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:

Solution: Option B is the right answer. Force is equal to us x m x g which is equal to ¾ x 0.04 x 10 = 0.3 N.

The force must always be equal to the static force between the coin and the rotating disc irrespective of any other factors

Answer 2

Answer:

0.5N

Explanation:

just before sliping

friction (k) = umg (u=mu)