JEE Advanced Mechanics (Equilibrium) 2

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

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