Learning Goal: apply principles of static friction and

circular motion to analyze the motion of a person on a

spinning carnival ride.

Macmillan Leaming

In a classic carnival ride, patrons stand against the wall

in a cylindrically shaped room. Once the room gets

spinning fast enough, the floor drops from the bottom of

the room! Friction between the walls of the room and

the people on the ride makes them "stick" to the wall so

they do not slide down.

In one ride, the radius of the cylindrical room is

R = 6.0 m and the room spins with a frequency of 23.5

revolutions per minute.

μs,min =

R

་

What is the minimum coefficient μs,min of friction needed

between the wall and the person?

To solve for the minimum coefficient of static friction (μₛ,min) needed to keep a person from sliding down the wall in a spinning cylindrical carnival ride, we need to apply principles of **circular motion** and **static friction**.

### Key Information:
- Radius of the room: \( R = 6.0 \, \text{m} \)
- Frequency of spinning: \( f = 23.5 \, \text{revolutions per minute (rpm)} \)
- Gravitational acceleration: \( g = 9.8 \, \text{m/s}^2 \)

### Approach:

1. **Convert frequency to angular velocity**:
   The angular velocity \( \omega \) in radians per second can be found from the frequency:
   \[
   \omega = 2\pi f
   \]
   where \( f \) is the frequency in revolutions per second (rps). First, convert rpm to rps:
   \[
   f = \frac{23.5 \, \text{rpm}}{60} = 0.3917 \, \text{rps}
   \]
   Now, calculate \( \omega \):
   \[
   \omega = 2\pi \times 0.3917 = 2.46 \, \text{rad/s}
   \]

2. **Calculate the centripetal force**:
   The centripetal force \( F_c \) acting on a person due to the spinning motion is:
   \[
   F_c = m \omega^2 R
   \]
   where \( m \) is the mass of the person. This force acts horizontally, pushing the person against the wall.

3. **Static friction force**:
   The static friction force \( F_f \) between the person and the wall keeps the person from sliding down. This frictional force must balance the weight of the person:
   \[
   F_f = mg
   \]
   where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.

4. **Friction condition**:
   The static friction is related to the normal force \( F_n \) (which is the centripetal force in this case) by:
   \[
   F_f = \mu_s F_n
   \]
   where \( F_n = F_c \). Therefore:
   \[
   mg = \mu_s m \omega^2 R
   \]
   Simplifying the equation:
   \[
   \mu_s = \frac{g}{\omega^2 R}
   \]

5. **Substitute known values**:
   Now, substitute the values for \( g \), \( \omega \), and \( R \):
   \[
   \mu_s = \frac{9.8}{(2.46)^2 \times 6.0}
   \]
   Calculate the denominator:
   \[
   (2.46)^2 \times 6.0 = 36.31
   \]
   Finally, solve for \( \mu_s \):
   \[
   \mu_s = \frac{9.8}{36.31} \approx 0.27
   \]

### Answer:
The minimum coefficient of static friction \( \mu_s \) needed is approximately **0.27**.

---

Would you like more details or have any questions on this? 

Here are 5 related questions to expand on this concept:
1. How would the minimum coefficient of friction change if the radius of the room were larger?
2. What would happen to the friction requirement if the ride spun faster?
3. How does the mass of the person affect the minimum friction coefficient?
4. What is the relationship between centripetal force and the radius of the spinning room?
5. How does static friction prevent slipping in scenarios involving circular motion?

**Tip**: Always ensure the angular velocity is correctly converted from rpm to radians per second when dealing with circular motion!

What is the minimum coefficient μs,min of friction needed between the wall and the person in a spinning cylindrical carnival ride?

Learn how to calculate the minimum coefficient of static friction for a person on a spinning carnival ride using principles of circular motion and static friction. The room has a radius of 6.0 m and spins at 23.5 rpm. Suitable for advanced physics students in Grades 11-12.

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