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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:\[\tau = 0.14157 \, \text{kg} \, \text{m}^2 \times (-3.13 \frac{\text{rad}}{\text{s}^2}) = -0.443 kg·m^2/s^2\]
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 \(\text{kg}·\text{m}^2/\text{s}^2\), 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’s 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

A 15.0 -m length of hose is wound around a reel, which is initially at rest. The moment of inertia of the reel is \(0.44 \mathrm{kg} \cdot \mathrm{m}^{2},\) and its radius is \(0.160 \mathrm{m} .\) When the reel is turning, friction at the axle exerts a torque of magnitude \(3.40 \mathrm{N} \cdot \mathrm{m}\) on the reel. If the hose is pulled so that the tension in it remains a constant \(25.0 \mathrm{N},\) how long does it take to completely unwind the hose from the reel? Neglect the mass and thickness of the hose on the reel, and assume that the hose unwinds without slipping.

The parallel axis theorem provides a useful way to calculate the moment of inertia \(I\) about an arbitrary axis. The theorem states that \(I=I_{\mathrm{cm}}+\) \(M h^{2},\) where \(I_{\mathrm{cm}}\) is the moment of inertia of the object relative to an axis that passes through the center of mass and is parallel to the axis of interest, \(M\) is the total mass of the object, and \(h\) is the perpendicular distance between the two axes. Use this theorem and information to determine an expression for the moment of inertia of a solid cylinder of radius \(R\) relative to an axis that lies on the surface of the cylinder and is perpendicular to the circular ends.

A clay vase on a potter's wheel experiences an angular acceleration of \(8.00 \mathrm{rad} / \mathrm{s}^{2}\) due to the application of a \(10.0-\mathrm{N} \cdot \mathrm{m}\) net torque. Find the total moment of inertia of the vase and potter's wheel.

A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provide a means for storing energy in the form of rotational kinetic energy and are being considered as a possible alternative to batteries in electric cars. The gasoline burned in a 300 -mile trip in a typical midsize car produces about \(1.2 \times 10^{9} \mathrm{J}\) of energy. How fast would a \(13-\mathrm{kg}\) flywheel with a radius of \(0.30 \mathrm{m}\) have to rotate to store this much energy? Give your answer in rev/min.

A man drags a 72 -kg crate across the floor at a constant velocity by pulling on a strap attached to the bottom of the crate. The crate is tilted \(25^{\circ}\) above the horizontal, and the strap is inclined \(61^{\circ}\) above the horizontal. The center of gravity of the crate coincides with its geometrical center, as indicated in the drawing. Find the magnitude of the tension in the strap.

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