Energy: What will be the minimum ratio of m/M here

[PhysicsBoi1908]

My sir just replied and our answer was indeed correct! Thank you very much https://www.physicsforums.com/members/delta.189563/ and Charles Link for your help. I really appreciate how I was led to the answer instead of it being given out to me. Thank you very much!

 

[Delta2]

Well thanks too but I didn't too much @Charles Link did all the job here.

The differential equations of motion for this problem would be like

for body of mass m:
$$mg-2T(t)\sin\theta_1(t)=L\frac{d^2\tan\theta_1(t)}{dt^2}$$

for body of mass M
$$2T(t)\sin\theta_2(t)-Mg=L\frac{d^2\tan\theta_2(t)}{dt^2}$$

$\theta_1(t)$ is the angle made by the horizontal and the strings that are connected to body of mass m. $\theta_2(t)$ is the angle by the horizontal and the strings that are connected to body of mass M.
Need one more equation that relates $\theta_1$ and $\theta_2$ but I don't seem to find an obvious one. Also not sure if the tension T(t) is constant or time varying...

 

[Charles Link]

My sir just replied and our answer was indeed correct! Thank you very much https://www.physicsforums.com/members/delta.189563/ and Charles Link for your help. I really appreciate how I was led to the answer instead of it being given out to me. Thank you very much!

@PhysicsBoi1908 I'm very glad we got the right answer. And thank you for posting the question. It really is one of the better ones we have encountered. At first sight, it almost looked like it was a problem with a very trivial solution of $\frac{m}{M} \geq 1$, but it turned out to be very interesting.

 

[Charles Link]

Well thanks too but I didn't too much @Charles Link did all the job here.

The differential equations of motion for this problem would be like

for body of mass m:
$$mg-2T(t)\sin\theta_1(t)=L\frac{d^2\tan\theta_1(t)}{dt^2}$$

for body of mass M
$$2T(t)\sin\theta_2(t)-Mg=L\frac{d^2\tan\theta_2(t)}{dt^2}$$

$\theta_1(t)$ is the angle made by the horizontal and the strings that are connected to body of mass m. $\theta_2(t)$ is the angle by the horizontal and the strings that are connected to body of mass M.
Need one more equation that relates $\theta_1$ and $\theta_2$ but I don't seem to find an obvious one. Also not sure if the tension T(t) is constant or time varying...

Using energy equations, I think you can write an equation for the velocity of each mass as a function of y . The equation for each mass would have the form $\frac{dy}{dt}=f(y)$, which could be written as $\frac{dy}{f(y)}=dt$. Perhaps it even has a closed form solution. I haven't worked it yet to see. $\\$ Edit: Letting $y$ the the distance that $m$ moves, and $y_1$ the distance that $M$ moves, and $s$ the cable length from $M$ to the pulley, we have $s=3L-\sqrt{L^2+y^2}$, and $y_1=\sqrt{3}L-\sqrt{s^2-L^2}$. $\\$ Finally, $K.E.=\frac{1}{2}m(\frac{dy}{dt})^2+\frac{1}{2}M(\frac{dy_1}{dt})^2=mgy-Mgy_1$. Let me work it a couple more steps to see if I can get a single equation of $\frac{dy}{dt}=f(y)$. $\\$ Additional edit: Yes, $y_1$ can be expressed completely in terms of $y$, but it's not a simple expression. Thereby, we have $g(\frac{dy}{dt})=f(y)$, but $f(y)$ is not a simple function, and $g(\frac{dy}{dt})$ is also not simple. $\\$ Additional edit: If anything at all useful comes out of this last bit of algebra and calculus, perhaps it is the almost obvious result that if $\frac{dy}{dt}=0$, then $\frac{dy_1}{dt}=0$ as well.

 

[Delta2]

yes I knew from start that no matter how you decide to make the ODEs for this problem they would be non linear. @Charles Link the functions g and f you mention are not linear, they contain squares and square roots of their arguments if I understood correctly.