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 coin exerted by the disc just before it starts slipping on the disc is:

Question

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 coin exerted by the disc just before it starts slipping on the disc is: <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-sjxqjhbxaw60h043.jpg" /> (A) 0.2 N (B) 0.3 N (C) 0.4 N (D) 0.5 N

Tardigrade Admin · Accepted Answer

The friction force on coin just before coin is to slip will be:

f = μ_{s} m g

Normal reaction on the coin;

N = m g

The resultant reaction by disk to the coin is

= N^{2} + f^{2} = m g^{2} + μ_{s} m g^{2} = m g 1 + μ_{s}^{2}

= 40 \times 1 0^{- 3} \times 10 \times 1 + \frac{9}{16} = 0.5 N

0.2 N

0.3 N

0.4 N

0.5 N

The friction force on coin just before coin is to slip will be: f=μs​mg Normal reaction on the coin; N=mg The resultant reaction by disk to the coin is =N2+f2​=mg2+μs​mg2​=mg1+μs2​​ =40×10−3×10×1+169​​=0.5N

The friction force on coin just before coin is to slip will be: $f = μ_{s} m g$
Normal reaction on the coin; $N = m g$
The resultant reaction by disk to the coin is
$= N^{2} + f^{2} = m g^{2} + μ_{s} m g^{2} = m g 1 + μ_{s}^{2}$
$= 40 \times 1 0^{- 3} \times 10 \times 1 + \frac{9}{16} = 0.5 N$