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 $\alpha =2{\text{ rad s}}^{-2}$ . 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 $r=1m$ 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 :

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

Answer 1

Answer: (D)

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
$=\sqrt{{\mathrm{N}}^2+{\mathrm{f}}^2}=\sqrt{(\mathrm{mg}{)}^2+{\left({\mu }_{\mathrm{s}}\mathrm{mg}\right)}^2}=\mathrm{mg}\sqrt{1+{\mu }_{\mathrm{s}}^2}$
$=40\times 10^{-3}\times 10\times \sqrt{1+\frac{9}{16}}=\text{0.5 N}$