The problem describes a four-bar linkage mechanism with links AB, BC, and CD. The lengths of the links are given as AB = 100 mm, BC = 150 mm, and CD = 100 mm. At the given instant, AB is vertical, CD is horizontal, and BC is inclined at 30 degrees to the horizontal. The angular velocity of link AB is 2 rad/s clockwise. We need to find the angular velocity and angular acceleration of link CD at this instant.

1. Problem Description

The problem describes a four-bar linkage mechanism with links AB, BC, and CD. The lengths of the links are given as AB = 100 mm, BC = 150 mm, and CD = 100 mm. At the given instant, AB is vertical, CD is horizontal, and BC is inclined at 30 degrees to the horizontal. The angular velocity of link AB is 2 rad/s clockwise. We need to find the angular velocity and angular acceleration of link CD at this instant.

2. Solution Steps

(i) Angular velocity of link CD.

First, we need to determine the loop closure equation in complex form. We will represent the position of each link as a complex number.

Let $r_{AB}$, $r_{BC}$, and $r_{CD}$ be the lengths of links AB, BC, and CD, respectively.

Let $\theta_{AB}$, $\theta_{BC}$, and $\theta_{CD}$ be the angles of links AB, BC, and CD, respectively, measured counter-clockwise from the horizontal axis.

The loop closure equation is:

$r_{AB}e^{i\theta_{AB}} + r_{BC}e^{i\theta_{BC}} = r_{AD} + r_{CD}e^{i\theta_{CD}}$

In this case, $r_{AD}$ is zero. We are given that $\theta_{AB} = 90^\circ = \pi/2$, $\theta_{BC} = 30^\circ = \pi/6$, and $\theta_{CD} = 0^\circ = 0$.

So,

$100e^{i\pi/2} + 150e^{i\pi/6} = 100e^{i0}$

$100(\cos(\pi/2) + i\sin(\pi/2)) + 150(\cos(\pi/6) + i\sin(\pi/6)) = 100(\cos(0) + i\sin(0))$

$100(0 + i) + 150(\sqrt{3}/2 + i/2) = 100(1 + 0i)$

$100i + 75\sqrt{3} + 75i = 100$

$75\sqrt{3} + 175i = 100$

This is not valid. So instead, let us take AD as fixed on the x axis. So the equation is

$r_{AB}e^{i\theta_{AB}} + r_{BC}e^{i\theta_{BC}} = r_{AD} + r_{CD}e^{i\theta_{CD}}$

$r_{AD} = r_{AB}e^{i\theta_{AB}} + r_{BC}e^{i\theta_{BC}} - r_{CD}e^{i\theta_{CD}}$

$r_{AD} = 100e^{i\pi/2} + 150e^{i\pi/6} - 100e^{i0}$

$r_{AD} = 100(i) + 150(\sqrt{3}/2 + i/2) - 100$

$r_{AD} = 75\sqrt{3} - 100 + 175i$

Now we differentiate the loop closure equation with respect to time:

$i r_{AB} \omega_{AB} e^{i\theta_{AB}} + i r_{BC} \omega_{BC} e^{i\theta_{BC}} = i r_{CD} \omega_{CD} e^{i\theta_{CD}}$

$100 (2) e^{i(\pi/2 + \pi/2)} + 150 \omega_{BC} e^{i(\pi/6 + \pi/2)} = 100 \omega_{CD} e^{i(\pi/2)}$

$100 (2) i + 150 \omega_{BC} e^{i(\pi/6 )} = 100 \omega_{CD}$

Here $\omega_{AB} = -2$, since AB rotates clockwise.

