Q. A gramophone disc is set revolving in a horizontal plane and reaches a steady state of motion of two revolutions per second. It is found that a small coin placed on the disc will remain there if its centre is not more than 5 cm from the axis of rotation; the coefficient of friction between the coin and the disc is
A) 0.2
B) 0.4
C) 0.6
D) 0.8
Explanation
(d) Here, $n = 2\ $rps, $r = 5\ cm = 0.05\ m,\ \mu = ?$
As $\frac{mv^{2}}{r} = F = \mu\ R = \mu\ mg$
$\mu = \frac{v^{2}}{rg} = \frac{(r2\pi n)^{2}}{rg} = \frac{4\pi^{2}n^{2}r}{g}$
$= 4 \times \frac{22}{7} \times \frac{22}{7} \times \frac{2^{2} \times 0.05}{9.8} = 0.8$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.