M and radius R is held up by a massless string, as shown in Figure. The surface of the disk is frictionless. What is the normal force per unit length the string applies to the disk?
A disk of mass
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@Keshav Tiwari Done!
Thanks a lot. Got it !
Why exactly is T=Mg/2 ?? And thank you
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Balancing forces in vertical gives that.
How did you get the sum as 2 T sin 2 d θ ? One is T and the other is T sin 2 d θ , so how does the sum become the one you mentioned?
Also, can you please also explain in a little deeper that why will they not be equal?
The question says per unit arc length that must b N/(pi)R right?
the integral of Ncos(x)dx = dmg from 0 to pi. so the result,2N=Mg N=MG/2
Normal Force per unit=N/R so,MG/2R
I did not understand your solution. How is normal force per unit N/R ?
What about the tension? Can we neglect that?
Can you please post a better solution ? @Pranjal Jain
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Consider the normal force, N d θ , on a small arc of the disk that subtends and angle d θ . The tension forces on each end of the corresponding small piece of string almost cancel, but they don’t exactly, because they point in slightly different directions. Their non-zero sum is what produces the normal force on the disk. From figure, we see that the two forces have a sum of 2 T sin 2 d θ , directed inward. Since d θ is small, we can use sin x = x to approximate this as T d θ . Therefore, N d θ = T d θ , and so N = T . The normal force per unit arc length, R N , then equals R T . Using T = 2 M g , we arrive at R N = 2 R M g .