Chapter 8: Problem 82
A torque of \(0.97 \mathrm{~N} \cdot \mathrm{m}\) is applied to a bicycle wheel of radius \(35 \mathrm{~cm}\) and mass \(0.75 \mathrm{~kg}\). Treating the wheel as a hoop, find its angular acceleration.
Short Answer
Expert verified
The angular acceleration is approximately 10.56 rad/sĀ².
Step by step solution
01
Understand the System as a Hoop
The bicycle wheel is treated as a hoop, and the formula for the moment of inertia (\(I\)) of a hoop is given by:\[I = m imes r^2\]where\(m = 0.75 \, \text{kg}\)is the mass of the tire and\(r = 0.35 \, \text{m}\)is the radius. We need to use these values to find the moment of inertia.
02
Calculate the Moment of Inertia
Substitute the values of the mass and radius into the formula for the moment of inertia:\[I = 0.75 \, \text{kg} \times (0.35 \, \text{m})^2 = 0.091875 \, \text{kg} \, \text{m}^2\]
03
Use Torque to Find Angular Acceleration
Use the relationship between torque, moment of inertia, and angular acceleration:\[\tau = I \alpha\]where:\(\tau = 0.97 \, \text{N} \, \text{m}\) is the torque,\(I\) is the moment of inertia, and \(\alpha\) is the angular acceleration. Rearrange this equation to solve for \(\alpha\):\[\alpha = \frac{\tau}{I}\]Substitute the known values:\[\alpha = \frac{0.97 \, \text{N} \, \text{m}}{0.091875 \, \text{kg} \, \text{m}^2} \approx 10.56 \, \text{rad/s}^2\]
04
Conclude the Angular Acceleration Value
Using the calculations, we determine the angular acceleration of the bicycle wheel to be approximately \(10.56 \, \text{rad/s}^2\). This is achieved by dividing the torque by the moment of inertia.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Torque
Torque is a crucial concept when discussing rotational movement. It can be thought of as the rotational equivalent of linear force. Imagine you are trying to open a door. The force you apply to the door handle, and its distance from the hinges, combined together create torque. In mathematical terms, torque \(\tau\) is calculated as:
\[\tau = F \times r\]where:
\[\tau = F \times r\]where:
- \(F\) is the force applied,
- \(r\) is the distance from the point of rotation.
Moment of Inertia
The moment of inertia, often represented by \(I\), is a measure of an object's resistance to changes in its rotation. Think of it as the rotational equivalent of mass in linear motion. Just as heavier objects require more force to accelerate in a straight line, objects with higher moments of inertia require more torque to start rotating.
For objects like a hoop, the moment of inertia is determined by the mass of the object and how that mass is distributed in relation to the axis of rotation. The formula is:
\[I = m \times r^2\]where:
This concept is important for designing objects that rotate, ensuring they are efficient and effective during operation.
For objects like a hoop, the moment of inertia is determined by the mass of the object and how that mass is distributed in relation to the axis of rotation. The formula is:
\[I = m \times r^2\]where:
- \(m\) is the mass, and
- \(r\) is the radius or distance from the axis of rotation.
This concept is important for designing objects that rotate, ensuring they are efficient and effective during operation.
Bicycle Wheel
A bicycle wheel is an everyday example of an object experiencing torque and angular acceleration. In this context, the wheel is treated like a hoop for simplicity. This model helps us analyze rotational motion calculations with ease. The wheel's mass and distribution of this mass affect how it behaves when rotated. By analyzing it as a hoop, we ignore complexities like spokes and only focus on the rim, making calculations straightforward and comparable to ideal situations.
Many practical applications use these ideal models to predict how real-world objects will behave. This approach is also useful for students trying to grasp the foundational principles of rotational dynamics. It allows you to understand how variables like torque, moment of inertia, and angular acceleration relate to each other in a straightforward system.
Many practical applications use these ideal models to predict how real-world objects will behave. This approach is also useful for students trying to grasp the foundational principles of rotational dynamics. It allows you to understand how variables like torque, moment of inertia, and angular acceleration relate to each other in a straightforward system.
Hoop Model
In physics, the hoop model is a simplified way to understand rotational dynamics. It treats the object, like a bicycle wheel, as a thin ring or hoop with mass concentrated along its circumference. This simplification is useful because it allows you to easily apply the formula for the moment of inertia without complicated calculations.
This model assumes that all of the bicycle wheel's mass is at a uniform distance from the center of rotation. This makes calculating rotational properties, like angular acceleration, convenient. By focusing solely on the rim's mass, students can better grasp the core principles of rotational motion.
This model assumes that all of the bicycle wheel's mass is at a uniform distance from the center of rotation. This makes calculating rotational properties, like angular acceleration, convenient. By focusing solely on the rim's mass, students can better grasp the core principles of rotational motion.
- It simplifies problem-solving, making it easier to analyze and predict motion.
- It helps visualize the effects of rotational forces in a real-world context.