$

What what mass do I need to take in conservation equation, Mass of string or mass of last bit of the string ?

If I take the mass of the last bit of the string then that mass would be ${M\over l}(dx)$ right ?
Use the whole mass. On the LHS calculate the change in vertical position of the CM of the whole string/chain. In the final position the whole string is hanging, so it is easy to find the CM. Initially only part of the string is hanging, so the CM is not so easy to locate. On the RHS you have to calculate the KE of the whole string. In the initial position none of it is moving, in the final position all of it is moving at the same speed (because it is inextensible).

The last bit of string to leave the table moves at the same speed as the rest of the string. There is no need to use $dx$ and no need to integrate.
Thanks, I will do the math and return with result. I am still wondering where my integration method went wrong, have any clue ?
Sorry I don't want to spend time trying to understand your solution.

Some problems are solved most easily using $F=ma$, some are solved most easily using conservation of energy (or the work-energy theorem), some there's not much difference. This problem cries out for conservation of energy to be applied.

Your solution method might be correct; perhaps it is just getting very complicated.