91Ó°ÊÓ

Interactive Solution 9.35 at presents a method for modeling this problem. MultipleConcept Example 10 offers useful background for problems like this. A cylinder is rotating about an axis that passes through the center of each circular end piece. The cylinder has a radius of \(0.0830 \mathrm{~m}\), an angular speed of \(76.0 \mathrm{rad} / \mathrm{s},\) and a moment of inertia of \(0.615 \mathrm{~kg} \cdot \mathrm{m}^{2}\). A brake shoe presses against the surface of the cylinder and applies a tangential frictional force to it. The frictional force reduces the angular speed of the cylinder by a factor of two during a time of \(6.40 \mathrm{~s}\). (a) Find the magnitude of the angular deceleration of the cylinder. (b) Find the magnitude of the force of friction applied by the brake shoe.

Step by step solution

01

Identify Initial and Final Angular Speed

The initial angular speed \( \omega_0 \) is given as \( 76.0 \, \text{rad/s} \). The problem states the angular speed is reduced by a factor of two, so the final angular speed \( \omega \) will be \( \frac{76.0}{2} = 38.0 \, \text{rad/s} \).

02

Calculate Angular Deceleration

The change in angular speed \( \Delta \omega \) is \( 38.0 - 76.0 = -38.0 \, \text{rad/s} \). Using the formula for angular deceleration, \( \alpha = \frac{\Delta \omega}{\Delta t} \), where \( \Delta t = 6.40 \, \text{s} \), we find \( \alpha = \frac{-38.0}{6.40} \approx -5.94 \, \text{rad/s}^2 \). The magnitude is \( 5.94 \, \text{rad/s}^2 \).

03

Calculate Torque

Torque \( \tau \) is related to angular deceleration by the equation \( \tau = I \alpha \), where \( I = 0.615 \, \text{kg} \cdot \text{m}^2 \). Substituting for \( \alpha \), we get \( \tau = 0.615 \times -5.94 \approx -3.65 \, \text{N} \cdot \text{m} \). The magnitude is \( 3.65 \, \text{N} \cdot \text{m} \).

04

Calculate Frictional Force

The frictional force \( F \) is related to torque by the formula \( \tau = rF \), where \( r = 0.0830 \, \text{m} \) is the radius of the cylinder. Solving for \( F \), we get \( F = \frac{3.65}{0.0830} \approx 43.98 \, \text{N} \).

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Key Concepts

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

Angular Deceleration

Angular deceleration refers to the rate at which an object's angular speed decreases over time. When a rotating object slows down, like in the instance with the cylinder, the change in its angular velocity can be quantified as angular deceleration.
An angular deceleration can be calculated using the formula:

• \( \alpha = \frac{\Delta \omega}{\Delta t} \)

Here, \( \alpha \) represents angular deceleration, \( \Delta \omega \) is the change in angular velocity, and \( \Delta t \) is the time taken for this change.
In our exercise, the angular speed of the cylinder changes from \( 76.0 \text{ rad/s} \) to \( 38.0 \text{ rad/s} \) over \( 6.40 \text{ s} \). Therefore, the angular deceleration magnitude is

• \( 5.94 \text{ rad/s}^2 \)

Understanding angular deceleration is crucial as it provides insight into how quickly an object stops rotating.

Moment of Inertia

The moment of inertia is a property of rotating objects that measures its resistance to changes in rotational motion. It depends not only on the mass of an object but also on how this mass is distributed relative to the axis of rotation.
In mathematical terms, the moment of inertia, denoted as \( I \), is involved in calculating the rotational effects of forces, such as torque and angular acceleration.
It’s crucial when analyzing rotational dynamics, as seen in this problem where the cylinder's moment of inertia is given as:

• \( 0.615 \text{ kg} \cdot \text{m}^2 \)

This implies that the cylinder has moderate resistance against changes in its rotation due to external forces. The concept is integral in calculating other dynamic quantities like torque, linking angular acceleration with applied force.

Frictional Force

Frictional force plays a key role in rotational dynamics by opposing motion and often acting as a mechanism for slowing objects down. When a brake shoe presses against a rotating cylinder, it applies a frictional force that reduces the angular speed.
In this exercise, the goal is to determine the magnitude of this force. Using the relationship between frictional force and torque:

• \( \tau = rF \)

where \( \tau \) is the torque and \( r \) is the radius, it becomes possible to compute the force. For our cylinder, with a calculated torque and the known radius, the force is found to be

• \( 43.98 \text{ N} \)

This interaction of forces is crucial to understanding how systems like brakes work to control rotational motion effectively.

Torque

Torque is a critical concept in rotational dynamics, representing the rotational equivalent of force in linear motion. It describes the effect of a force that causes an object to rotate around an axis.
In mathematical terms, torque \( \tau \) is calculated as:

• \( \tau = I \alpha \)
• \( \tau = rF \)

For the braking cylinder scenario, torque can be derived using the moment of inertia and the angular deceleration. With known values for these parameters, torque is calculated to be approximately

• \( 3.65 \text{ N} \cdot \text{m} \)

Torque is an essential parameter for determining how effectively a force can influence rotational motion, factoring into not just the magnitude of a force but also the distance from the axis at which it is applied.