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Question
A disc revolves with a speed of `33 1/3` rev/min, and has a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the co-efficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record?
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Solution 1
If the coin is to revolve with the record, then the force of friction must be enough to provide the necessary centripetal force
`:. mr omega^2 <= mu_s mg` or `r <= (mu_s mg)/(m omega^2)` or `r <= (mu_sg)/omega^2`
frequency = `33 1/2 "rpm"` = `100/3 "rpm" = 100/(3xx60) "rps"`
The Problems in which centripetal force is obtained fromforce of friction, start with the follwoing equation:
`m romega^2 <= mu_s mg`
`omega = 2pi xx 100/(3xx60)"rad s"^(-1) = 10/9 pi rad s^(-1)`
`(mu_sg)/omega^2 = (0.15xx10)/(10/9 pi)^2 m = 0.12 m = 12 cm`
The condition (r < = 12 cm) is satified by the coin placed at 4 cm from the centre of record. So the coin at 4 cm will revolve with the record.
Solution 2
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, v = `33 1/3` rev/min = `100/ (3xx60) = 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 = 2πν = `2 xx 22/7 xx 5/9 = 3.49 s^(-1)`
Frictional force, f = μmg = 0.15 × m × 10 = 1.5m N
Centripetal force on the coin:
`F_"cent" = mr^'omega^2`
`= m xx 0.04 xx (3.49)^2`
= 0.49 m N
Since f > Fcent, the coin will revolve along with the record.
Coin placed at 14 cm:
Radius, `r^n` = 14 cm = 0.14 m
Angular frequency, ω = 3.49 s–1
Frictional force, f' = 1.5m N
Centripetal force is given as:
`F_"cent" = mr^"` `omega^2`
= m × 0.14 × (3.49)2
= 1.7m N
Since f < Fcent., the coin will slip from the surface of the record.
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