Given that

Frequency of the disk f = 33 \frac{1}{3} rev/min = \frac{100}{3} rev/min

v = \frac{100}{3* 60} = \frac{5}{9} rev/sec

Radius of the disk R = 15 cm = 0.15 m

Distance between the first coin and centre r_1 = 4 cm =0.04 m

Distance between the second coin and centre r_2 = 14 cm =0.14 m

Co-efficient of the friction \mu = 0.15

Acceleration due to gravity g = 10 m/s^2

Step 1: Make a note of the forces acting on the first coin

Normal force is balanced by the weight of the coin

N = mg

Centrifugal force and centripetal force is balanced by the frictional force.

Centrifugal force = Centrifugal force F_c = \frac{mv^2}{r}

Where, m-mass of the coin, v- velocity of the coin and r -radius of the coin

Velocity v = \omega r

Centrifugal force F_c = \frac{m(\omega r)^2}{r} = mr \omega^2

Angular velocity \omega = 2 \pi f

\omega = 2 * 3.14 * \frac{5}{9}

\omega = 3.49

Step 2: Calculating the frictional and centrifugal force acting on the first coin F_c = mr \omega^2

Friction force of the first coin F_r= \mu N

F_r = \mu mg

F_r = 0.15 * m* 10

F_r = 1.5 m

Centrifugal force F_c = mr \omega^2

F_c = m * 0.04 * (3.49)^2

F_c = 0.49m N

Frictional force is greater than the centrifugal force. So first coin is revolve with record.

Step 3: Centrifugal forces acting on the second coin

Centrifugal force F_c = mr \omega^2

F_c = m * 0.14 * (3.49)^2

F_c = 1.7m N

The centrifugal force is stronger than the frictional force. As a result, the second coin will fall from the record's surface.