The circle's radius in Case A is twice that of Case B. The period (time for one circle) in Case A is three times that of Case B. How does the speed in Case A compare to the speed in Case B? Case A Case B 3-т T R 2-R O 3(3.14)/2 O 2(3.14)/3 О 3/2 2/3
Displacement, Velocity and Acceleration
In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.
Linear Displacement
The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.
The circle’s radius in Case A is twice that of Case B. The period (time for one circle) in Case A is three times that of Case B. How does the speed in Case A compare to the speed in Case B?
![### Understanding the Relationship Between Radius and Speed in Circular Motion
**Problem Description:**
The circle's radius in Case A is twice that of Case B. The period (time for one circle) in Case A is three times that of Case B. How does the speed in Case A compare to the speed in Case B?
**Diagram Explanation:**
**Case A:**
- Radius (R): \( 2R \)
- Period (Time for one circle): \( 3T \)
**Case B:**
- Radius (R): \( R \)
- Period: \( T \)
**Options for Comparison:**
1. \( \frac{3(3.14)}{2} \)
2. \( \frac{2(3.14)}{3} \)
3. \( \frac{3}{2} \)
4. \( \frac{2}{3} \)
Let's analyze the speed in both cases to determine the correct comparison.
### Explanation and Solution
Speed (v) in circular motion can be calculated using the formula:
\[ v = \frac{2\pi r}{T} \]
**Case A:**
- Radius = \( 2R \)
- Period = \( 3T \)
\[ v_A = \frac{2\pi (2R)}{3T} = \frac{4\pi R}{3T} \]
**Case B:**
- Radius = \( R \)
- Period = \( T \)
\[ v_B = \frac{2\pi R}{T} \]
Now, compare \( v_A \) and \( v_B \):
\[ \frac{v_A}{v_B} = \frac{\frac{4\pi R}{3T}}{\frac{2\pi R}{T}} = \frac{4\pi R}{3T} \times \frac{T}{2\pi R} = \frac{4}{3} \times \frac{1}{2} = \frac{4}{6} = \frac{2}{3} \]
Therefore, the speed in Case A is \(\frac{2}{3}\) of the speed in Case B.
### Correct Answer:
\(\frac{2}{3}\)
**Educational Insight:**
This problem helps in understanding the relationship between the radius of a circle, the period of rotation, and the speed in uniform circular motion. It demonstrates](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3f0787bb-1f84-49a3-ae5b-f5fa6271d00e%2F1daa622d-5cad-4011-955b-e3e180e55e70%2F9hf33ym_processed.png&w=3840&q=75)
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