The coin can only revolve with the record when the force of friction is enough to provide the centripetal force. If this force is not enough, then the coin slips on the record.
Here, R = mg, and the centripetal force is given by mrω2
To prevent slipping, the condition should be
μ mg ≥ mrω2
or, μ g ≥ rω2 ...(i)
For first coin:
r = 4 cm = 4/100 m
v = 331/3 rev/min = {100}/{3 x 60}rev/sec
ω = 2πv = 2π x {100}/{3 x 60} = 3.49 ms-1
rω2 = 4/100 x (3.49)2 = 0.49 ms-2
and μg = 0.15 x 9.8 = 1.47 ms-2
As μg > rω2, therefore this coin will revolve with the record
For 2nd coin:
r = 14 cm = 14/100 m
ω = 3.49 s-1
Hence, rω2 = 14/100 x (3.49)2 = 1.705 ms-2
and μg = 1.47 ms-2
As μg ≥ rω2 is not satisfied, so this coin will not rotate with the record.