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 fri

Question

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 :

TestHub · Accepted Answer

The friction force on coin just before coin is to slip will be : $f = μ_{s} mg$
Normal reaction on the coin ; N = mg
The resultant reaction by disk to the coin is

$= \sqrt{N^{2} + f^{2}} = \sqrt{{(mg)}^{2} + {(μ_{s} mg)}^{2}} = mg \sqrt{1 + μ_{s}^{2}}$

$= 40 \times 1 0^{- 3} \times 10 \times \sqrt{1 + \frac{9}{16}} = 0.5 N$

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