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Statement

$1.5.15^*.$ Plot an approximate graph of the speed of point $B$ as a function of time, if the speed $v_A$ of point $A$ is constant. Find the formula for this relationship if $x(0) = 0$.

 For problem $1.5.15^*$
For problem $1.5.15^*$

Solution

 Velocity distribution on threads
Velocity distribution on threads
Velocity distribution on threads

NO: Before viewing the solution to this problem, I advise you to familiarize yourself with the solution 1.5.14

At time $t$, the height to which the point dropped

$$ x = v_A t\quad(1) $$

Let's consider the change in the length of the thread over a small period of time $dt$

$$ dl = \sqrt{L^2 + (x+dx)^2}-\sqrt{L^2 + x^2} $$ $$ dl = \sqrt{L^2 + x^2}\cdot \left(\sqrt{1 + \frac{2xdx}{L^2 + x^2}}-1\right) $$

We will use the formula for small quantities $(1+x)^\alpha\approx 1+\alpha x$, where $x\rightarrow 0$:

$$ dl = \frac{xdx}{\sqrt{L^2 + x^2}} $$

Given that $v_B = \frac{dl}{dt}$ and $v_A = \frac{dx}{dt}$

$$ v_B = \frac{x}{\sqrt{L^2 + x^2}} \frac{dx}{dt} $$ $$ v_B = v_A\frac{x}{\sqrt{L^2 + x^2}} $$

Substitute $(1):$

$$ \fbox{$v_B = \frac{v_A^2t}{\sqrt{L^2 + v_A^2t^2}}$} $$

NO: A more detailed and beautiful problem with a similar idea can be found in "Very Long Physics Problems" by A.I. Slobodyanyuk (Problem 2)

Answer:

$$v_B = \frac{v_A^2t}{\sqrt{L^2 + v_A^2t^2}}$$