Question

A playing record of radius 15.0 cm revolves with 100/3 rev/min. Two
coins are placed at 5 cm and 10 cm away from the centre of record. If
the coefficient of friction between the coin and record is 0.15, so that
coins revolve with the record. Find the position of coins.​

Answer 1

Answer:

A coin placed at 4 cm from the center

Mass of each coin =m

Radius of the disc,r=15cm=0.15m

Frequency of revolution, =100/3rev/min=100/(3x60)=5/9rev/s

Coefficient of friction, =0.15

In the given situation, the coin having a force of friction greater than or equal to the centripetal force provided by the rotation of the disc will revolve with the disc. If this is not the case, then the coin will slip from the disc.Coin placed at 4 cm:Radius of revolution, r

′

=4cm=0.04m

Angular frequency, =2πN=2×(22/7)×(5/9)=3.49rad/s

Frictional force, f=μmg=0.15m×10=1.5mN

Centripetal force on the coin:F

cent

.=mr

′

ω

2

=m×0.04×(3.49)

2

=0.49mN

Since f>F

cent

, the coin will revolve along with the record.

Coin placed at 14 cm:

Radius, r"=14cm=0.14m

Angular frequency, =3.49s

−1

Frictional force, f

′

=1.5mN

Centripetal force is given as:

F

cent

=mr"ω

2

=m×0.14×(3.49)

2

=1.7mN

Since f<F

cent

., the coin will slip from the surface of the record.