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Chapter 9: Problem 45

A stationary bicycle is raised off the ground, and its front wheel \((m=1.3 \mathrm{kg})\) is rotating at an angular velocity of \(13.1 \mathrm{rad} / \mathrm{s}\) (see the drawing). The front brake is then applied for \(3.0 \mathrm{s},\) and the wheel slows down to \(3.7 \mathrm{rad} / \mathrm{s} .\) Assume that all the mass of the wheel is concentrated in the rim, the radius of which is \(0.33 \mathrm{m} .\) The coefficient of kinetic friction between each brake pad and the rim is \(\mu_{\mathrm{k}}=0.85 .\) What is the magnitude of the normal force that each brake pad applies to the rim?

Short Answer

Expert verified
The normal force each brake pad applies is 1.58 N.

Step by step solution

01

Calculate Initial and Final Angular Velocities in Revolution

First, note the initial (\(\omega_i = 13.1 \frac{\text{rad}}{\text{s}}\) ) and final (\(\omega_f = 3.7 \frac{\text{rad}}{\text{s}}\) ) angular velocities given in the problem.
02

Determine the Angular Deceleration

The angular deceleration, \( \alpha \), can be found using the formula:\[\alpha = \frac{\omega_f - \omega_i}{t}\]Substituting the values gives:\[\alpha = \frac{3.7 \frac{\text{rad}}{\text{s}} - 13.1 \frac{\text{rad}}{\text{s}}}{3.0\,\text{seconds}} = -3.13 \frac{\text{rad}}{\text{s}^2}\]
03

Calculating Torque Due to Friction

The torque \(\tau\) due to friction is related to the angular deceleration by:\[\tau = I \cdot \alpha\]where \(I\) is the moment of inertia of the wheel, i.e., \(I = m \cdot r^2\) since the mass is concentrated at the rim. Given \(m = 1.3 \, \text{kg}\) and \(r = 0.33 \, \text{m}\), we have:\[I = 1.3 \, \text{kg} \times (0.33 \, \text{m})^2 = 0.14157 \, \text{kg} \, \text{m}^2\]Therefore:\[\tau = 0.14157 \, \text{kg} \, \text{m}^2 \times (-3.13 \frac{\text{rad}}{\text{s}^2}) = -0.443 kg路m^2/s^2\]
04

Determine Force of Friction

Torque can also be defined by the force associated with friction:\[\tau = r \cdot F_{friction}\]Solving for \(F_{friction}\), we get:\[F_{friction} = \frac{\tau}{r} = \frac{-0.443}{0.33} \, \text{N} \, \text{m} = -1.34 \text{N}\]
05

Calculate Normal Force Applied by Brake Pad

Since \(F_{friction} = \mu_k \cdot F_{normal}\), we can solve for \(F_{normal}\):\[F_{normal} = \frac{F_{friction}}{\mu_k} = \frac{1.34 \text{N}}{0.85} = 1.58 \text{N}\]

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Key Concepts

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

Angular Velocity
Angular velocity is the rate at which an object rotates around an axis. It tells us how fast something like a wheel is spinning. Measured in radians per second (\(\text{rad/s}\)), it's crucial for understanding how quickly a wheel can change speed.

In the exercise, we're given an initial angular velocity of 13.1 rad/s and a final angular velocity after the brakes are applied, of 3.7 rad/s. This shows that the wheel slows down quite a bit. Think of it like how fast the hands of a clock move, but in this case, it's a wheel slowing down its spin.

When working with angular motion, comparing how much these angular velocities change helps determine other factors like angular deceleration and torque, which we'll explore next. The slowing down process, due to the brake, shows us the effectiveness of friction acting on the wheel.
Friction
Friction is a force that opposes motion between two surfaces in contact. It acts to slow down the movement of objects. When you apply brakes to a bicycle, you're creating friction against the rotating wheel to make it stop. The coefficient of kinetic friction (\(\mu_k\)) quantifies this force.

  • In our exercise, \(\mu_k = 0.85\)
This value tells us about the intensity of the friction applied by the brake pads against the wheel's rim.

