Consider the double Atwood machine in the image
Using conservation of string, how would one prove mathematically that $a_1$ = $-\frac{a_2 + a_3}{2}$?
There is a video trying to prove this by taking the derivative of the string's length two times to get the acceleration relationship.
It is as follows:
$l_2 = x_2 – l$
$l_3 = x_3 – l$
$l_2 + l_3 = C$ because the string is conserved
$(x_2 – l) + (x_3 – l) = C$
taking the derivative two times we get:
$(a_2 – a_p) + (a_3 – a_p) = 0$
where $a_p$ is the acceleration of the bottom pulley
and because $a_p$ = $-a_1$
$a_2 + a_3 + 2 a_1 = 0$
$a_1$ = $-\frac{a_2 + a_3}{2}$
But isn't this wrong? Doesn't taking the derivative of $(x_2 – l) + (x_3 – l) = C$
two times result in $((a_2 + a_p) – a_p) + ((a_3 + a_p) – a_p) = 0$ since $x_2$ and $x_3$ both depends on $l$ and $l_2$ & $l_3$ respectively?
This just gives the not so interesting result $a_2 = -a_3$.
Am I missing something here? Any help would be greatly appreciated!
Best Answer
How to obtain the kinematic equations.
I use the Instant center of rotation ICO to obtain the velocities equations.
$$\omega\,L=v_2\\ \omega\,(L+R)=v_1\\ \omega\,(L+2\,R)=v_3$$
thus $$v_2=\frac{L}{L+R}\,v_1\\ v_3=\frac{L+2\,R}{L+R}\,v_1\quad\Rightarrow\\ v_1=\frac{v_2+v_3}{2}$$