A disc revolves with a speed of 33 frac{1}{3} | vedclass

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(B) For a coin to revolve with the disc,the frictional force must be greater than or equal to the required centripetal force: $\mu mg \geq mr\omega^2$

Vedclass · Accepted Answer

Only the coin at $4 \; cm$

Vedclass · Answer

(B) For a coin to revolve with the disc,the frictional force must be greater than or equal to the required centripetal force: $\mu mg \geq mr\omega^2$,or $\mu g \geq r\omega^2$.Given:Frequency $
u = 33 \frac{1}{3} \; rev/min = \frac{100}{3 	imes 60} = \frac{5}{9} \; rev/s$.Angular velocity $\omega = 2\pi
u = 2 	imes \pi 	imes \frac{5}{9} = \frac{10\pi}{9} \approx 3.49 \; rad/s$.Maximum acceleration provided by friction $a_{max} = \mu g = 0.15 	imes 10 = 1.5 \; m/s^2$.For the coin at $r_1 = 4 \; cm = 0.04 \; m$:Required centripetal acceleration $a_1 = r_1\omega^2 = 0.04 	imes (3.49)^2 \approx 0.49 \; m/s^2$.Since $a_1 For the coin at $r_2 = 14 \; cm = 0.14 \; m$:Required centripetal acceleration $a_2 = r_2\omega^2 = 0.14 	imes (3.49)^2 \approx 1.70 \; m/s^2$.Since $a_2 > a_{max}$ $(1.70 > 1.5)$,the coin at $14 \; cm$ will slip.

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