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Chapter 5: Problem 111

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 \(/ \mathrm{s}\) , 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 station 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 side down the walls to the floor.)

Short Answer

Expert verified
(a) Forces are gravity (down) and friction (up); (b) Minimum coefficient is 0.275; (c) Coefficient does not depend on mass.

Step by step solution

01

Draw the Force Diagram

In this scenario, the forces acting on the person are the gravitational force (\(F_g\)), acting downward, and the normal force (\(F_n\)), acting radially inward. Additionally, the force of static friction acts upward to counteract gravity. This static friction is what prevents the person from sliding downward. The frictional force is equal to \(\mu_s \cdot F_n\), where \(\mu_s\) is the coefficient of static friction.
02

Express Forces in Terms of Known Quantities

The frictional force required to hold the person up is equal to their weight, so \(\mu_s \cdot F_n = mg\), where \(m\) is the mass of the person and \(g\) is the acceleration due to gravity. The normal force is the centripetal force required to keep the person moving in a circle, \(F_n = m\cdot \omega^2 \cdot R\), where \(R = 2.5\,m\) and \(\omega\) is the angular velocity.
03

Calculate Angular Velocity

Convert the angular velocity from revolutions per second to radians per second by using the conversion factor \(2\pi\,rad/rev\). Thus, \(\omega = 0.60\,rev/s \times 2\pi\,rad/rev = 3.77\,rad/s\).
04

Solve for Minimum Coefficient of Friction

Combine the expressions for frictional force and normal force: \(\mu_s \cdot m \cdot \omega^2 \cdot R = mg\). Solving for \(\mu_s\), we have \(\mu_s = \frac{g}{\omega^2 \cdot R}\). Substituting in values, \(\mu_s = \frac{9.8}{(3.77)^2 \cdot 2.5}\), which simplifies to approximately \(\mu_s = 0.275\).
05

Evaluate Mass Dependency

Notice that the mass \(m\) cancels out from the previously derived equation for \(\mu_s\): \(\mu_s = \frac{g}{\omega^2 \cdot R}\). Therefore, the minimum coefficient of static friction does not depend on the mass of the passenger.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Velocity
In the context of circular motion, angular velocity is a crucial concept. It describes how quickly an object rotates around a particular point.
For the amusement park ride "Spindletop," angular velocity (\( \omega \)) is initially given as revolutions per second. However, it's often more useful to express \( \omega \) in radians per second because radians are a standard mathematical unit in angular measurements.
To convert from revolutions to radians, we use the conversion factor \( 2\pi \) (as there are \( 2\pi \) radians in one revolution). This conversion helps in calculating other forces like centripetal force.
The formula for conversion is: \( \omega = N \times 2\pi \), where \( N \) stands for the number of revolutions per second. This means for this ride, \( \omega = 0.60 \times 2\pi = 3.77 \, rad/s \).
Centripetal Force
Centripetal force is necessary to keep an object moving in a circular path. It's the force that acts towards the center of the circle.
On "Spindletop," the walls of the cylinder exert this radially inward force on the riders.
The formula for centripetal force (\( F_c \)) is given by \( F_c = m \omega^2 R \), where \( m \) is mass, \( \omega \) is angular velocity, and \( R \) is the radius of the circular path.
This force is crucial for ensuring that the riders are pinned against the wall during rotation. Without sufficient centripetal force, objects wouldn't maintain their circular path.
Frictional Force
In this scenario, frictional force prevents the riders from sliding down when the floor drops. This static friction acts opposite to the gravitational force.
The role of frictional force on the ride is significant; it provides the upward force that supports the riders' weight.
Frictional force (\( F_f \)) can be calculated using \( F_f = \mu_s \cdot F_n \), where \( \mu_s \) is the coefficient of static friction and \( F_n \) is the normal force.
In our ride, the normal force is equivalent to the centripetal force exerted by the walls.
Coefficient of Static Friction
The coefficient of static friction (\( \mu_s \)) defines how much frictional force resists the sliding motion between two static surfaces.
For the riders on "Spindletop," a certain level of static friction is necessary to remain pinned against the wall without sliding.
Derived from the equations, \( \mu_s = \frac{g}{\omega^2 \cdot R} \), the coefficient depends on gravity, the angular velocity, and the radius.
The calculation gives us an approximate value of \( \mu_s = 0.275 \) for the ride, indicating a specific frictional requirement for safety.
Force Diagram
A force diagram (also known as a free-body diagram) is a visual representation of all the forces acting on an object.
For a person on "Spindletop" after the floor drops, the force diagram includes:
  • The gravitational force (\( mg \)) acting downwards.
  • The normal force (\( F_n \)) from the cylinder's wall acting radially inward.
  • The static frictional force acting upwards, counteracting gravity.
Understanding this diagram simplifies the analysis of forces and helps solve physics problems related to motion, providing a clearer picture of how different forces interact to maintain equilibrium.

