Problem 47
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 \(0.60 \mathrm{rev} / \mathrm{s},\) the floor on which the people were standing dropped about 0.5 \(\mathrm{m}\) . The people remained pinned against the wall. (a) Draw a free-body 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
Verified Answer
The minimum coefficient of static friction is approximately 0.276, and this result does not depend on passenger mass.
1Step 1: Understand the problem setup
In this problem, you have a rotating cylinder where people stand against the inner wall. The problem describes a scenario where the floor drops and people remain pinned due to the rotation. We need to analyze the forces acting on a person at this stage.
2Step 2: Draw the free-body diagram
The cylinder rotates, and the person experiences a centripetal force that keeps them against the wall. There are two main forces to consider:1. Gravitational Force (\(F_g = mg\)), acting downward.2. Normal force (\(F_N\)), acting radially inward from the wall.3. Frictional Force (\(F_f\)), acting vertically upward to prevent sliding.Since the person is not moving vertically, the frictional force equals the gravitational force in magnitude.
3Step 3: Write the equation for the normal force
The normal force is provided by the centripetal force required for rotation:\[ F_N = m \cdot a_c, \text{ where } a_c \text{ is the centripetal acceleration.} \]The centripetal acceleration is given by:\[ a_c = \frac{v^2}{r}, \]where \( v \) is the linear speed. For a cylinder, \( v = 2\pi r f \), where \( f \) is the frequency in revolutions per second. Plug in the values:\[ v = 2\pi \times 2.5 \times 0.6 \text{ m/s}. \]
4Step 4: Calculate the centripetal acceleration
First, calculate the linear speed \( v \):\[ v = 2\pi \times 2.5 \times 0.6 \approx 9.42 \text{ m/s}. \]Now substitute in the formula for centripetal acceleration:\[ a_c = \frac{9.42^2}{2.5} = \frac{88.7364}{2.5} \approx 35.49 \text{ m/s}^2. \]
5Step 5: Determine the frictional force equation
The frictional force has to balance the gravitational force to keep the person from sliding down:\[ F_f = \mu F_N = mg, \]where \( \mu \) is the coefficient of static friction. Replace \( F_N \) with the expression from centripetal force:\[ \mu (m \cdot a_c) = mg. \]
6Step 6: Solve for the coefficient of static friction
Rearrange the equation for \( \mu \):\[ \mu = \frac{g}{a_c}. \]Substitute the known values (with \( g = 9.8 \text{ m/s}^2 \)):\[ \mu = \frac{9.8}{35.49} \approx 0.276. \]
7Step 7: Consider the dependence on mass
The equation for \( \mu \) does not depend on the mass \( m \) of the passenger, as both sides of the equation for frictional force contain the mass term, which cancels out. \( \mu = 0.276 \) is independent of \( m \).
Key Concepts
Free-Body DiagramFriction CoefficientCentripetal Force
Free-Body Diagram
A free-body diagram is a vital tool in physics that helps us visualize and analyze the forces acting on an object. For the amusement park ride described, drawing a free-body diagram involves illustrating the forces at play once the floor drops and the passenger stays against the rotating wall.
- **Gravity Force**: Known as the weight of the person, this force acts downward. It is represented by the equation \( F_g = mg \), where \( m \) is the passenger's mass and \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \).
- **Normal Force**: This force acts radially inward, toward the center of the cylinder. It is perpendicular to the surface on which the person is pressed against.
- **Frictional Force**: This force acts vertically upward to counteract gravity and prevents the person from sliding downward. Its magnitude matches that of the gravitational force, keeping the person stationary in the vertical direction.
By identifying and understanding these forces in the free-body diagram, we can analyze the conditions required for equilibrium, particularly in a rotating system.
- **Gravity Force**: Known as the weight of the person, this force acts downward. It is represented by the equation \( F_g = mg \), where \( m \) is the passenger's mass and \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \).
- **Normal Force**: This force acts radially inward, toward the center of the cylinder. It is perpendicular to the surface on which the person is pressed against.
- **Frictional Force**: This force acts vertically upward to counteract gravity and prevents the person from sliding downward. Its magnitude matches that of the gravitational force, keeping the person stationary in the vertical direction.
