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Consider the following system in equilibrium. The disk has radius $R$ and the rod has length $2R$. The rod is attached to the disk and the block is attached in place to the rod (it does not move radially).

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The forces present are the tension T on the string and the 3 gravity forces (on the block, rod and disk).

The condition that the net torque vanishes results in

$$TR=(m_{block}2R+m_{rod}R)g\sin\theta_{0}$$

First question: What is the result of requiring all forces to vanish? It seems that there are 3 vertical gravity forces and one horizontal force only (tension T) so how can those forces cancel each other?

Second question: How is the linear velocity of the block related to the angular acceleration of the disk once we cut the string?

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    $\begingroup$ That's a very good question, about how horizontal and vertical forces are balancing. But what is being balanced is not the forces, no no. Those horizontal and vertical forces are balanced by their respective constraint forces too (for example, the rod wants to go down so there is a force on the disk due to the rod, the disk pushes back just like the ground pushes us with reaction force). What is being balanced is the torque, and they are all in same direction, perpendicular to the forces (cross product) in the negative and positive directions. So the equation is balance of torque, not forces. $\endgroup$ – Kalpak Gupta Jan 11 '17 at 4:33
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  1. All forces must cancel each other out because the system is stationary. What you are missing is the reaction forces on the pin at the center of the disk. The horizontal force at the pin must counteract the tension $T$.
  2. Since the block is attached to the rod and the rod to the disk, they all share the same angular velocity and acceleration. The angular acceleration experienced is purely a result of the net torque applied, and it does not depend on the velocity conditions of the disk/rod system. That is because all centrifugal forces do not apply torque in this case. That being said, the fact is that the combined center of mass is not at the disk center, and thus part of the reaction forces from the pin go towards accelerating the center of mass.
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