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Imagine that there is a cyllinder which can rotate about an axis $O$ without friction. There is also an inextendible, massless string attached to it (we assume that the string does not slip on the surface of the cyllinder). In the first case we pull the string with a constant force $F$ - in result, the cyllinder will rotate with a constant angular acceleration $\alpha$, because there is torque $\tau=rT$, where $T$ is tension in the string. In the second case, we attach an object of mass $m$ to the string, such that $mg=F$. The question is if angular acceleration $\alpha$ of the cyllinder will be the same in both situations?
It seems to me, the the answer in negative, since in the second case we have $ma=mg-T \Rightarrow T=m(g-a) \Rightarrow \tau=rm(g-a) < rF$. The only problem is, I don't know how to proof that tension in the string in the first case is equal to the force applied to it, e.g. $F=T$. In the second case, considering the above reasoning, tension MUST be less than the weight of the object.

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    $\begingroup$ Please provide a diagram. $\endgroup$
    – Gert
    Oct 27, 2019 at 14:06

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In the first case, since the string is assumed to be massless, you can apply Newton's law to it to see that forces on both ends of the string must be equal.

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  • $\begingroup$ Thanks a lot! Very neat - a non-zero force would cause infinite acceleration to the string! $\endgroup$ Oct 27, 2019 at 20:25

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