Question

A small coin of mass $40g$ is placed on the horizontal surface of a rotating disc. The disc starts from rest and is given a constant angular acceleration $\alpha =2rad/s^2.$ The coefficient of static friction between the coin and the disc is ${\mu }_s=3/4$ and coefficient of kinetic friction is ${\mu }_k=\mathrm{0.5.}$ 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: (Take $g=10m/s^2$)

• A. $0.2N$
• B. $0.3N$
• C. $0.4N$
• D. $0.5N$

Answer 1

Answer: (D)

$0.5N$

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:
$F=\sqrt{N^2+f^2}=\sqrt{{\left(mg\right)}^2+{\mu }_s^2{\left(mg\right)}^2}$ $=mg\sqrt{1+{\mu }^2}$ $=40\times {10}^{-3}\times 10\times \sqrt{1+\frac{9}{16}}=0.5N$