Question

Question 5.37 A disc revolves with a speed of 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?

Chapter Laws Of Motion Page 113

Answer 1

angular frequency = 331/3 rev/min
                                = 100/3*60 rev/sec
                                =5/9 rev/sec 
angular velocity = 2π*angular frequency 
                            =2π*5/9 rad/sec
radius of disc = 15 cm 
distance of 1st coin from the center  = 4 cm
distance of the 2nd coin from the center = 14 cm 
coefficient of friction between coin and record = 0.15 

TO prevent slipping the force of friction≥centripetal force
          μmg ≥ mω²r
           μg≥rω²

FOR 1st COIN A
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                  here ,
            r = distance of coin A from the center = 4 cm 
            ω = 10π/9 rad/sec 
            rω² = 4/100 * (10π/9)² 
                   =0.488 m/s²
and    μg  = 0.15 * 9.8 = 1.47 m/s²
here μg >rω² hence coin A will revolve with record .
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FOR 2nd coin B
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        rω² = 14/100 * (10π/9)²
               =1.707 m/s²
        μg = 0.15*9.8 = 1.47 m/s²
here μg <rω² therefore centripetal force won't be obtained from the force of friction,hence this coin won't revolve with the record