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 $\alpha =2rads^{- 2}$ . 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. $0.2 \, N$
• B. $0.3N$
• C. $0.4N$
• D. $0.5 \, N$

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
$= \sqrt{\left(\text{N}\right)^{2} + \left(\text{f}\right)^{2}} = \sqrt{\left(\text{mg}\right)^{2} + \left(\left(\mu \right)_{\text{s}} \text{mg}\right)^{2}} = \text{mg} \sqrt{1 + \mu _{\text{s}}^{2}}$
$= 4 0 \times 1 0^{- 3} \times 1 0 \times \sqrt{1 + \frac{9}{1 6}} = \text{0.5 N}$