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 a=2 rad/s2.the cofficient of static friction between the coin and the disc is us=3/4 and cofficientof kinetic friction is uk=0.5 the coin is placed at a distance r=1 from the centre of the disc. the magnitude of the resutanat 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 with explanation

Answer 1

Answer:

b ) 0.3

Explanation:

F = us * m * g

  = 3/4 * 0.04 *10

  = 0.3

Answer 2

Explanation:

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

Normal reaction on the coin ; N=mg

The resultant reaction by disk to the coin:

F=N2+f2=(mg)2+μs2(mg)2 =mg1+μ2 =40×10−3×10×1+169=0.5 N