$15\, cm$ त्रिज्या का कोई बड़ा ग्रामोफोन रिकार्ड $33 \frac{1}{3} rev / min$ की चाल से घूर्णन कर रहा है। रिकार्ड पर उसके केंद्र से $4 \,cm$ तथा $14\, cm$ की दूरियों पर दो सिक्के रखे गए हैं। यदि सिक्के तथा रिकार्ड के बीच घर्षण गुणांक $0.15$ है तो कौन सा सिक्का रिकार्ड के साथ परिक्रमा करेगा ?

Coin placed at $4\, cm$ from the centre

Mass of each coin $=m$

Radius of the disc, $r=15 \,cm =0.15\, m$

Frequency of revolution, $v=33 \frac{1}{3}$ rev/min $=\frac{100}{3 \times 60}=\frac{5}{9}$ $rev/s$

Coefficient of friction, $\mu=0.15$

In the given situation, the coin having a force of friction greater than or equal to the centripetal force provided by the rotation of the disc will revolve with the disc. If this is not the case, then the coin will slip from the disc.

Coin placed at $4 \,cm$ :

Radius of revolution, $r^{\prime}=4 \,cm =0.04 \,m$

Angular frequency, $\omega=2 \pi v^{ }=2 \times \frac{22}{7} \times \frac{5}{9}=3.49 \,s ^{-1}$

Frictional force, $f=\mu \,m g=0.15 \times m \times 10=1.5\, m\, N$

Centripetal force on the coin:

$F_{\text {cent }}=m r^{\prime} \omega^{2}$

$=m \times 0.04 \times(3.49)^{2}$

$=0.49 \,m\, N$

since $f\,>\, F _{\text {cent }},$ the coin will revolve along with the record.

coin placed at $14\, cm$ :

Radius, $r^{\prime \prime}=14 \,cm =0.14 \,m$

Angular frequency, $\omega=3.49 \,s ^{-1}$

Frictional force, $f^{\prime}=1.5 \,m\, N$

Centripetal force is given as:

$F_{\text {cent. }}=m r'' \omega^{2}$

$=m \times 0.14 \times(3.49)^{2}$

$=1.7 \,m \,N$

since $f\, \,< F$ cent, the coin will slip from the surface of the record.