A long playing record revolves with a speed of `33 (1/4)` 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 coefficients of friction between the coin and the record is 0.15, which of the two coins will revolve with record ?

Explanation

To solve the problem step by step, we will follow the process of calculating the angular velocity, the outward centripetal force acting on each coin, and then comparing it with the maximum static friction force to determine which coin will revolve with the record.

### Step 1: Convert revolutions per minute to radians per second
Given the speed of the record is \( 33 \frac{1}{4} \) revolutions per minute (rev/min), we first convert this to radians per second.

\[ \text{Speed in rev/min} = 33.25 \text{ rev/min} \]

To convert to radians per second, we use the formula:
\[ \omega = \text{rev/min} \times \frac{2\pi \text{ radians}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ seconds}} \]

Calculating:
\[ \omega = 33.25 \times \frac{2\pi}{60} \approx 3.48 \text{ radians/second} \]

### Step 2: Calculate the outward centripetal force for each coin
The outward centripetal force (\( F_{\text{out}} \)) on an object in circular motion is given by:
\[ F_{\text{out}} = m \omega^2 r \]

Where:
- \( m \) is the mass of the coin (which will cancel out later),
- \( \omega \) is the angular velocity,
- \( r \) is the radius from the center of the record.

#### For Coin 1 (at 4 cm from the center):
Convert radius to meters:
\[ r_1 = 4 \text{ cm} = 0.04 \text{ m} \]
Now calculate \( F_{\text{out1}} \):
\[ F_{\text{out1}} = m (3.48)^2 (0.04) \approx m \times 0.48 \text{ N} \]

#### For Coin 2 (at 14 cm from the center):
Convert radius to meters:
\[ r_2 = 14 \text{ cm} = 0.14 \text{ m} \]
Now calculate \( F_{\text{out2}} \):
\[ F_{\text{out2}} = m (3.48)^2 (0.14) \approx m \times 1.69 \text{ N} \]

### Step 3: Calculate the maximum static friction force
The maximum static friction force (\( F_{\text{max}} \)) that can act on the coins is given by:
\[ F_{\text{max}} = \mu_s m g \]
Where:
- \( \mu_s = 0.15 \) (coefficient of friction),
- \( g = 10 \text{ m/s}^2 \) (acceleration due to gravity).

Calculating:
\[ F_{\text{max}} = 0.15 \times m \times 10 = 1.5 m \text{ N} \]

### Step 4: Compare the forces
Now we compare \( F_{\text{out}} \) for each coin with \( F_{\text{max}} \):

- For Coin 1:
\[ F_{\text{out1}} = 0.48 m \text{ N} < 1.5 m \text{ N} \quad \text{(Coin 1 will revolve with the record)} \]

- For Coin 2:
\[ F_{\text{out2}} = 1.69 m \text{ N} > 1.5 m \text{ N} \quad \text{(Coin 2 will slip)} \]

### Conclusion
Coin 1 (at 4 cm) will revolve with the record, while Coin 2 (at 14 cm) will slip.

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