JEE Advanced Mechanics (Equilibrium) 2

A disk of mass M M and radius R 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?

This problem is a part of My picks for JEE Advanced 1
M g 2 R \dfrac{Mg}{2R} M g π R \dfrac{Mg}{\pi R} M g 2 π R \dfrac{Mg}{2\pi R} M g R \dfrac{Mg}{R}

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2 solutions

Pranjal Jain
Apr 7, 2015

Consider the normal force, N d θ N~d\theta , on a small arc of the disk that subtends and angle d θ d\theta . 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 d θ 2 2T \sin \dfrac{d\theta}{2} , directed inward. Since d θ d\theta is small, we can use sin x = x \sin x=x to approximate this as T d θ T~d\theta . Therefore, N d θ = T d θ N d\theta = T d\theta , and so N = T N = T . The normal force per unit arc length, N R \dfrac{N}{R} , then equals T R \dfrac{T}{R} . Using T = M g 2 T = \dfrac{Mg}{2} , we arrive at N R = M g 2 R \dfrac{N}{R} =\dfrac{Mg}{2R} .

@Keshav Tiwari Done!

Pranjal Jain - 6 years, 2 months ago

Thanks a lot. Got it !

Keshav Tiwari - 6 years, 2 months ago

Why exactly is T=Mg/2 ?? And thank you

Tala Al Saleh - 5 years, 10 months ago

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Balancing forces in vertical gives that.

Pranjal Jain - 5 years, 10 months ago

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Yeah right, thanks!!

Tala Al Saleh - 5 years, 10 months ago

How did you get the sum as 2 T sin d θ 2 2T \sin \dfrac{d\theta}{2} ? One is T T and the other is T sin d θ 2 T \sin \dfrac{d\theta}{2} , 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?

Kushashwa Ravi Shrimali - 5 years, 5 months ago

The question says per unit arc length that must b N/(pi)R right?

Ankan Dutta - 5 years, 1 month ago

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 ?

Arpan Banerjee - 6 years, 3 months ago

What about the tension? Can we neglect that?

Hem C - 3 years, 1 month ago

Can you please post a better solution ? @Pranjal Jain

Keshav Tiwari - 6 years, 2 months ago

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