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Most popular questions from this chapter

A model airplane with mass 2.20 \(\mathrm{kg}\) moves in the \(x y\) -plane such that its \(x-\) and \(y\) -coordinates vary in time according to \(x(t)=\alpha-\beta t^{3}\) and \(y(t)=y t-\delta t^{2},\) where \(\alpha=1.50 \mathrm{m}, \beta=\) \(0.120 \mathrm{m} / \mathrm{s}^{3}, \gamma=3.00 \mathrm{m} / \mathrm{s},\) and \(\delta=1.00 \mathrm{m} / \mathrm{s}^{2} .\) (a) Calculate the \(x-\) and \(y\) -components of the net force on the plane as functions of time. (b) Sketch the trajectory of the airplane between \(t=0\) and \(t=3.00 \mathrm{s}\) , and draw on your sketch vectors showing the net force on the airplane at \(t=0, t=1.00 \mathrm{s}, t=2.00 \mathrm{s},\) and \(t=3.00 \mathrm{s}\) . For each of these times, relate the direction of the net force to the direction that the airplane is turning, and to whether the airplane is speeding up or slowing down (or neither). (c) What are the magnitude and direction of the net force at \(t=3.00 \mathrm{s} ?\)

A box with mass \(m\) is dragged across a level fioor having a coefficient of kinctic friction \(\mu_{k}\) by a rope that is pulled upward at an angle \(\theta\) above the horizontal with a force of magnitude \(F .(\text { a) In }\) terms of \(m, \mu_{k}, \theta,\) and \(g,\) obtain an expression for the magnitude of force required to move the box with constant speed. (b) Knowing that you are studying physics, a CPR instructor asks you how much force it would take to slide a \(90-\mathrm{kg}\) patient across a floor at constant speed by pulling on him at an angle of \(25^{\circ}\) above the borizontal. By dragging some weights wrapped in an old pair of pants down the hall with a spring balance, you find that \(\mu_{k}=0.35 .\) Use the result of part (a) to answer the instructor's question.

You observe a 1350 -kg sports car rolling along flat pavement in a straight line. The only horizontal forces acting on it are a constant rolling friction and air resistance (proportional to thesquare of its speed). You take the following data during a time interval of \(25 \mathrm{s} :\) When its speed is 32 \(\mathrm{m} / \mathrm{s}\) , the car slows down at a rate of \(-0.42 \mathrm{m} / \mathrm{s}^{2},\) and when its speed is decreased to \(24 \mathrm{m} / \mathrm{s},\) it slows down at \(-0.30 \mathrm{m} / \mathrm{s}^{2} .\) (a) Find the coefficient of rolling friction and the air drag constant \(D .(b)\) At what constant speed will this car move down an incline that makes a \(2.2^{\circ}\) angle with the horizontal? (c) How is the constant speed for an incline of angle \(\beta\) related to the terminal speed of this sports car if the car drops off a high cliff? Assume that in both cases the air resistance force is proportional to the square of the speed, and the air drag constant is the same.

The Monkey and Bananas Problem. A \(20-k g\) monkey has a firm hold on a light rope that passes over a frictionless pulley and is attached to a \(20-\mathrm{kg}\) bunch of bananas (Fig. 5.77\()\) . The monkey looks up, sees the bananas, and starts to climb the rope to get them. (a) As the monkey climbs, do the bananas move up, down, or remain at rest? (b) As the monkey climbs, does the distance between the monkey and the bananas decrease, increase, or remain constant? (c) The monkey releases her hold on the rope. What happens to the distance between the monkey and the bananas while she is falling?(d) Before reaching the ground, the monkey grabs the rope to stop her fall. What do the bananas do? figure can't copy

Block \(A\) , with weight 3\(w\) , slides down an inclined plane \(S\) of slope angle \(36.9^{\circ}\) at a constant speed while plank \(B\) , with weight \(w\) , rests on top of \(A\) . The plank is attached by a cord to the wall \((\text { Fig. } 5.75) .\) (a) Draw a diagram of all the forces acting on block \(A .\) (b) If the coefficient of kinetic friction is the same between \(A\) and \(B\) and between \(S\) and \(A\) , determine its value. figure can't copy
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