Problem 115 - University Physics with Modern Physics

Question

On the ride "Spindletop" at the amusement park Six Flags Over Texas, people stood against the inner wall of a hollow vertical cylinder with radius 2.5 \(\mathrm{m} .\) The cylinder started to rotate, and when it reached a constant rotation rate of 0.60 rev/s, the floor on which people were standing dropped about 0.5 \(\mathrm{m}\) . The people remained pinned against the wall. (a) Draw a force diagram for a person on this ride, after the floor has dropped. (b) What minimum coefficient of static friction is required if the person on the ride is not to slide downward to the new position of the floor? (c) Does your answer in part (b) depend on the mass of the passenger? (Note: When the ride is over, the cylinder is slowly brought to rest. As it slows down, people slide down the walls to the floor.)

Step-by-Step Solution

Answer

(b) Minimum coefficient of static friction is 0.276. (c) It is independent of the passenger's mass.

Step 1: Understanding Forces Acting on the Person

After the floor drops, the person is subject to three main forces: the gravitational force (weight), the normal force exerted by the wall, and the frictional force. The gravitational force acts downward with magnitude \( F_g = mg \), where \( m \) is the mass of the person and \( g \) is the acceleration due to gravity. The normal force \( F_n \) acts radially outward from the center of the cylinder. The frictional force \( F_f \) acts upward, opposing the gravitational force, thus preventing the person from sliding down.

Step 2: Drawing the Force Diagram

In the force diagram:- Draw a vertical downward arrow representing gravitational force \( F_g = mg \).- Draw a horizontal arrow pointing outward from the person's center towards the cylinder's wall representing the normal force \( F_n \).- Draw a vertical upward arrow representing frictional force \( F_f \).

Step 3: Calculating the Required Frictional Force

For the person to not slide downward, the static frictional force \( F_f \) must equal the gravitational force \( mg \). Thus, \( F_f = mg \). The static frictional force is also given by \( F_f = \mu_s F_n \), where \( \mu_s \) is the coefficient of static friction and \( F_n = m \cdot a_c \) with \( a_c \) being the centripetal acceleration.

Step 4: Determining the Centripetal Acceleration

The centripetal acceleration \( a_c \) is given by \( a_c = r \cdot \omega^2 \), where \( r = 2.5 \) m is the radius, and \( \omega = 0.60 \times 2\pi \) rad/s (converting revolutions per second to radians per second). Calculating \( \omega \), we have \( \omega = 3.77 \) rad/s.

Step 5: Calculating the Normal Force

Using the formula for centripetal acceleration \( a_c = r \cdot \omega^2 \), we find \( a_c = 2.5 \times (3.77)^2 = 35.5 \) m/s². Then, the normal force \( F_n = m \cdot a_c = m \cdot 35.5 \) N.

Step 6: Finding the Minimum Coefficient of Static Friction

Equate the frictional force to gravitational force: \( \mu_s F_n = mg \). Substituting for \( F_n \), we get \( \mu_s (m \cdot 35.5) = mg \). Solving for \( \mu_s \), we have \( \mu_s = \frac{g}{35.5} \approx \frac{9.8}{35.5} \approx 0.276 \).

Step 7: Analyzing Dependency on Mass

The minimum coefficient of static friction \( \mu_s \) is \( \frac{g}{a_c} \) and does not include mass \( m \) in its final form after cancellations. Thus, the value of \( \mu_s \) is independent of the mass of the passengers.

Key Concepts

Understanding the Static Friction Coefficient

When you are on a ride like the "Spindletop," static friction is what prevents you from sliding down the wall once the floor drops away. Think of static friction as the glue that keeps you pinned in place. It acts between your back and the wall of the spinning cylinder.

The static friction coefficient, denoted as \( \mu_s \), is a measure of this 'glue's' strength. In this context, \( \mu_s \) determines how well the wall can hold you up against gravity's pull. Calculations show:

• Static frictional force \( F_f = \mu_s F_n \), where \( F_n \) is the normal force.
• For friction to just counteract gravity, \( F_f \) must equal your weight, \( mg \).

To prevent sliding, the static friction needs to satisfy \( \mu_s \geq \frac{g}{a_c} \). Here, \( g \) is gravitational acceleration (9.8 m/s²) and \( a_c \) is centripetal acceleration. In our case, the minimum \( \mu_s \) works out to approximately 0.276, creating enough 'holding power' to keep you securely against the wall.

Exploring Centripetal Acceleration

Centripetal acceleration is key to how rides like "Spindletop" work. It describes how your velocity changes direction as you move along the circular path inside the spinning cylinder.

The formula for centripetal acceleration \( a_c \) is:

• \( a_c = r \cdot \omega^2 \)
• \( r \) is the cylinder's radius (2.5 m here).
• \( \omega \) is the angular velocity, converted from revolutions per second to radians per second.

For example, if the cylinder rotates at 0.60 rev/s, conversion gives \( \omega = 3.77 \) rad/s. Plugging this into the formula, we find:
Centripetal acceleration \( a_c = 2.5 \times (3.77)^2 = 35.5 \) m/s².

This value of \( a_c \) is essential as it determines the normal force, directly impacting the frictional force needed to counteract gravity and keep you in place.

Deciphering Force Diagrams in Physics

Force diagrams are essential tools to visualize the different forces acting on an object. In our "Spindletop" scenario, these diagrams help us understand how forces interact to keep you pinned against the wall.

In a force diagram:

• The weight or gravitational force \( F_g = mg \) is drawn as a downward arrow.
• The normal force \( F_n \) is depicted as an arrow pointing outward from the center, perpendicular to the wall.
• The frictional force \( F_f \) points upward, opposing the gravitational pull.

Each arrow represents the magnitude and direction of a force, helping you see how equilibrium is maintained. The frictional force equals the gravitational force when you are at rest relative to the wall, meaning that it effectively stops you from sliding down. This visual breakdown makes complex interactions clearer and aids in better understanding mechanical balances.