A disc revovles with a speed of `33 (1)/(3) rev//min` and has a radius of 15 cm Two coins are palaced at 4 cm and 14 cm away from the center of the record If the coefficient of friction between the coins and the record is 0.5 which of the coins will revolve with the road ?

Options

A. A
B. B
C. Both A and B
D. Neither A nor B

Explanation

To solve the problem, we need to determine whether each coin will revolve with the disc based on the centripetal force and the frictional force acting on them.

### Step-by-Step Solution:

1. **Convert the speed of the disc to revolutions per second:**
The speed of the disc is given as \( 33 \frac{1}{3} \) revolutions per minute.
\[ \text{Revolutions per second} = \frac{33 \frac{1}{3}}{60} = \frac{100}{3 \times 60} = \frac{5}{9} \text{ rev/s} \]

2. **Calculate the angular velocity (\(\omega\)):**
The angular velocity in radians per second is given by:
\[ \omega = 2\pi \nu = 2\pi \left(\frac{5}{9}\right) = \frac{220}{63} \text{ rad/s} \]

3. **Determine the centripetal force condition:**
The centripetal force required for an object to revolve in a circle is given by:
\[ F_c = m \frac{v^2}{r} \]
where \(v = r \omega\). Thus, we can express the centripetal force as:
\[ F_c = m \omega^2 r \]

4. **Determine the frictional force:**
The maximum frictional force that can act on the coins is given by:
\[ F_f = \mu mg \]
where \(\mu\) is the coefficient of friction and \(g\) is the acceleration due to gravity (approximately \(10 \, \text{m/s}^2\)).

5. **Set up the inequality for each coin:**
For the coin to revolve with the disc, the centripetal force must be less than or equal to the frictional force:
\[ m \omega^2 r \leq \mu mg \]
This simplifies to:
\[ \omega^2 r \leq \mu g \]
or
\[ r \leq \frac{\mu g}{\omega^2} \]

6. **Calculate \(\frac{\mu g}{\omega^2}\):**
Substituting the values:
\[ \frac{\mu g}{\omega^2} = \frac{0.5 \times 10}{\left(\frac{220}{63}\right)^2} \]
First, calculate \(\left(\frac{220}{63}\right)^2\):
\[ \left(\frac{220}{63}\right)^2 = \frac{48400}{3969} \]
Then,
\[ \frac{\mu g}{\omega^2} = \frac{5}{\frac{48400}{3969}} = \frac{5 \times 3969}{48400} \approx 0.41 \text{ m} \approx 41 \text{ cm} \]

7. **Compare the distances of the coins from the center:**
- Coin A is at \(4 \, \text{cm}\) (which is less than \(41 \, \text{cm}\)).
- Coin B is at \(14 \, \text{cm}\) (which is also less than \(41 \, \text{cm}\)).

8. **Conclusion:**
Since both coins are within the limit of \(41 \, \text{cm}\), they will both revolve with the disc.