A motorcycle accelerates uniformly from rest and reaches a linear speed of 22.4 m/s in a time of 9.06 s. T radius of each tire is 0.287 m. What is the magnitude of the angular acceleration of each tire?
A motorcycle accelerates uniformly from rest and reaches a linear speed of 22.4 m/s in a time of 9.06 s. T radius of each tire is 0.287 m. What is the magnitude of the angular acceleration of each tire?
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Angular speed, acceleration and displacement
Angular acceleration is defined as the rate of change in angular velocity with respect to time. It has both magnitude and direction. So, it is a vector quantity.
Angular Position
Before diving into angular position, one should understand the basics of position and its importance along with usage in day-to-day life. When one talks of position, it’s always relative with respect to some other object. For example, position of earth with respect to sun, position of school with respect to house, etc. Angular position is the rotational analogue of linear position.
Question
![### Physics Problem: Calculating Angular Acceleration
---
**Problem Statement:**
A motorcycle accelerates uniformly from rest and reaches a linear speed of 22.4 m/s in a time of 9.06 s. The radius of each tire is 0.287 m. What is the magnitude of the angular acceleration of each tire?
---
To solve this problem, we need to use the relationship between linear acceleration, angular acceleration, and the radius of the tires.
Let's break down the problem step-by-step:
1. **Linear Speed and Time:**
- Initial linear speed (u) = 0 m/s (since it starts from rest)
- Final linear speed (v) = 22.4 m/s
- Time (t) = 9.06 s
2. **Linear Acceleration:**
Using the formula for linear acceleration (a):
\[
a = \frac{v - u}{t}
\]
Substituting the given values:
\[
a = \frac{22.4\, \text{m/s} - 0\, \text{m/s}}{9.06\, \text{s}}
\]
\[
a = \frac{22.4}{9.06} \approx 2.473\, \text{m/s}^2
\]
3. **Radius of Each Tire (r):**
Radius (r) = 0.287 m
4. **Angular Acceleration:**
Angular acceleration (\(\alpha\)) can be found using the formula:
\[
\alpha = \frac{a}{r}
\]
Substituting the values of linear acceleration and the radius:
\[
\alpha = \frac{2.473\, \text{m/s}^2}{0.287\, \text{m}}
\]
\[
\alpha \approx 8.61\, \text{rad/s}^2
\]
**Conclusion:**
The magnitude of the angular acceleration of each tire is approximately \(8.61\, \text{rad/s}^2\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F70495d9e-d834-4a3c-919e-f23d4c42494e%2Ffd12f6d1-5393-49a3-b6d8-01e465439e23%2Fbp53x35_processed.png&w=3840&q=75)
Transcribed Image Text:### Physics Problem: Calculating Angular Acceleration
---
**Problem Statement:**
A motorcycle accelerates uniformly from rest and reaches a linear speed of 22.4 m/s in a time of 9.06 s. The radius of each tire is 0.287 m. What is the magnitude of the angular acceleration of each tire?
---
To solve this problem, we need to use the relationship between linear acceleration, angular acceleration, and the radius of the tires.
Let's break down the problem step-by-step:
1. **Linear Speed and Time:**
- Initial linear speed (u) = 0 m/s (since it starts from rest)
- Final linear speed (v) = 22.4 m/s
- Time (t) = 9.06 s
2. **Linear Acceleration:**
Using the formula for linear acceleration (a):
\[
a = \frac{v - u}{t}
\]
Substituting the given values:
\[
a = \frac{22.4\, \text{m/s} - 0\, \text{m/s}}{9.06\, \text{s}}
\]
\[
a = \frac{22.4}{9.06} \approx 2.473\, \text{m/s}^2
\]
3. **Radius of Each Tire (r):**
Radius (r) = 0.287 m
4. **Angular Acceleration:**
Angular acceleration (\(\alpha\)) can be found using the formula:
\[
\alpha = \frac{a}{r}
\]
Substituting the values of linear acceleration and the radius:
\[
\alpha = \frac{2.473\, \text{m/s}^2}{0.287\, \text{m}}
\]
\[
\alpha \approx 8.61\, \text{rad/s}^2
\]
**Conclusion:**
The magnitude of the angular acceleration of each tire is approximately \(8.61\, \text{rad/s}^2\).
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