The force of friction plays an essential part in acting against the wheel's rotation, thus leading to its deceleration. Friction is needed to bring about changes in angular velocity and ensure the wheel slows down effectively when brakes are applied. Balancing these forces correctly is what ensures safe and effective braking.
Torque
Torque is a measure of the force that causes an object to rotate about an axis. Think of it as the 'twist' or rotational force applied to an object. It's one of the essential concepts in understanding rotational dynamics. Torque (\(\tau\)) is calculated as the product of the moment of inertia and the angular acceleration (\(\alpha\)).

  • \(\tau = I \cdot \alpha\)
In our bicycle exercise, the torque results from the frictional force caused by the brake pads on the wheel, managing the wheel's deceleration.

With a calculated torque of -0.443 \(\text{kg}路\text{m}^2/\text{s}^2\), we see how strongly the brake pads can change the wheel's rotation. Negative torque indicates the direction is opposite to the initial rotation, emphasizing the deceleration aspect.
Moment of Inertia
Moment of inertia (\(I\)) represents how mass is distributed in a rotating object and its resistance to changes in rotation. It鈥檚 like the rotational equivalent of mass for linear motion. For our bicycle wheel, all mass is concentrated at the rim, calculated as mass times the radius squared (\(I = m \cdot r^2\)).

  • Given: \(m = 1.3 \text{kg}\) and \(r = 0.33 \text{m}\)
This results in \(I = 0.14157 \text{kg}\cdot \text{m}^2\), showing how mass and distance affect the object's ability to resist angular acceleration.

The larger the moment of inertia, the more torque needed for the same angular acceleration, illustrating why heavier or larger objects rotate more slowly or require more effort to speed up or slow down.

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Most popular questions from this chapter

Two identical wheels are moving on horizontal surfaces. The center of mass of each has the same linear speed. However, one wheel is rolling, while the other is sliding on a frictionless surface without rolling. Each wheel then encounters an incline plane. One continues to roll up the incline, while the other continues to slide up. Eventually they come to a momentary halt, because the gravitational force slows them down. Each wheel is a disk of mass \(2.0 \mathrm{kg} .\) On the horizontal surfaces the center of mass of each wheel moves with a linear speed of \(6.0 \mathrm{m} / \mathrm{s}\). (a) What is the total kinetic energy of each wheel? (b) Determine the maximum height reached by each wheel as it moves up the incline.

Two thin rods of length \(L\) are rotating with the same angular speed \(\omega\) (in \(\mathrm{rad} / \mathrm{s}\) ) about axes that pass perpendicularly through one end. Rod \(\mathrm{A}\) is massless but has a particle of mass \(0.66 \mathrm{kg}\) attached to its free end. Rod B has a mass of 0.66 kg, which is distributed uniformly along its length. The length of each rod is \(0.75 \mathrm{m},\) and the angular speed is \(4.2 \mathrm{rad} / \mathrm{s}\). Find the kinetic energies of rod \(A\) with its attached particle and of rod \(B\).

A rod is lying on the top of a table. One end of the rod is hinged to the table so that the rod can rotate freely on the tabletop. Two forces, both parallel to the tabletop, act on the rod at the same place. One force is directed perpendicular to the rod and has a magnitude of \(38.0 \mathrm{N}\). The second force has a magnitude of \(55.0 \mathrm{N}\) and is directed at an angle \(\theta\) with respect to the rod. If the sum of the torques due to the two forces is zero, what must be the angle \(\theta ?\)

The drawing shows a rectangular piece of wood. The forces applied to corners \(\mathrm{B}\) and \(\mathrm{D}\) have the same magnitude of \(12 \mathrm{N}\) and are directed parallel to the long and short sides of the rectangle. The long side of the rectangle is twice as long as the short side. An axis of rotation is shown perpendicular to the plane of the rectangle at its center. A third force (not shown in the drawing) is applied to corner A, directed along the short side of the rectangle (either toward \(\mathrm{B}\) or away from \(\mathrm{B}\) ), such that the piece of wood is at equilibrium. Find the magnitude and direction of the force applied to corner A.

A hiker, who weighs \(985 \mathrm{N}\), is strolling through the woods and crosses a small horizontal bridge. The bridge is uniform, weighs \(3610 \mathrm{N}\), and rests on two concrete supports, one at each end. He stops one-fifth of the way along the bridge. What is the magnitude of the force that a concrete support exerts on the bridge (a) at the near end and (b) at the far end?
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