A) Coin A will revolve but B will not revolve
B) B will revolve but A will not revolve
C) None of the coins will revolve
D) Both coins will revolve
Correct Answer: A
Solution :
[a]| Frequency of revolution \[n=33\frac{1}{3}=\frac{100}{3}\operatorname{rev}/min=\frac{100}{3\times 60\operatorname{rev}/min}\]\[=\frac{5}{9}rev/s\] |
| \[\therefore \] Angular velocity \[(\omega )=2\pi n\]\[=2\times \frac{22}{7}\times \frac{5}{9}=\frac{220}{63}\operatorname{rad}/s\] |
 |
| Radius of the disc \[(r)=15\text{ }\operatorname{cm}.\] |
| Distance of first coin A from the centre \[({{x}_{1}})=4cm\] |
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| Distance of the second coin B from the Centre \[({{x}_{1}})=\text{ }14cm\] |
| Coefficient of friction between the coins and the record\[~=0.15\] |
| If force of friction between the coin and the record is sufficient to provide the centripetal force, then coil will revolve with the record. |
| \[\therefore \]To prevent slipping (or to revolve the Coin along with record) the force of friction \[f\ge \]Centripetal force\[({{f}_{c}})\] |
| \[\Rightarrow \mu \operatorname{mg}\ge \operatorname{mr}{{\omega }^{2}}\operatorname{or}\,\mu g\ge r{{\omega }^{2}}\] |
| For first coin A \[r{{\omega }^{2}}=\frac{4}{100}\times {{\left( \frac{220}{63} \right)}^{2}}=\frac{4\times 220\times 220}{100\times 63\times 63}\]\[=0.488\operatorname{m}/{{s}^{2}}\] and \[\mu g=0.15\times 9.8=1.47m/{{s}^{2}}\] |
| Here, \[\mu g\ge r{{\omega }^{2}}\] therefore this coin will revolve with the record. |
| For second coin B \[r{{\omega }^{2}}==\frac{14}{100}\times {{\left( \frac{220}{63} \right)}^{2}}\]\[=\frac{14\times 220\times 220}{100\times 63\times 63}=1.707\operatorname{m}/s\] and \[\mu g=1.47m/s\] |
| Here, \[\mu g<r{{\omega }^{2}},\] therefore centripetal force will not be obtained from the force of friction, hence this coin will not revolve with the record. |
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