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I'm solving a problem involving oberbeck pendulum

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We are given $M$, $m$, $r$ (radius of pulley), $R$ (distance from center of rotation to each of the 4 small weights). The problem says that the weight descends from a height $h$, reaches the lowest point and then begins to rise.

We are asked to find the tension $T$ during the time that the weight is changing directions.

This are the equations of motion for the system: $$I=4mR^2$$ $$I\frac{dw}{dt}=Tr$$ $$Ma=Mg-T$$ $$\frac{dw}{dt}=\frac{a}{r}$$

I managed to find acceleration $a_0$ and tension $T_0$ during the time that the weight is descending/ascending, but I I'm stuck at finding the tension during the time that it is changing directions. $$a_0=\frac{Mg}{\frac{I}{r^2}+M}$$ $$T_0=\frac{Mg}{1+\frac{r^2M}{I}}$$

I found a solution that says this: during the time of the "yank", the pulley makes a $180^\circ$ turn, not changing its angular velocity (given the high moment of intertia). So we get: $$(T-Mg)\Delta t=2Mv$$ $\Delta t = \frac{\pi r}{v}$, and $v$ is the velocity when it reaches the lowest point.

As I understand, the pendulum keeps rotating, since the torque created by this change in tension $\Delta T$ is small, taking into account the moment of intertia of the system. But why is it that the tension $T=T_0+\Delta T$ acts only during this $180^\circ$ rotationi and then it goes back to being $T_0$? It kind of makes sense at an intuitive level, but I don't know what physics law backs it up.

It's my first time posting here, thank you in advance for your help!

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