Question

Coefficient of static friction between
a coin and a gramophone disc is 0.5.

Radius of the disc is 8 cm. Initially the

centre of the coin is 2 cm away from

the centre of the disc. At what minimum

frequency will it start slipping from there ? by what factor will answer Change if the coin is almost at the rim ( use g= π^2)​

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Answer 1

Answer :-

1)Minimum frequency with which the coin start slipping from the gramophone disc if coin is at 2cm away f1 = 2.5 cycles/ sec

2)factor of 1/2 i.e f2  = ½( f1  )

Explanation :-

Given data :-  

1)Coefficient of static friction μ = 0.5 .

2) Radius of disc = 8 cm = 8 x 10-2 m

3) Center of coin is 2 cm away from the center of disc.

What we have to find out:-

1)Minimum frequency with which the coin start slipping from the gramophone disc if coin is at 2cm away f1 = ?

2)By what factor will answer Change if the coin is almost at the rim

Solution

Since coin is at distance of r = 2cm from center on rotating disc so it experience a centrifugal force away from center which is balanced by force of friction. Hence we can write,

 centrifugal force = force of friction

mrω2  = μN          ----------- N = mg is normal reaction

mrω2  = μmg            

mrω2  = μmg                                          

rω2  = μg

 ω2  = μg / r

(2 π f1 )2 = (μg / r)

4 π2 f12 = ( 0.5 π2 / 0.02 )    ------- coin at distance r = 2cm = 0.02m  

 f12  = 25/4

f1  = 5/2 = 2.5 cycles / sec

2) if the coin is almost at the rim i.e now r = radius of disc = 8cm = 0.08 m

Using same formula,

 (2 π f2 )2 = (μ g / r)                                                          

4 π2 f22 = ( 0.5 π2 / 0.08 )  

f2 2 = 5/3.2 = 1.5625

f2  = 1.25 cycles / sec

Factor of change = f2 / f1

         = 1.25 / 2.5

         = 1/2  

 f2  = ½( f1  )