$i r_{AB} \omega_{AB} e^{i\theta_{AB}} + i r_{BC} \omega_{BC} e^{i\theta_{BC}} = i r_{CD} \omega_{CD} e^{i\theta_{CD}}$

$i 100 (-2) e^{i(\pi/2)} + i 150 \omega_{BC} e^{i(\pi/6)} = i 100 \omega_{CD} e^{i(0)}$

$-200 i (i) + i 150 \omega_{BC} (\sqrt{3}/2 + i/2) = 100 i\omega_{CD}$

$200 + 150 i\omega_{BC} (\sqrt{3}/2 + i/2) = 100 \omega_{CD}i$

Equating real and imaginary parts:

$200 + 150 \omega_{BC} (-\frac{1}{2}) = 0$

$150 \omega_{BC} \frac{\sqrt{3}}{2} = 100 \omega_{CD}$

$\omega_{BC} = 200/75 = 8/3$ rad/s

$150 (\frac{8}{3}) \frac{\sqrt{3}}{2} = 100 \omega_{CD}$

$400\sqrt{3} = 200\omega_{CD}$

$\omega_{CD} = 2\sqrt{3}/2 = 2\sqrt{3}$ rad/s.

(ii) Angular acceleration of link CD.

Differentiating the velocity equation with respect to time:

$i r_{AB} \alpha_{AB} e^{i\theta_{AB}} - r_{AB} \omega_{AB}^2 e^{i\theta_{AB}} + i r_{BC} \alpha_{BC} e^{i\theta_{BC}} - r_{BC} \omega_{BC}^2 e^{i\theta_{BC}} = i r_{CD} \alpha_{CD} e^{i\theta_{CD}} - r_{CD} \omega_{CD}^2 e^{i\theta_{CD}}$

Since the angular velocity of AB is constant, $\alpha_{AB} = 0$. So,

$- r_{AB} \omega_{AB}^2 e^{i\theta_{AB}} + i r_{BC} \alpha_{BC} e^{i\theta_{BC}} - r_{BC} \omega_{BC}^2 e^{i\theta_{BC}} = i r_{CD} \alpha_{CD} e^{i\theta_{CD}} - r_{CD} \omega_{CD}^2 e^{i\theta_{CD}}$

$-(100) (-2)^2 e^{i\pi/2} + i(150)\alpha_{BC} e^{i\pi/6} - (150)(\frac{8}{3})^2 e^{i\pi/6} = i(100)\alpha_{CD} - (100)(2\sqrt{3})^2$

$-400i + 150i\alpha_{BC} (\sqrt{3}/2 + i/2) - 150(\frac{64}{9}) (\sqrt{3}/2 + i/2) = 100i\alpha_{CD} - 1200$

$-400i + 75\sqrt{3}i \alpha_{BC} - 75\alpha_{BC} - (\frac{4800}{9}) (\sqrt{3}/2 + i/2) = 100i\alpha_{CD} - 1200$

$-400i + 75\sqrt{3}i \alpha_{BC} - 75\alpha_{BC} - \frac{2400\sqrt{3}}{9} - \frac{2400}{9} i = 100i\alpha_{CD} - 1200$

Equating real parts:

$-75\alpha_{BC} - \frac{800\sqrt{3}}{3} = -1200$

$75\alpha_{BC} = 1200 - \frac{800\sqrt{3}}{3} = \frac{3600 - 800\sqrt{3}}{3}$

$\alpha_{BC} = \frac{3600 - 800\sqrt{3}}{225} = \frac{144-32\sqrt{3}}{9}$

Equating imaginary parts:

$-400 + 75\sqrt{3}\alpha_{BC} - \frac{800}{3} = 100\alpha_{CD}$

$75\sqrt{3}\alpha_{BC} - \frac{2000}{3} = 100\alpha_{CD}$

$75\sqrt{3}(\frac{144-32\sqrt{3}}{9}) - \frac{2000}{3} = 100\alpha_{CD}$

$25\sqrt{3}(\frac{144-32\sqrt{3}}{3}) - \frac{2000}{3} = 100\alpha_{CD}$

$\frac{3600\sqrt{3} - 2400}{3} - \frac{2000}{3} = 300\alpha_{CD}$

$\frac{3600\sqrt{3} - 4400}{3} = 300\alpha_{CD}$

$100\alpha_{CD} = \frac{3600\sqrt{3} - 4400}{3} \implies$

$\alpha_{CD} = \frac{36\sqrt{3} - 44}{3} \approx 1.71 \frac{rad}{s^2}$

3. Final Answer

(i) The angular velocity of link CD is $2\sqrt{3}$ rad/s.

(ii) The angular acceleration of link CD is $\frac{36\sqrt{3} - 44}{3}$ rad/s².