__Question 5.37__:

__Solution__:
Coin
placed at 4 cm from the centre

Mass
of each coin = *m*

Radius of the disc,

*r*= 15 cm = 0.15 m

Frequency of revolution,

*ν*= 100 / 3 rev/min = 100 / (3 X 60) = 5 / 9 rev/s

Coefficient of friction,

*μ*= 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*' = 4 cm = 0.04 m

Angular frequency,

*ω*= 2π

*ν = 2 X (22/7) X (5/9) = 3.49 s*

^{-1}Frictional force,

*f*=

*μ*

*m*g = 0.15 ×

*m*× 10 = 1.5

*m*N

Centripetal force on the coin:

*F*

_{cent.}= mr'ω

^{2}

= m X 0.04 X (3.49)

^{2}

= 0.49m N

Since

*f >*F

_{cent}, the coin will revolve along with the record.

__Coin placed at 14 cm:__

Radius, r" = 14 cm = 0.14 m

Angular frequency,

*ω*= 3.49 s

^{–1}

Frictional force,

*f*' = 1.5

*m*N

Centripetal force is given as:

*F*

_{cent.}= mr"ω

^{2}

=

*m*× 0.14 × (3.49)

^{2}

= 1.7

*m*N

Since

*f*<

*F*

_{cent.}, the coin will slip from the surface of the record.

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