All collars slide freely. A = 5.0 ft/s downward at the instant shown. Using ICR, determine @AB and wCD. 8 dABZ: 24 in dABy: 18 in dac 12 in dacy=9 in dc12.in dBcy=9.in dcpr 4 in dcDy=dBcy B -12 in- 9 in 9 mm

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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### Determining Angular Velocities Using Instantaneous Center of Rotation (ICR)

In this exercise, we are given a system involving two sliding collars, A and D, connected by rigid bars AB and CD. The objective is to determine the angular velocities of bars AB (\(\omega_{AB}\)) and CD (\(\omega_{CD}\)) using the concept of the Instantaneous Center of Rotation (ICR).

**Given Data:**
- Collar A is moving downward with a velocity (\(v_A\)) of 5.0 ft/s.
- All collars are assumed to slide freely.

The distances between various points of interest are as follows:
- \(v_{A} = 5 \, \text{ft/s}\)
- \(d_{ABx} = 24 \, \text{in}\)
- \(d_{ABy} = 18 \, \text{in}\)
- \(d_{ACx} = 12 \, \text{in}\)
- \(d_{ACy} = 9 \, \text{in}\)
- \(d_{BCx} = 12 \, \text{in}\)
- \(d_{BCy} = 9 \, \text{in}\)
- \(d_{CDx} = -4 \, \text{in}\)

### Diagram

The provided diagram shows the configuration of the system:
- A vertical track at the right side with collar A sliding freely.
- A horizontal track at the bottom housing collar D, which also slides freely.
- A rigid bar AB, inclined at an angle and connecting collar A (at the top right) to point B on another rigid bar CD.
- A smaller bar CD is connected to AB via point C, with point D at the bottom on the horizontal track. 

Key measurements from the diagram:
- The vertical distance between collars A and C is 9 inches.
- The horizontal distance between collars A and B is 12 inches.
- The horizontal distance between points B and C is 4 inches.
- The horizontal distance between points C and D is 8 inches.

### Goals

Using Integrated Control Room (ICR) methodology, we aim to find:
- The angular velocity of bar AB, denoted as \(\omega_{AB}\).
- The angular velocity of bar CD, denoted as \(\omega_{CD}\).

### Approach
1. **Identify ICR for the Bars
Transcribed Image Text:### Determining Angular Velocities Using Instantaneous Center of Rotation (ICR) In this exercise, we are given a system involving two sliding collars, A and D, connected by rigid bars AB and CD. The objective is to determine the angular velocities of bars AB (\(\omega_{AB}\)) and CD (\(\omega_{CD}\)) using the concept of the Instantaneous Center of Rotation (ICR). **Given Data:** - Collar A is moving downward with a velocity (\(v_A\)) of 5.0 ft/s. - All collars are assumed to slide freely. The distances between various points of interest are as follows: - \(v_{A} = 5 \, \text{ft/s}\) - \(d_{ABx} = 24 \, \text{in}\) - \(d_{ABy} = 18 \, \text{in}\) - \(d_{ACx} = 12 \, \text{in}\) - \(d_{ACy} = 9 \, \text{in}\) - \(d_{BCx} = 12 \, \text{in}\) - \(d_{BCy} = 9 \, \text{in}\) - \(d_{CDx} = -4 \, \text{in}\) ### Diagram The provided diagram shows the configuration of the system: - A vertical track at the right side with collar A sliding freely. - A horizontal track at the bottom housing collar D, which also slides freely. - A rigid bar AB, inclined at an angle and connecting collar A (at the top right) to point B on another rigid bar CD. - A smaller bar CD is connected to AB via point C, with point D at the bottom on the horizontal track. Key measurements from the diagram: - The vertical distance between collars A and C is 9 inches. - The horizontal distance between collars A and B is 12 inches. - The horizontal distance between points B and C is 4 inches. - The horizontal distance between points C and D is 8 inches. ### Goals Using Integrated Control Room (ICR) methodology, we aim to find: - The angular velocity of bar AB, denoted as \(\omega_{AB}\). - The angular velocity of bar CD, denoted as \(\omega_{CD}\). ### Approach 1. **Identify ICR for the Bars
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