A disc revolve with a speed of 33.5 revolution per minute and has a radius of 15 cm .two coin are placed at 4 cm and 14 cm away from the center of disc. Coefficient of friction between coin and disc is 0.5 cm .which of the coin will revolve with the disc?

Answer 1

force friction on coin=mg*mu
centripetla force= m w^2 r

set them equal. Examle, the inner coin
so to stay, the mu can be found..
mg*mu=m w^2 r
mu=(33.5*2pi/60)^2*.04
mu=12.3*.04= .49 but the actual coefficient is greater than this, so that xoin stays on the disk.
Now do this for the outer coin.

Answer 2

Give the answer completely i don't understand it