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I am told that a massless string is holding up a disk of mass $M$ and radius $R$. I want to find out the value of the tension $T$ in the string.

The textbook does this trivially by stating that $2T=Mg$, but I am not convinced. The way I see it, the only part of the string that should exert a force on the disk should be that part of the string that is in contact with it, and then the net upwards force should come by integrating the vertical component of tension around the disk.

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If I compute $Tsin\theta$, which is the vertical component of the tension, around the disk by:

$$2\int^{\pi/2}_{0}Tsin\theta \ \ d\theta$$ I obtain the desired answer of $2T$, so I can now say that $2T=Mg$. However, I cannot justify where the $d\theta$ comes from. Usually, in these sorts of problems we consider an infinitesimal section of the string, consider the equation of motion and then from that come up with an integral, but I am struggling to do this, and would appreciate some guidance.

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Try to sum up the vertical components of normal reactions exerted on the disk by all the infinitesimal elements of the string.

To do this, consider an element of the string subtending an angle of dtheta (infinitesimally small) at the centre of the disk. Since the element is massless, net force on it must be zero. So equate the normal reaction on the element to the component of tension acting towards the centre. $$ N = 2T\sin\left(\frac{d\theta}{2}\right) \approx Td\theta $$ By Newton's third law, this is also the normal reaction exerted by the element on the disk.

So, now take an element at angle theta with the vertical. It exerts normal reaction of $Td\theta$, whose vertical component is $Td\theta \cos(\theta)$. Integrate this from $-\pi/2$ to $+\pi/2$. You get $2T$, which is the net vertical normal reaction on the disk. Equate this to $Mg$ and you have your answer.

The textbook you have referred to is also perfectly correct. What they have done is, they have considered forces on the entire part of the string supporting the disk instead of an infinitesimal element. Thus there is a net force of $2T$ upward on this part of the string ($T$ at each end) which must balance the normal reaction of the disk on the string. And the normal reaction of the string on the disk, in turn, balances it's weight $Mg$. Now since both exert equal and opposite forces on each other, $2T = Mg$.

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Think about it ..... Tension is an internal force of the string and hence it should not appear in the Free Body Diagram of disc. Moreover there is no friction anywhere and hence the only interaction between the string and the disc should arise from NORMAL forces only.

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  • $\begingroup$ In that case, you will be integrating the vertical component of the normal force as opposed to that of the tension. Either way, where would the $d\theta$ come from in the infinitesimal equation? $\endgroup$ – Pancake_Senpai Mar 13 at 13:45
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The text book probably solves for the tension by drawing a free body diagram for the disc, as shown in Fig A below. That will give us $T+T-Mg=0$.

But in my opinion there is nothing wrong with your approach either since it sums up all the upward normal contact force contributions of the string on the frictionless disc, as I attempted to show in Fig B, and which would need to add up to 2T.

I don’t understand why you would question the justification for $dθ$. It is needed to account for all the differential contributions of the vertical tension force.

Hope this helps enter image description here

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