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  4. A small coin of mass 40 g is placed on the horizontal surface of a rotating disc. The disc starts from rest and is given a constant angular acceleration α =2 rad s- 2 . The coefficient of static friction between the coin and the disc is μ s = 3 / 4 and the coefficient of kinetic friction is μ k = 0.5 . The coin is placed at a distance r=1 m from the centre of the disc. The magnitude of the resultant force on the coin exerted by the disc just before it starts slipping on the disc is: <img class=img-fluid question-image alt=Question src=https://cdn.tardigrade.in/q/nta/p-sjxqjhbxaw60h043.jpg />

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Q. A small coin of mass $40\,g$ is placed on the horizontal surface of a rotating disc. The disc starts from rest and is given a constant angular acceleration $\alpha =2\,rad\,s^{- 2}$ . The coefficient of static friction between the coin and the disc is $\mu _{s} = 3 / 4$ and the coefficient of kinetic friction is $\mu _{k} = 0.5$ . The coin is placed at a distance $r=1\,m$ from the centre of the disc. The magnitude of the resultant force on the coin exerted by the disc just before it starts slipping on the disc is :
Question

NTA AbhyasNTA Abhyas 2022

Solution:

Solution
Solution
The friction force on coin just before coin is to slip will be: $f =\mu_{ s } mg$
Normal reaction on the coin; $N = mg$
The resultant reaction by disk to the coin is
$=\sqrt{ N ^{2}+ f ^{2}}=\sqrt{ mg ^{2}+\mu_{ s } mg ^{2}}= mg \sqrt{1+\mu_{ s }^{2}} $
$=40 \times 10^{-3} \times 10 \times \sqrt{1+\frac{9}{16}}=0.5\, N$