By identifying and understanding these forces in the free-body diagram, we can analyze the conditions required for equilibrium, particularly in a rotating system.
Friction Coefficient
The friction coefficient is a measure of how much frictional force exists between two surfaces. In this case, it determines the force needed to prevent the passenger from sliding down the wall of the spinning cylinder.
- **Static vs. Kinetic Friction**: Here, the static friction coefficient (\( \mu \)) comes into play because the surfaces do not move relative to each other. Static friction is typically greater than kinetic friction, meaning it takes more force to start moving an object than to keep it moving.
For the "Spindletop" ride, the required frictional force \( F_f \) must balance the gravitational force \( mg \). This equilibrium condition is written as:\[ F_f = \mu F_N = mg \]
- **Solving for the Friction Coefficient**: Rearranging the equation gives \( \mu = \frac{mg}{F_N} \). However, because \( F_N \) equals the centripetal force \( m \cdot a_c \), the mass \( m \) cancels out, simplifying to \[ \mu = \frac{g}{a_c} \].
This equation shows that the friction coefficient relies solely on the ratio of gravity to centripetal acceleration. For the given values, \( \mu \approx 0.276 \). This independence from mass highlights a key characteristic of the problem's physics: the experience is the same regardless of the person's mass.
- **Static vs. Kinetic Friction**: Here, the static friction coefficient (\( \mu \)) comes into play because the surfaces do not move relative to each other. Static friction is typically greater than kinetic friction, meaning it takes more force to start moving an object than to keep it moving.
For the "Spindletop" ride, the required frictional force \( F_f \) must balance the gravitational force \( mg \). This equilibrium condition is written as:\[ F_f = \mu F_N = mg \]
- **Solving for the Friction Coefficient**: Rearranging the equation gives \( \mu = \frac{mg}{F_N} \). However, because \( F_N \) equals the centripetal force \( m \cdot a_c \), the mass \( m \) cancels out, simplifying to \[ \mu = \frac{g}{a_c} \].
This equation shows that the friction coefficient relies solely on the ratio of gravity to centripetal acceleration. For the given values, \( \mu \approx 0.276 \). This independence from mass highlights a key characteristic of the problem's physics: the experience is the same regardless of the person's mass.
Centripetal Force
Centripetal force is crucial in circular motion, particularly for maintaining the path of an object moving around a center. In the case of the amusement ride, this force keeps the passenger pinned against the rotating wall.
- **Definition and Direction**: Centripetal force is always directed towards the center of the circle around which an object is moving. It's a result of the object's need to constantly change direction to maintain its circular path.
- **Formula**: The standard formula for centripetal force is \( F_c = m \cdot a_c \), where \( a_c \) is the centripetal acceleration given by \( a_c = \frac{v^2}{r} \). Here, \( v \) is the object's linear velocity, and \( r \) is the radius of the circle.
- **Application in the Ride**: For the spinning cylinder, the rotation gives passengers the centripetal acceleration necessary to provide the normal force. Using the problem's specifications, \( a_c \approx 35.49 \text{ m/s}^2 \), ensuring passengers remain firmly against the wall as the floor falls away.
Understanding centripetal force is essential for grasping how objects move in a circle and remain in that circular path, making it a fundamental concept in the physics of circular motion.
- **Definition and Direction**: Centripetal force is always directed towards the center of the circle around which an object is moving. It's a result of the object's need to constantly change direction to maintain its circular path.
- **Formula**: The standard formula for centripetal force is \( F_c = m \cdot a_c \), where \( a_c \) is the centripetal acceleration given by \( a_c = \frac{v^2}{r} \). Here, \( v \) is the object's linear velocity, and \( r \) is the radius of the circle.
- **Application in the Ride**: For the spinning cylinder, the rotation gives passengers the centripetal acceleration necessary to provide the normal force. Using the problem's specifications, \( a_c \approx 35.49 \text{ m/s}^2 \), ensuring passengers remain firmly against the wall as the floor falls away.
Understanding centripetal force is essential for grasping how objects move in a circle and remain in that circular path, making it a fundamental concept in the physics of circular motion.
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