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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 text rad s- 2 . The coefficient of static friction between the coin and the disc is μ texts = 3 / 4 and the coefficient of kinetic friction is μ textk = 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-p4oajs1emur3.png />

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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 $\alpha =2\text{ rad s}^{- 2}$ . The coefficient of static friction between the coin and the disc is $\mu _{\text{s}} = 3 / 4$ and the coefficient of kinetic friction is $\mu _{\text{k}} = 0.5$ . 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 2020Laws of Motion

Solution:

Solution
Solution
The friction force on coin just before coin is to slip will be : $\text{f} = \mu _{\text{s}} \text{mg}$
Normal reaction on the coin ; N = mg
The resultant reaction by disk to the coin is
$=\sqrt{\mathrm{N}^2+\mathrm{f}^2}=\sqrt{(\mathrm{mg})^2+\left(\mu_{\mathrm{s}} \mathrm{mg}\right)^2}=\mathrm{mg} \sqrt{1+\mu_{\mathrm{s}}^2}$
$= 4 0 \times 1 0^{- 3} \times 1 0 \times \sqrt{1 + \frac{9}{1 6}} = \text{0.5 N}$