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 SolutionSpeed (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

Name: The circle’s radius in Case A is twice that of Case B. The period (tim
Uploaded: 2024-03-05T11:56:34.000Z
Channel: Bartleby
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? ### Understanding the Relationship Between Radius and Speed