Q5.37 A disc revolves with a speed of 33\frac{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 coefficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record?

Answers (1)
D Devendra Khairwa

Frequency of revolution is : 

                                                 =\ \frac{100}{ 3\times60}\ =\ \frac{5}{9}\ rev/sec

(i) Case 1 :- When coin is placed at 4 cm :  

                                         Radius  =  0.04 m

                                    Angular frequency : 

                                                                    \omega\ =\ 2 \pi v

or                                                                       =\ 2 \pi \times \frac{5}{9}\ =\ 3.49\ s^{-1}

The frictional force is given by : 

                                                f\ =\ \mu mg 

 or                                                   =\ 0.15\times m\times 10\ =\ 1.5m\ N

Thus the centripetal force will be :

                                               F_c \ =\ mr\omega^2

or                                                    =\ m\times 0.04 (0.39)^2\ =\ 0.49m\ N

Since frictional force is greater than the centripetal force so coin will revolve around the record.

 

(ii) Case 2:-  When the coin is placed at 14 cm  :

                                           Radius  =  0.14  m

The centripetal force will be :

                                               F_c \ =\ mr\omega^2

or                                                    =\ m\times 0.14 (0.39)^2\ =\ 1.7m\ N

Since frictional force is lesser than the centripetal force so the coin will slip from the record.

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