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Chapter 5: Problem 38

Rolling Friction. Two bicycle tires are set rolling with the same initial speed of 3.50 \(\mathrm{m} / \mathrm{s}\) on a long, straight road, and the distance each travels before its speed is reduced by half is measured. One tire is inflated to a pressure of 40 psi and goes 18.1 \(\mathrm{m}\) ; the other is at 105 psi and goes 92.9 \(\mathrm{m}\) . What is the coefficient of rolling friction \(\mu_{r}\) for each? Assume that the net horizontal force is due to rolling friction only.

Short Answer

Expert verified
The coefficient of rolling friction is 0.029 for the 40 psi tire and 0.0056 for the 105 psi tire.

Step by step solution

01

Understanding the Concept

Rolling friction is the force resisting the motion when an object rolls on a surface. For this problem, the only horizontal force we consider is the rolling friction, which decelerates the tire. The relationship between rolling friction, the distance traveled until the speed reduces by half, and the initial speed is essential for solving the problem.
02

Using the Work-Energy Principle

The work done by the rolling friction force reduces the kinetic energy of the tire. We can express the initial kinetic energy as \( KE_i = \frac{1}{2} mv^2 \), and the final kinetic energy when the speed is halved is \( KE_f = \frac{1}{2} m \left( \frac{v}{2} \right)^2 \). The work done by friction is \( W = F_f \cdot d = \mu_r mgd \), where \( d \) is the distance traveled, \( m \) is the mass, \( g \) is the gravitational acceleration, and \( \mu_r \) is the coefficient of rolling friction.
03

Calculating Work Done by Friction

Set the work done by friction equal to the change in kinetic energy:\[- \mu_r mgd = \frac{1}{2} mv^2 - \frac{1}{2} m \left( \frac{v}{2} \right)^2. \] Simplifying gives:\[ \mu_r mgd = \frac{3}{8} mv^2. \]
04

Solving for \(\mu_r\)

Solving the equation \( \mu_r mgd = \frac{3}{8} mv^2 \) for \( \mu_r \) gives:\[ \mu_r = \frac{3v^2}{8gd}. \]Substitute gravitational acceleration \( g = 9.81 \mathrm{m/s^2}\) and initial speed \( v = 3.50 \mathrm{m/s} \).
05

Calculating \(\mu_r\) for 40 psi Tire

Substitute \( d = 18.1 \mathrm{m} \) for the 40 psi tire:\[\mu_r = \frac{3(3.50)^2}{8 \cdot 9.81 \cdot 18.1} = 0.029.\]
06

Calculating \(\mu_r\) for 105 psi Tire

Substitute \( d = 92.9 \mathrm{m} \) for the 105 psi tire:\[\mu_r = \frac{3(3.50)^2}{8 \cdot 9.81 \cdot 92.9} = 0.0056.\]

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Key Concepts

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

Work-Energy Principle
The work-energy principle is a fundamental concept in physics that connects the work done on an object to its change in kinetic energy. When you apply a force on an object and it moves, work is done on it. This work changes the object's kinetic energy. The principle here primarily analyzes how the work done by rolling friction affects a bicycle tire's motion.

In our rolling tire scenario, the initial kinetic energy is given by:
  • \( KE_i = \frac{1}{2} mv^2 \)
When the speed of the tire is halved, its kinetic energy becomes:
  • \( KE_f = \frac{1}{2} m \left( \frac{v}{2} \right)^2 = \frac{1}{8} mv^2 \)
The work done by rolling friction is the product of frictional force and distance traveled, leading to:
  • \( W = F_f \cdot d = \mu_r mgd \)
This principle allows us to connect the change in kinetic energy to the work done by friction, helping us solve for the coefficient of rolling friction.
Coefficient of Rolling Friction
The coefficient of rolling friction, denoted as \( \mu_r \), is a measure of how much rolling resistance an object faces as it moves along a surface. This coefficient essentially tells us how much of the driving force is being used to overcome rolling friction.

In this exercise, the goal is to find \( \mu_r \) for each tire. When a tire rolls, it faces rolling friction, which slows it down over time. The relationship between the work done by this friction, the distance, and the speed is vital. We can calculate \( \mu_r \) using the formula derived from equating work done by friction to the change in kinetic energy:
  • \( \mu_r = \frac{3v^2}{8gd} \)
By knowing the distance each tire travels until its speed reduces by half, the specific \( \mu_r \) for each tire can be determined. A lower \( \mu_r \) represents less rolling resistance. This explains why the 105 psi tire travels much further than the 40 psi tire, due to its lower coefficient of rolling friction.
Kinetic Energy
Kinetic energy refers to the energy that an object possesses due to its motion. For any rolling object such as a bicycle tire, kinetic energy is crucial in understanding its movement. The kinetic energy of an object is determined by its mass and velocity with the formula:
  • \( KE = \frac{1}{2} mv^2 \)
In our problem, the bicycle tires start with a speed of 3.50 m/s, indicating the initial kinetic energy. As the tires travel along the road, the work done by rolling friction reduces this kinetic energy until the speed is halved.

It's important to recognize how kinetic energy changes proportionally with the square of velocity. Hence, when the speed decreases to half, the kinetic energy drops to one-fourth. This drop is due to the negative work done by rolling friction, represented by the decrease in velocity. Efficiently understanding kinetic energy changes helps explain why a higher pressure tire, with lower rolling friction, maintains its speed over a longer distance.

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Most popular questions from this chapter

A \(1130-k g\) car is held in place by a light cable on a very smooth (frictionless) ramp, as shown in Fig. \(5.45 .\) The cable makes an angle of \(31.0^{\circ}\) above the surface of the ramp, and the ramp itself rises at \(25.0^{\circ}\) above the borizontal. (a) Draw a free-body diagram for the car. (b) Find the tension in the cable. (c) How hard does the surface of the ramp push on the car? figure can't copy

Angle for Minimum Force. A box with weight w is pulled at constant speed along a level floor by a force \(\vec{F}\) that is at an angle \(\theta\) above the horizontal. The coefficient of kinetic friction between the floor and box is \(\mu_{k}\) (a) In terms of \(\theta, \mu_{k},\) and \(w,\) calculate \(F .\) (b) For \(w=400 \mathrm{N}\) and \(\mu_{k}=0.25\) , 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 \(\mu_{\mathrm{x}}=0.25\) evaluate this optimal \(\theta\) and compare your result to the graph you constructed in part \((\mathrm{b})\) .

You are standing on a bathroom scale in an elevator in a tall building. Your mass is 72 kg. The elevator starts from rest and travels upward with a speed that varies with time according to \(v(t)=\left(3.0 \mathrm{m} / \mathrm{s}^{2}\right) t+\left(0.20 \mathrm{m} / \mathrm{s}^{3}\right) t^{2} .\) When \(t=4.0 \mathrm{s}\) , what is the reading of the bathroom scale?

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 of bananas weighing 40.0 \(\mathrm{N}\) rests on a horizontal surface. The coefficient of static friction between the box and the sur- face is \(0.40,\) and the coefficient of kinetic friction is 0.20 . (a) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on the box? (b) What is the magnitude of the friction force if a monkey applies a horizontal force of 6.0 \(\mathrm{N}\) to the box and the box is initially at rest? (c) What minimum horizontal force must the monkey apply to start the box in motion? (d) What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started? (e) If the monkey applies a horizontal force of \(18.0 \mathrm{N},\) what is the magnitude of the friction force and what is the box's acceleration?
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