Question

A small coin of mass `80g` 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 cofficient of kinetic friction is `mu_(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

Correct Answer - D
A small …………..

The friction force 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 is
`= sqrt(N^(2)+f^(2))`
`= sqrt((mg)^(2)+mu_(s)^(2)(mg)^(2)) = mg sqrt(1+mu^(2))`
`= 40xx10^(-3)xx10xx sqrt(1+(9)/(16)) = 0.5 N`