In the figure, you see a particle moving on a circular path. If the angle is given as 0 = 3t2 in ra where t is in seconds, what is the speed of the particle at t = 1 s? 0 0.5 m
In the figure, you see a particle moving on a circular path. If the angle is given as 0 = 3t2 in ra where t is in seconds, what is the speed of the particle at t = 1 s? 0 0.5 m
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: Speed of a Particle on a Circular Path**
In the figure, you see a particle moving on a circular path. If the angle is given as \( \theta = 3t^2 \) in radians where \( t \) is in seconds, what is the speed of the particle at \( t = 1 \) second?
**Diagram Explanation:**
The diagram shows a particle moving in a circular path with a radius of 0.5 meters. The angle \( \theta \) made by the particle at any point is measured from the positive \( x \)-axis.
- \( r \) represents the radius of the circular path.
- \( \theta \) is the angle in radians that the particle makes in its circular motion.
**Calculation:**
To find the speed \( v \) of the particle, we need to determine the angular velocity \( \omega \) first:
1. Given:
\[ \theta = 3t^2 \]
2. The angular velocity \( \omega \) is the time derivative of \( \theta \):
\[ \omega = \frac{d\theta}{dt} = \frac{d(3t^2)}{dt} = 6t \]
3. At \( t = 1 \text{ second} \):
\[ \omega = 6 \times 1 = 6 \text{ rad/sec} \]
4. The linear speed \( v \) of the particle is related to the angular velocity by the formula:
\[ v = \omega \times r \]
5. Given \( r = 0.5 \text{ m} \):
\[ v = 6 \text{ rad/sec} \times 0.5 \text{ m} = 3 \text{ m/s} \]
**Answer Options:**
- \( \) \( 6 \text{ m/s} \)
- \(\) \( 1.5 \text{ m/s} \)
- \(\) \( 3 \text{ m/s} \)
- \(\) \( 9 \text{ m/s} \)
**Correct Answer:**
\[ 3 \text{ m/s} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F57026d2c-4d86-446e-bb74-abb5fa8f33cd%2F7e20d03f-cf9e-4951-8add-31d2f9c5de3c%2F6qms1d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem: Speed of a Particle on a Circular Path**
In the figure, you see a particle moving on a circular path. If the angle is given as \( \theta = 3t^2 \) in radians where \( t \) is in seconds, what is the speed of the particle at \( t = 1 \) second?
**Diagram Explanation:**
The diagram shows a particle moving in a circular path with a radius of 0.5 meters. The angle \( \theta \) made by the particle at any point is measured from the positive \( x \)-axis.
- \( r \) represents the radius of the circular path.
- \( \theta \) is the angle in radians that the particle makes in its circular motion.
**Calculation:**
To find the speed \( v \) of the particle, we need to determine the angular velocity \( \omega \) first:
1. Given:
\[ \theta = 3t^2 \]
2. The angular velocity \( \omega \) is the time derivative of \( \theta \):
\[ \omega = \frac{d\theta}{dt} = \frac{d(3t^2)}{dt} = 6t \]
3. At \( t = 1 \text{ second} \):
\[ \omega = 6 \times 1 = 6 \text{ rad/sec} \]
4. The linear speed \( v \) of the particle is related to the angular velocity by the formula:
\[ v = \omega \times r \]
5. Given \( r = 0.5 \text{ m} \):
\[ v = 6 \text{ rad/sec} \times 0.5 \text{ m} = 3 \text{ m/s} \]
**Answer Options:**
- \( \) \( 6 \text{ m/s} \)
- \(\) \( 1.5 \text{ m/s} \)
- \(\) \( 3 \text{ m/s} \)
- \(\) \( 9 \text{ m/s} \)
**Correct Answer:**
\[ 3 \text{ m/s} \]
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