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Q5.37:
A disc revolves with a speed of $33\frac{1}{3}$ rev/min, and has a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the co-efficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record ?

Solution :

Step 1: Conversion of Physical Quantities to SI Units

First, we convert the given parameters into Standard International (SI) units to ensure dimensional consistency [Per the SI system of measurement].

  • Angular velocity ($\omega$): The frequency $f = 33\frac{1}{3} \text{ rev/min} = \frac{100}{3 \times 60} \text{ rev/s} = \frac{10}{18} \text{ rev/s} = \frac{5}{9} \text{ Hz}$.
  • Calculation of $\omega$: $\omega = 2\pi f = 2 \times \pi \times \frac{5}{9} = \frac{10\pi}{9} \text{ rad/s}$.
  • Radius of disc ($R$) = $0.15 \text{ m}$.
  • Coefficient of static friction ($\mu_s$) = $0.15$.
  • Acceleration due to gravity ($g$) $\approx 9.8 \text{ m/s}^2$.

Step 2: Derivation of the Condition for Equilibrium

For a coin to revolve with the disc without slipping, the centripetal force required for circular motion must be provided by the static frictional force [By Newton’s Second Law of Motion].

$F_{centripetal} \le F_{friction(max)}$

$m r \omega^2 \le \mu_s m g$

Simplifying for the critical radius ($r_c$):

$r \le \frac{\mu_s g}{\omega^2}$

Step 3: Calculating the Maximum Possible Radius ($r_c$)

Using the derived formula to find the boundary of the safe zone for placement:

$r_c = \frac{0.15 \times 9.8}{(\frac{10\pi}{9})^2}$

$r_c = \frac{1.47}{(\frac{100 \times 9.8696}{81})} = \frac{1.47}{12.1847} \approx 0.1206 \text{ m}$

$r_c \approx 12.06 \text{ cm}$

Step 4: Comparative Analysis of Coin Positions

Coin Distance from Center ($r$) Condition ($r \le 12.06 \text{ cm}$) Status
Coin A $4 \text{ cm}$ $4 < 12.06$ Revolves
Coin B $14 \text{ cm}$ $14 > 12.06$ Slips

Visual Representation of Forces

Center Coin A (4cm) Coin B (14cm) Static Friction

Final Solution: Since the maximum radius for which friction can provide sufficient centripetal force is approximately 12.06 cm, the coin placed at 4 cm will revolve with the record, whereas the coin placed at 14 cm will slip.


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