https://file.aiquickdraw.com/mathbot/file/1727892789422op6j868j.pdf

The exercise 1 from the document asks:

If the roller **R** of the inclined connecting rod moves to the left at a constant speed of \( V_R = 1.2 \, \text{m/s} \), determine:
1. The angular velocity of the link.
2. The velocity of the roller **L** when the angle \( \theta = 20^\circ \).

### Given Data:
- \( V_R = 1.2 \, \text{m/s} \)
- \( LR = 2.3 \, \text{m} \)
- \( \theta = 20^\circ \)

### Solution:

To solve this problem, we will use concepts of relative velocity in mechanisms, specifically in a 4-bar mechanism involving a sliding element.

#### Step 1: Express the velocity of point **R**
The velocity of point **R** is horizontal, as given. We know that the velocity of any point on a rigid body in plane motion can be expressed as:

\[
V_R = V_L + \omega \times LR
\]

Where:
- \( V_L \) is the velocity of point **L**,
- \( \omega \) is the angular velocity of the link,
- \( LR \) is the distance between **L** and **R**.

Since \( V_R \) is horizontal, we can project the velocities along the direction of motion.

#### Step 2: Break down into components
Let’s decompose the velocity and angular relationships, considering that \( \theta = 20^\circ \) affects the angle of the link.

The relationship between the angular velocity \( \omega \) and the linear velocity \( V_R \) is:

\[
V_R = \omega \times LR \times \sin(\theta)
\]

#### Step 3: Solve for \( \omega \)
\[
1.2 \, \text{m/s} = \omega \times 2.3 \, \text{m} \times \sin(20^\circ)
\]

\[
\omega = \frac{1.2}{2.3 \times \sin(20^\circ)}
\]

Using \( \sin(20^\circ) \approx 0.342 \):

\[
\omega \approx \frac{1.2}{2.3 \times 0.342} \approx \frac{1.2}{0.787} \approx 1.525 \, \text{rad/s}
\]

#### Step 4: Find the velocity of roller **L**
The velocity of point **L** can be found using the same relationship:

\[
V_L = \omega \times LR \times \cos(\theta)
\]

\[
V_L = 1.525 \times 2.3 \times \cos(20^\circ)
\]

Using \( \cos(20^\circ) \approx 0.94 \):

\[
V_L \approx 1.525 \times 2.3 \times 0.94 \approx 3.3 \, \text{m/s}
\]

### Final Answer:
1. The angular velocity \( \omega \) of the link is approximately \( 1.525 \, \text{rad/s} \).
2. The velocity of the roller **L** is approximately \( 3.3 \, \text{m/s} \).

Let me know if you would like more details on any part of the process!

---

**Next questions:**
1. How does the velocity analysis change if the angle \( \theta \) is different?
2. What if the roller moves in the opposite direction?
3. How would angular acceleration affect the results?
4. What happens if \( LR \) is increased?
5. How can we apply the relative velocity method for different configurations?

**Tip:** Understanding vector decomposition is essential when working with velocities in mechanisms.

Learn how to determine the angular velocity of a link and the velocity of a roller in a sliding rod mechanism. This detailed solution uses the relative velocity method and trigonometric relationships, with key formulas provided. Suitable for engineering students.

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