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(D) The centripetal force required for the coin to rotate with the disc is provided by the static frictional force.For the coin to remain at rest rela

Vedclass · Accepted Answer

$0.6$

Vedclass · Answer

(D) The centripetal force required for the coin to rotate with the disc is provided by the static frictional force.For the coin to remain at rest relative to the disc,the frictional force must be equal to or greater than the required centripetal force: $f = mr\omega^2$.The maximum static friction is given by $f_{max} = \mu mg$.Equating the two,we get $\mu mg = mr\omega^2$,which simplifies to $\mu = \frac{r\omega^2}{g}$.Given:Frequency $n = 3.5\,rev/s$.Angular velocity $\omega = 2\pi n = 2 	imes \pi 	imes 3.5 = 7\pi\,rad/s$.Radius $r = 1.25\,cm = 1.25 	imes 10^{-2}\,m$.Acceleration due to gravity $g = 10\,m/s^2$.Substituting the values:$\mu = \frac{(1.25 	imes 10^{-2}) 	imes (7\pi)^2}{10}$.Using $\pi \approx \frac{22}{7}$,we have $(7\pi)^2 = (7 	imes \frac{22}{7})^2 = 22^2 = 484$.$\mu = \frac{1.25 	imes 10^{-2} 	imes 484}{10} = \frac{1.25 	imes 4.84}{10} = 0.605 \approx 0.6$.

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