A four-bar linkage mechanism is given. The lengths of the links are $AB = 100$ mm, $BC = 150$ mm, and $CD = 100$ mm. At the given instant, link $AB$ is vertical, link $CD$ is horizontal, and link $BC$ is inclined at $30^\circ$ to the horizontal. Link $AB$ rotates with a constant angular speed of 2 rad/s in the clockwise direction. We need to find the angular velocity and the angular acceleration of link $CD$ at the given instant.

1. Problem Description

A four-bar linkage mechanism is given. The lengths of the links are $AB = 100$ mm, $BC = 150$ mm, and $CD = 100$ mm. At the given instant, link $AB$ is vertical, link $CD$ is horizontal, and link $BC$ is inclined at $30^\circ$ to the horizontal. Link $AB$ rotates with a constant angular speed of 2 rad/s in the clockwise direction. We need to find the angular velocity and the angular acceleration of link $CD$ at the given instant.

2. Solution Steps

First, we convert all the lengths to meters: $AB = 0.1$ m, $BC = 0.15$ m, $CD = 0.1$ m.

Since $AB$ rotates at a constant angular velocity, the angular acceleration of $AB$ is zero, i.e., $\alpha_{AB} = 0$. The angular velocity of $AB$ is $\omega_{AB} = -2$ rad/s (clockwise is negative).

Velocity Analysis:

We can write the velocity equation as:

$v_B = v_A + \omega_{AB} \times r_{B/A}$

$v_C = v_B + \omega_{BC} \times r_{C/B}$

$v_D = v_C + \omega_{CD} \times r_{D/C}$

Since $A$ and $D$ are fixed points, $v_A = 0$ and $v_D = 0$.

$v_B = \omega_{AB} \times r_{B/A} = -2k \times (0.1j) = -0.2i$ m/s

$v_C = v_B + \omega_{BC} \times r_{C/B} = \omega_{BC} \times (0.15\cos(30^\circ)i + 0.15\sin(30^\circ)j) - 0.2i$

$v_C = -0.2i + 0.15\omega_{BC}\cos(30^\circ)j - 0.15\omega_{BC}\sin(30^\circ)i = (-0.2 - 0.075\omega_{BC})i + 0.1299\omega_{BC}j$

$v_D = v_C + \omega_{CD} \times r_{D/C} = 0$

$0 = (-0.2 - 0.075\omega_{BC})i + 0.1299\omega_{BC}j + \omega_{CD}k \times (-0.1i)$

$0 = (-0.2 - 0.075\omega_{BC})i + 0.1299\omega_{BC}j - 0.1\omega_{CD}j$

Equating the $i$ and $j$ components:

$-0.2 - 0.075\omega_{BC} = 0 \implies \omega_{BC} = -\frac{0.2}{0.075} = -2.6667$ rad/s

$0.1299\omega_{BC} - 0.1\omega_{CD} = 0 \implies \omega_{CD} = 1.299\omega_{BC} = 1.299(-2.6667) = -3.464$ rad/s

Acceleration Analysis:

$a_B = a_A + \alpha_{AB} \times r_{B/A} - \omega_{AB}^2 r_{B/A} = 0 + 0 - (-2)^2(0.1j) = -0.4j$ m/s$^2$

$a_C = a_B + \alpha_{BC} \times r_{C/B} - \omega_{BC}^2 r_{C/B}$

$a_D = a_C + \alpha_{CD} \times r_{D/C} - \omega_{CD}^2 r_{D/C} = 0$

$a_C = -0.4j + \alpha_{BC}k \times (0.15\cos(30^\circ)i + 0.15\sin(30^\circ)j) - (-2.6667)^2(0.15\cos(30^\circ)i + 0.15\sin(30^\circ)j)$

$a_C = -0.4j + 0.1299\alpha_{BC}j - 0.075\alpha_{BC}i - 1.0(0.15\cos(30^\circ)i + 0.15\sin(30^\circ)j) = -0.1299i - 0.075j$

$a_C = (-0.1299 - 0.075\alpha_{BC})i + (-0.4 + 0.1299\alpha_{BC} - 0.533)j = (-0.1299 - 0.075\alpha_{BC})i + (-0.933 + 0.1299\alpha_{BC})j$

$a_D = 0 = a_C + \alpha_{CD}k \times (-0.1i) - \omega_{CD}^2 (-0.1i)$

$0 = (-0.1299 - 0.075\alpha_{BC})i + (-0.933 + 0.1299\alpha_{BC})j - 0.1\alpha_{CD}j - (-3.464)^2(-0.1i)$

$0 = (-0.1299 - 0.075\alpha_{BC})i + (-0.933 + 0.1299\alpha_{BC})j - 0.1\alpha_{CD}j + 1.2i$

$0 = (1.2 - 0.1299 - 0.075\alpha_{BC})i + (-0.933 + 0.1299\alpha_{BC} - 0.1\alpha_{CD})j$

$0 = (1.0701 - 0.075\alpha_{BC})i + (-0.933 + 0.1299\alpha_{BC} - 0.1\alpha_{CD})j$

$1.0701 - 0.075\alpha_{BC} = 0 \implies \alpha_{BC} = \frac{1.0701}{0.075} = 14.268$ rad/s$^2$

$-0.933 + 0.1299\alpha_{BC} - 0.1\alpha_{CD} = 0$

$-0.933 + 0.1299(14.268) - 0.1\alpha_{CD} = 0$

$-0.933 + 1.8535 - 0.1\alpha_{CD} = 0$

$0.9205 - 0.1\alpha_{CD} = 0$

$\alpha_{CD} = \frac{0.9205}{0.1} = 9.205$ rad/s$^2$

3. Final Answer

The angular velocity of link CD is $\omega_{CD} = -3.464$ rad/s.

The angular acceleration of link CD is $\alpha_{CD} = 9.205$ rad/s$^2$.