Four-bar angular velocities by IC method

Instantaneous centres of the four links (1 = AD, 2 = AB, 3 = BC, 4 = CD)

Number of ICs $= \dfrac{4 \times 3}{2} = 6$. Four are pin joints:

$I_{12} = A\,(0, 0)$, $I_{23} = B\,(50.0, 86.6)$, $I_{34} = C\,(273.7, 198.3)$, $I_{14} = D\,(300, 0)$ mm.

By Kennedy's theorem the remaining two are:

• $I_{13}$ is the intersection of line AB (through $I_{12}$, $I_{23}$) and line DC (through $I_{14}$, $I_{34}$).
• $I_{24}$ is the intersection of line BC (through $I_{23}$, $I_{34}$) and line AD (through $I_{12}$, $I_{14}$).

Solving the line equations:

$$I_{13} = (243.9,\ 422.5)\ \text{mm}, \qquad I_{24} = (-123.5,\ 0)\ \text{mm}$$

   I13 *  (243.9, 422.5)
        \  .
         \    C
          B  /  \
         /      \
 I24 *--A---------D------

(Not to scale; $I_{24}$ lies on AD produced beyond A.)

Angular velocity of rocker CD

$I_{24}$ is a point common to links 2 and 4, so its velocity is the same whether taken on AB or on CD:

$$\omega_2 \cdot I_{12}I_{24} = \omega_4 \cdot I_{14}I_{24}$$ $$I_{12}I_{24} = 123.5\ \text{mm}, \qquad I_{14}I_{24} = 300 + 123.5 = 423.5\ \text{mm}$$ $$\omega_4 = 10 \times \frac{123.5}{423.5} = 2.916\ \text{rad/s}$$

$I_{24}$ lies outside A and D on the same side of both, so $\omega_4$ has the same sense as $\omega_2$: anticlockwise.

Angular velocity of coupler BC

$I_{13}$ is the IC of links 1 and 3. The pin B is common to links 2 and 3:

$$v_B = \omega_2 \cdot AB = 10 \times 0.100 = 1.0\ \text{m/s}$$ $$\omega_3 = \frac{v_B}{I_{13}B} = \frac{1.0}{387.8/1000} = 2.578\ \text{rad/s}$$

B lies between $I_{12}$ and $I_{13}$ on the same line, so the senses of $\omega_2$ and $\omega_3$ are opposite: $\omega_3$ is clockwise.

Velocity of C

$$v_C = \omega_4 \cdot CD = 2.916 \times 0.200 = 0.583\ \text{m/s}$$

perpendicular to CD (check: $v_C = \omega_3 \cdot I_{13}C$ gives the same value).

Answer: $\omega_3 = 2.578$ rad/s (clockwise), $\omega_4 = 2.916$ rad/s (anticlockwise), $v_C = 0.583$ m/s.