5.37 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 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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Vishal Baghel | Contributor-Level 10

8 months ago

Speed of revolution of the disc, n = $33 \frac{1}{3}$ rev/min = 100/3 rpm = 100/ (3 $* 60) = 0.56 r p s$

Angular acceleration $ω$ = 2 $π n$ = 2 $* \frac{22}{7} * 0.56$ = 3.492 rad/s

The coins revolve with the disc. The centripetal force is provided by the frictional force $m v^{2} / r \leq μ m g$ …. (1)

As v = r $ω$ , the above equation becomes mr $ω^{2} / r \leq μ m g$

r $\leq μ g / ω^{2}$

r $\leq$ (0.15 $* 10) / {3.492}^{2}$ = 12 cm

For coin A, r = 4 cm

The con

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physics ncert solutions class 11th 2023

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5.37 A disc revolves with a speed of 3313 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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