NTA Abhyas JEE Main2020PhysicsLaws of MotionPractice
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
Options
- A0 . 2 N
- B0 . 3   N
- C0 . 4   N
- D0 . 5 N
Correct answer
D. 0 . 5 N
Step-by-step solution
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 = N 2 + f 2 = mg 2 + μ s mg 2 = mg 1 + μ s 2 = 4 0 × 1 0 - 3 × 1 0 × 1 + 9 1 6 = 0.5 N