Question: User: 16-110. Determine the angular acceleration of link AB if link CD has the angular velocity and angular deceleration shown.
Prob. 16-110
0.6 m
0.6 m
0.3 m
\alpha_{CD} = 4 \text{ rad/s}^2
\omega_{CD} = 2 \text{ rad/s}
Graph shape: a four-bar linkage diagram with two vertical links AB and CD, connected by a horizontal top link BC; A and D are ground pivots at the bottom, B is on the left link, C is on the right link, and the dimensions shown are 0.6 m across the top, 0.6 m from the top down to B/C level on the left, and 0.3 m from B down to A on the left. position_hint=top-left
Graph shape: the blue curved arrows near link CD indicate clockwise rotation with angular acceleration \alpha_{CD} and angular velocity \omega_{CD}. position_hint=bottom-right
Solve this problem in this image.
1. **Problem Statement:**
Determine the angular acceleration $\alpha_{AB}$ of link AB given that link CD has angular velocity $\omega_{CD} = 2$ rad/s and angular acceleration $\alpha_{CD} = 4$ rad/s$^2$.
The linkage dimensions are:
- Length BC = 0.6 m
- Length CD = 0.6 m
- Length AB = 0.3 m
2. **Understanding the Mechanism:**
This is a four-bar linkage with vertical links AB and CD connected by horizontal link BC.
3. **Known values:**
$$\omega_{CD} = 2 \text{ rad/s (clockwise)}$$
$$\alpha_{CD} = 4 \text{ rad/s}^2 \text{ (clockwise)}$$
4. **Goal:**
Find $$\alpha_{AB}$$.
5. **Approach:**
Use relative angular velocity and acceleration relations for the four-bar linkage.
6. **Step 1: Angular velocity relation**
Since BC is common to both links AB and CD, the velocity of point B and C must be consistent.
Velocity of point B on AB:
$$v_B = \omega_{AB} \times AB$$
Velocity of point C on CD:
$$v_C = \omega_{CD} \times CD$$
Since BC is rigid and horizontal, velocities at B and C must satisfy:
$$v_B = v_C$$
Given lengths:
$$AB = 0.3 \text{ m}, \quad CD = 0.6 \text{ m}$$
Using velocity ratio:
$$\omega_{AB} \times 0.3 = \omega_{CD} \times 0.6$$
Solve for $\omega_{AB}$:
$$\omega_{AB} = \frac{0.6}{0.3} \times 2 = 2 \times 2 = 4 \text{ rad/s}$$
7. **Step 2: Angular acceleration relation**
Angular acceleration of AB relates to that of CD by:
$$\alpha_{AB} \times 0.3 = \alpha_{CD} \times 0.6$$
Solve for $\alpha_{AB}$:
$$\alpha_{AB} = \frac{0.6}{0.3} \times 4 = 2 \times 4 = 8 \text{ rad/s}^2$$
8. **Step 3: Direction**
Since $\omega_{CD}$ and $\alpha_{CD}$ are clockwise, and the linkage is rigid, $\alpha_{AB}$ is also clockwise.
**Final answer:**
$$\boxed{\alpha_{AB} = 8 \text{ rad/s}^2 \text{ (clockwise)}}$$