Question
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?
Answers
bobpursley
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.
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.
Give the answer completely i don't understand it