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Chapter 5: Problem 115

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.)

Short Answer

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

Step by step solution

01

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.
02

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 \( F_f \).
03

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.
04

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.
05

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/s虏. Then, the normal force \( F_n = m \cdot a_c = m \cdot 35.5 \) N.
06

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 \( \mu_s = \frac{g}{35.5} \approx \frac{9.8}{35.5} \approx 0.276 \).
07

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.

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Key Concepts

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

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/s虏) 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/s虏.

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.

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Most popular questions from this chapter

A physics major is working to pay his college tuition by performing in a traveling carnival. He rides a motorcycle inside a hollow, transparent plastic sphere. After gaining sufficient speed, he travels in a vertical circle with a radius of 13.0 \(\mathrm{m} .\) The physics major has mass \(70.0 \mathrm{kg},\) and his motorcycle has mass 40.0 \(\mathrm{kg}\) . (a) What minimum speed must he have at the top of the circle if the tires of the motorcycle are not to lose contact with the sphere? (b) At the bottom of the circle, his speed is twice the value calculated in part (a). What is the magnitude of the normal force exerted on the motorcycle by the sphere at this point?

A 15.0 -kg load of bricks hangs from one end of a rope that passes over a small, frictionless pulley. A 28.0 kg counterweight is suspended from the other end of the rope, as shown in Fig. E5.15. The system is released from rest. (a) Draw two free-body diagrams, one for the load of bricks and one for the counterweight. (b) What is the magnitude of the upward acceleration of the load of bricks? (c) What is the tension in the rope while the load is moving? How does the tension compare to the weight of the load of bricks? To the weight of the counterweight?

A box with weight \(w\) is pulled at constant speed along a level floor by a force \(\vec{\boldsymbol{F}}\) that is at an angle \(\theta\) above the horizontal. The coefficient of kinetic friction between the floor and box is \(\mu_{\mathrm{k}}\) (a) In terms of \(\theta, \mu_{\mathrm{k}},\) and \(w\) calculate \(F .\) (b) For \(w=400 \mathrm{N}\) and \(\mu_{\mathrm{k}}=0.25,\) calculate \(F\) for \(\theta\) ranging from \(0^{\circ}\) to \(90^{\circ}\) in increments of \(10^{\circ} .\) Graph \(F\) versus \(\theta\) .(c) From the general expression in part (a), calculate the value of \(\theta\) for which the value of \(F\) , required to maintain constant speed, is a minimum. (Hint: At a point where a function is minimum, what are the first and second derivatives of the function? Here \(F\) is a function of \(\theta . )\) For the special case of \(w=400 \mathrm{N}\) and \(\mu_{\mathrm{k}}=0.25\) evaluate this optimal \(\theta\) and compare your result to the graph you constructed in part (b).

A 2.00 -kg box is moving to the right with speed 9.00 \(\mathrm{m} / \mathrm{s}\) on a horizontal, frictionless surface. At \(t=0\) a horizontal force is applied to the box. The force is directed to the left and has magnitude \(F(t)=\left(6.00 \mathrm{N} / \mathrm{s}^{2}\right) t^{2}\) . (a) What distance does the box move from its position at \(t=0\) before its speed is reduced to zero? (b) If the force continues to be applied, what is the speed of the box at \(t=3.00 \mathrm{s} ?\)

A flat (unbanked) curve on a highway has a radius of 220.0 \(\mathrm{m} .\) A car rounds the curve at a speed of 25.0 \(\mathrm{m} / \mathrm{s}\) . (a) What is the minimum coefficient of friction that will prevent sliding? (b) Suppose the highway is icy and the coefficient of friction between the tires and pavement is only one-third what you found in part (a). What should be the maximum speed of the car so it can round the curve safely?
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