0
$\begingroup$

I am training to solve the following physical system via lagrangian equations for equilibrium Points: double Atwood machine I have got the following boundary condition: $\frac{1}{2}(y_1+y_2) = l_2-y_3$ $\Rightarrow$ $y_3 = l_2 - \frac{1}{2} (y_1 - y_2)$. $y_1, y_2, y_3$ are the vertical positions of $m_1, m_2, m_3$, and $l_1$ and $l_2$ are the lengths of the ropes on the small and the big wheel. I get the following equations for the kinetic energy $$ T = \frac{1}{2} \left( m_1\dot{y_1} + m_2 \dot{y_2} + m_3 \dot{y_3} \right) = \frac{1}{2} \left( \left( m_1 + \frac{m_3}{4}\right)\dot{y_1} + \left(m_2 + \frac{m_3}{4} \right) \dot{y_2} - \frac{m_3}{2} \dot{y_1}\dot{y_2}\right) $$ and the potential Energy $$ V = mg(h-y_1) + m_2g(h-y_2) + m_3g(h-y_3) = \left( \frac{m_3}{2} - m_1\right)y_1g + \left( \frac{m_3}{2} - m_2\right) y_2g + const. $$

Which leads to the following 2 lagrangian equations: $$ (i) \quad \frac{d}{dt} \frac{\partial L}{\partial \dot{y_1}} - \frac{\partial L}{\partial y_1} = \left( m_1 + \frac{m_3}{4}\right)\ddot{y_1} - \frac{m_3}{4}\ddot{y_2} + \left(\frac{m_3}{2} - m_1\right) g = 0 $$ and $$ (ii) \quad \frac{d}{dt}\frac{\partial L}{\partial \dot{y_2}} - \frac{\partial L}{\partial y_2} = \left( m_2 + \frac{m_3}{4} \right) \ddot{y_2} - \frac{m_3}{4}\ddot{y_1} + \left( \frac{m_3}{2} - m_2 \right)g = 0. $$

Now for one of the equibilirium points i guessed the following condition: $m_1 = m_2 \Rightarrow$

$$ (i) - (ii) = m_1\ddot{y_1} - m_2\ddot{y_2} + \frac{m_3}{4} \left( \ddot{y_1} - \ddot{y_2}\right) + (m_2 - m_1) g = 0 $$ with $m_1 = m_2 = m$ $$ \Rightarrow m \ddot{y_1} - m \ddot{y_2} + \frac{m_3}{4} ( \ddot{y_1} - \ddot{y_2}) = 0\\ \Rightarrow m(\ddot{y_1} - \ddot{y_2}) = -\frac{m_3}{4} ( \ddot{y_1} - \ddot{y_2}) $$ $$ \Rightarrow m = -\frac{m_3}{4} $$ Where is my error? Please help me. D:

$\endgroup$
3
  • 1
    $\begingroup$ "equibilirium" $\to$ "equilibrium" $\endgroup$
    – Jean Marie
    Sep 10, 2017 at 3:55
  • $\begingroup$ write the lenghts and the vertical positions on the picture. $\endgroup$
    – alexjo
    Sep 10, 2017 at 14:12
  • $\begingroup$ Your error $y_3 = l_2 - \frac{1}{2} (y_1 -y_2)$ must be $y_3 = l_2 - \frac{1}{2} (y_1 +y_2)$ $\endgroup$
    – Eli
    Jan 18 at 7:48

1 Answer 1

0
$\begingroup$

Okay i think i just figured it out myself, and the answer is rather obvious. From the condition: $$ m(\ddot{y_1} - \ddot{y_2}) = -\frac{m_3}{4}(\ddot{y_1} - \ddot{y_2}) $$ it follows that $$ \ddot{y_1} - \ddot{y_2} = 0. $$ Which is the expected result, because this means, that $m_1$ doesn't experience any acceleration relative to $m_2$.

Am i right?

$\endgroup$
2
  • $\begingroup$ actually you must find $\ddot y_1=-\ddot y_2$. $\endgroup$
    – alexjo
    Sep 10, 2017 at 18:24
  • $\begingroup$ Hm, can you explain why. I thought this is the case for the equilibrium $m_1+m_2=m_3$? $\endgroup$
    – iqopi
    Sep 10, 2017 at 19:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.