Movable pulley

Relevant wiki: Newton's Second Law

The problem has one string which goes over frictionless pulleys, therefore, the tension is same throughout the string. The two movable pulleys in the problems will have different accelerations. The relation between their accelerations can be found as follows. Let the smaller pulley moves down by a distance $x_1$ and the bigger pulley moves up by a distance $x_2$ .

The work done by the tension on smaller pulley will be, $- (2T-T) x_1$ and on the bigger pulley is $2T x_2$ . The net work done by the tension force has to be zero. Therefore, $2T x_2- (2T-T) x_1 = 0 \\ 2x_2 = x_1 \\$ Differentiating the equation twice with respect to time, we get. $a_1 = 2 a_2$

Now, writing the Newton's second law equations for smaller pulley, $mg +T -2T = m a_1$

Newton's second law for bigger pulley, $2T -mg = m a_2$

On solving the above equations we get, $a_1 = 4 \text{ m/s}^2$ and $a_2 = 2 \text{ m/s}^2$ . Their sum equals, $6 \text{ m/s}^2$ .

Relevant Wiki: Lagrangian formulation of mechanics

Let $x$ , $y$ be variables and $c$ be some constant, and also the length of rope be $l$ , which is also a constant.

$l=x+(x-c)+(y-c)+(y-x)=x+2y-2c$

$x=l+2c-2y$

$\dot{x}=-2\dot{y}----------(1)$

I only take straight line one but no semicircle because the length of semicircles remain same.

Compute $P.E.$

$P.E.=-mgx-mgy=-mgl-2mgc+2mgy-mgy=mgy-(2c+l)mg$

Compute $K.E.$

$K.E.=\frac{1}{2}m\dot{x}^2+\frac{1}{2}m\dot{y}^2=\frac{5}{2}m\dot{y}^2$

Compute $L$

$L=\frac{5}{2}m\dot{y}^2-mgy+(2c+l)mg$

Using Lagrangian Equation

$\frac{d}{dt}(\frac{\partial L}{\partial \dot{y}})=\frac{\partial L}{\partial y}$

$5m\ddot {y}=-mg$

$\ddot{y}=-\frac{1}{5}g=-2ms^{-2}$

By using $(1)$ ,

$\ddot{x}=-2\ddot{y}$

$\ddot{x}=\frac{2}{5}g=4ms^{-2}$

Sum of their magnitude be $2+4=\boxed{6}$