A particle is on a circular path as shown in the figure. At position shown, the particle has speed v = 5 m/s, which is increasing at a rate of 7.5 m/s². What is the magnitude of its acceleration for the same position? v = 5 m/s
A particle is on a circular path as shown in the figure. At position shown, the particle has speed v = 5 m/s, which is increasing at a rate of 7.5 m/s². What is the magnitude of its acceleration for the same position? v = 5 m/s
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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![### Problem Statement
A particle is on a circular path as shown in the figure. At the position shown, the particle has a speed \( v = 5 \ \text{m/s} \), which is increasing at a rate of \( 7.5 \ \text{m/s}^2 \). What is the magnitude of its acceleration for the same position?
### Diagram Description
The diagram presents a circle with a radius of \( 2.5 \ \text{m} \). A particle is indicated by a blue dot on the circumference of the circle. The velocity vector (\( v \)) at the point where the particle is located is directed tangentially to the circle upwards, with \( v = 5 \ \text{m/s} \).
### Choices for the Solution
- \( a = 10 \ \text{m/s}^2 \)
- \( a = 5 \ \text{m/s}^2 \)
- \( a = 12.5 \ \text{m/s}^2 \)
- \( a = 7.5 \ \text{m/s}^2 \)
### Explanation
To find the magnitude of the total acceleration \( a \), we need to consider both the tangential acceleration \( a_t \) and the centripetal (radial) acceleration \( a_r \).
1. **Tangential Acceleration (\( a_t \)):**
Given as \( 7.5 \ \text{m/s}^2 \).
2. **Centripetal Acceleration (\( a_r \)):**
This can be calculated using the formula:
\[
a_r = \frac{v^2}{r}
\]
Where:
- \( v = 5 \ \text{m/s} \)
- \( r = 2.5 \ \text{m} \)
Substituting the values:
\[
a_r = \frac{(5)^2}{2.5} = \frac{25}{2.5} = 10 \ \text{m/s}^2
\]
3. **Total Acceleration (\( a \)):**
The magnitude of the total acceleration \( a \) is the vector sum of the tangential and centripetal accelerations:
\[
a = \sqrt{a_t^2 + a_r^](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2b7dd8f1-dbab-4607-8b33-bde9974f18c9%2F2a7d365f-6702-445c-a32f-3e709fa231a2%2Firlladm_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
A particle is on a circular path as shown in the figure. At the position shown, the particle has a speed \( v = 5 \ \text{m/s} \), which is increasing at a rate of \( 7.5 \ \text{m/s}^2 \). What is the magnitude of its acceleration for the same position?
### Diagram Description
The diagram presents a circle with a radius of \( 2.5 \ \text{m} \). A particle is indicated by a blue dot on the circumference of the circle. The velocity vector (\( v \)) at the point where the particle is located is directed tangentially to the circle upwards, with \( v = 5 \ \text{m/s} \).
### Choices for the Solution
- \( a = 10 \ \text{m/s}^2 \)
- \( a = 5 \ \text{m/s}^2 \)
- \( a = 12.5 \ \text{m/s}^2 \)
- \( a = 7.5 \ \text{m/s}^2 \)
### Explanation
To find the magnitude of the total acceleration \( a \), we need to consider both the tangential acceleration \( a_t \) and the centripetal (radial) acceleration \( a_r \).
1. **Tangential Acceleration (\( a_t \)):**
Given as \( 7.5 \ \text{m/s}^2 \).
2. **Centripetal Acceleration (\( a_r \)):**
This can be calculated using the formula:
\[
a_r = \frac{v^2}{r}
\]
Where:
- \( v = 5 \ \text{m/s} \)
- \( r = 2.5 \ \text{m} \)
Substituting the values:
\[
a_r = \frac{(5)^2}{2.5} = \frac{25}{2.5} = 10 \ \text{m/s}^2
\]
3. **Total Acceleration (\( a \)):**
The magnitude of the total acceleration \( a \) is the vector sum of the tangential and centripetal accelerations:
\[
a = \sqrt{a_t^2 + a_r^
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