The problem describes a four-bar linkage mechanism. Link AB is vertical, BC is horizontal, and CD is inclined at 60 degrees to the horizontal. We are given AB = CD = 150 mm and BC = 200 mm. Link AB rotates at a constant angular velocity of 2 rad/s counterclockwise. We need to find the angular velocity and angular acceleration of link CD.

1. Problem Description

The problem describes a four-bar linkage mechanism. Link AB is vertical, BC is horizontal, and CD is inclined at 60 degrees to the horizontal. We are given AB = CD = 150 mm and BC = 200 mm. Link AB rotates at a constant angular velocity of 2 rad/s counterclockwise. We need to find the angular velocity and angular acceleration of link CD.

2. Solution Steps

First, establish the loop closure equation:

$\vec{AB} + \vec{BC} = \vec{AD} = \vec{AC} + \vec{CD}$

Differentiating with respect to time gives the velocity equation:

$\vec{V_B} + \vec{V_{C/B}} = \vec{V_D}$

Where:

$\vec{V_B} = \omega_{AB} \times \vec{AB}$

$\vec{V_{C/B}} = \omega_{BC} \times \vec{BC}$

$\vec{V_D} = \omega_{CD} \times \vec{CD}$

Also, the acceleration equation is:

$\vec{A_B} + \vec{A_{C/B}} = \vec{A_D}$

Where:

$\vec{A_B} = \alpha_{AB} \times \vec{AB} - \omega_{AB}^2 \vec{AB}$

$\vec{A_{C/B}} = \alpha_{BC} \times \vec{BC} - \omega_{BC}^2 \vec{BC}$

$\vec{A_D} = \alpha_{CD} \times \vec{CD} - \omega_{CD}^2 \vec{CD}$

Since AB is vertical and rotating counter-clockwise, we have:

$\vec{V_B} = \omega_{AB} \times \vec{AB} = 2 \hat{k} \times 0.15 \hat{j} = -0.3 \hat{i} \ m/s$

Let $\omega_{BC} = \omega_{BC} \hat{k}$ and $\omega_{CD} = \omega_{CD} \hat{k}$. Since BC is horizontal:

$\vec{V_{C/B}} = \omega_{BC} \times \vec{BC} = \omega_{BC} \hat{k} \times 0.2 \hat{i} = 0.2 \omega_{BC} \hat{j} \ m/s$

The link CD is inclined at $60^{\circ}$, so

$\vec{CD} = 0.15 \cos(60^{\circ}) \hat{i} + 0.15 \sin(60^{\circ}) \hat{j} = 0.075 \hat{i} + 0.1299 \hat{j} \ m$

$\vec{V_D} = \omega_{CD} \times \vec{CD} = \omega_{CD} \hat{k} \times (0.075 \hat{i} + 0.1299 \hat{j}) = -0.1299 \omega_{CD} \hat{i} + 0.075 \omega_{CD} \hat{j} \ m/s$

Substituting these into the velocity equation:

$-0.3 \hat{i} + 0.2 \omega_{BC} \hat{j} = -0.1299 \omega_{CD} \hat{i} + 0.075 \omega_{CD} \hat{j}$

Equating components:

$-0.3 = -0.1299 \omega_{CD}$

$0.2 \omega_{BC} = 0.075 \omega_{CD}$

Solving for $\omega_{CD}$:

$\omega_{CD} = \frac{0.3}{0.1299} = 2.309 \ rad/s$

Solving for $\omega_{BC}$:

$\omega_{BC} = \frac{0.075}{0.2} \omega_{CD} = 0.375(2.309) = 0.866 \ rad/s$

Now, find the acceleration.

$\vec{A_B} = \alpha_{AB} \times \vec{AB} - \omega_{AB}^2 \vec{AB} = 0 - (2^2) (0.15 \hat{j}) = -0.6 \hat{j} \ m/s^2$

$\vec{A_{C/B}} = \alpha_{BC} \times \vec{BC} - \omega_{BC}^2 \vec{BC} = \alpha_{BC} \hat{k} \times 0.2 \hat{i} - (0.866)^2 (0.2 \hat{i}) = 0.2 \alpha_{BC} \hat{j} - 0.15 \hat{i} \ m/s^2$

$\vec{A_D} = \alpha_{CD} \times \vec{CD} - \omega_{CD}^2 \vec{CD} = \alpha_{CD} \hat{k} \times (0.075 \hat{i} + 0.1299 \hat{j}) - (2.309)^2 (0.075 \hat{i} + 0.1299 \hat{j}) = -0.1299 \alpha_{CD} \hat{i} + 0.075 \alpha_{CD} \hat{j} - 0.3997 \hat{i} - 0.6924 \hat{j} \ m/s^2$

Substituting these into the acceleration equation:

$-0.6 \hat{j} - 0.15 \hat{i} + 0.2 \alpha_{BC} \hat{j} = -0.1299 \alpha_{CD} \hat{i} + 0.075 \alpha_{CD} \hat{j} - 0.3997 \hat{i} - 0.6924 \hat{j}$

Equating components:

$-0.15 = -0.1299 \alpha_{CD} - 0.3997$

$-0.6 + 0.2 \alpha_{BC} = 0.075 \alpha_{CD} - 0.6924$

Solving for $\alpha_{CD}$:

$0.1299 \alpha_{CD} = -0.3997 + 0.15$

$\alpha_{CD} = \frac{-0.2497}{0.1299} = -1.922 \ rad/s^2$

3. Final Answer

(i) Angular velocity of link CD: 2.309 rad/s

(ii) Angular acceleration of link CD: -1.922 rad/s^2