A rotating carnival ride is 8.0 m in diameter.
The coefficient of static friction between the
riders and the outer walls is μs = 0.24. What is

the minimum speed with which a 75 kg passen-
ger must be rotating to be supported only by

friction against the wall?

We can solve this problem using the concept of centripetal force and the role of friction in preventing the passenger from falling.

### Key concepts:
- The ride rotates, creating a centripetal force that pushes the rider outward against the wall.
- The friction between the rider and the wall is what prevents the rider from sliding down.
- The frictional force must equal the gravitational force to support the rider.

### Step-by-step approach:

1. **Known values**:
   - Diameter of the ride, \( D = 8.0 \, \text{m} \), so the radius, \( R = \frac{D}{2} = 4.0 \, \text{m} \).
   - Coefficient of static friction, \( \mu_s = 0.24 \).
   - Mass of the passenger, \( m = 75 \, \text{kg} \).
   - Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \).

2. **Forces acting on the rider**:
   - The gravitational force pulling the rider down: \( F_{\text{gravity}} = mg \).
   - The frictional force keeping the rider from sliding down: \( F_{\text{friction}} = \mu_s F_{\text{normal}} \), where \( F_{\text{normal}} \) is the normal force exerted by the wall (equal to the centripetal force in this case).
   - The centripetal force is given by \( F_{\text{centripetal}} = \frac{mv^2}{R} \), where \( v \) is the speed of the rider and \( R \) is the radius of the ride.

3. **Equilibrium condition**:
   For the rider to stay supported by friction, the frictional force must be equal to the gravitational force:
   \[
   F_{\text{friction}} = F_{\text{gravity}}
   \]
   \[
   \mu_s F_{\text{normal}} = mg
   \]
   Since \( F_{\text{normal}} = F_{\text{centripetal}} \), we substitute:
   \[
   \mu_s \left( \frac{mv^2}{R} \right) = mg
   \]
   Simplify by canceling out the mass \( m \) from both sides:
   \[
   \mu_s \frac{v^2}{R} = g
   \]
   Solving for \( v^2 \):
   \[
   v^2 = \frac{gR}{\mu_s}
   \]
   Now, take the square root to find \( v \):
   \[
   v = \sqrt{\frac{gR}{\mu_s}}
   \]

4. **Substitute the known values**:
   \[
   v = \sqrt{\frac{(9.8 \, \text{m/s}^2)(4.0 \, \text{m})}{0.24}}
   \]
   \[
   v = \sqrt{\frac{39.2}{0.24}} = \sqrt{163.33} \approx 12.78 \, \text{m/s}
   \]

### Final answer:
The minimum speed with which the passenger must be rotating to be supported only by friction is approximately **12.78 m/s**.

Would you like more details or have any questions?

Here are 5 related questions:
1. How does increasing the coefficient of friction affect the required speed?
2. What if the radius of the ride was halved—how would that affect the speed?
3. How would the problem change if the passenger's mass was different?
4. What role does centripetal acceleration play in this scenario?
5. How could you calculate the force the wall exerts on the passenger?

**Tip**: Friction depends on the normal force, which in circular motion is provided by the centripetal force.

A rotating carnival ride is 8.0 m in diameter. The coefficient of static friction between the riders and the outer walls is μs = 0.24. What is the minimum speed with which a 75 kg passenger must be rotating to be supported only by friction against the wall?

Learn how to calculate the minimum speed for a 75 kg passenger to stay supported by friction in a rotating carnival ride with an 8.0 m diameter and a static friction coefficient of μs = 0.24. This physics problem explores centripetal force, circular motion, and friction. Ideal for Grades 11-12.

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