A disc revolves with a speed of $$33 \dfrac{1}{3}$$ rpm 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 record is 0.15, which of the coins will revolve with the record?

Asked by Abhisek | 1 year ago |  190

##### Solution :-

Speed of revolution of the disc = 33 $$\dfrac{1}{3}$$ rpm

$$\dfrac{100}{3}$$ rpm= $$\dfrac{100}{3\times 60}$$ rps = $$\dfrac{5}{9}$$ rps

ω = 2πv = 2 x ($$\dfrac{22}{7}$$)x ($$\dfrac{5}{9}$$) = $$\dfrac{220}{63}$$ rad/s

The coins revolve with the disc, the centripetal force is provided by

the frictional force $$\dfrac{mv^2}{r ≤ μmg}$$ ——–(1)

As v = rω, equation (1) becomes $$\dfrac{ mrω^2}{r ≤ μmg}$$

r ≤ μg/ω2 = $$\dfrac{ (0.15 \times 10)}{( \dfrac{220}{63})^2}$$ = 12 cm

For coin A, r = 4 cm

The condition (r≤ 12) is satisfied for coin placed at 4cm, so coin A will revolve with the disc

For coin B, r = 14 cm

The condition (r≤ 12) is not satisfied for the coin placed at 14cm, so coin B will not revolve with the disc.

Answered by Pragya Singh | 1 year ago

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