Velocity analysis of slider-crank mechanism

Given: $OA = r = 0.2$ m, $AB = l = 0.8$ m, $N = 300$ rpm counter-clockwise, $\theta = 60^\circ$.

        A (pin)
       /\
      /  \  AB = 0.8
     /60  \
   O-------B-------> slider (horizontal guide)

Crank speed

$$\omega_{OA} = \frac{2\pi \times 300}{60} = 31.416\ \text{rad/s} \qquad v_A = \omega_{OA}\,r = 6.283\ \text{m/s}$$

$v_A$ is perpendicular to OA. Its components: $v_{Ax} = -v_A\sin60^\circ = -5.441$, $v_{Ay} = v_A\cos60^\circ = 3.142$ m/s.

Geometry of the rod

$\sin\phi = \dfrac{r\sin60^\circ}{l} = \dfrac{0.1732}{0.8} = 0.2165$, so the rod angle $\phi = 12.50^\circ$. Horizontal projection of AB: $d = l\cos\phi = 0.7810$ m.

Relative velocity method

$\mathbf{v}_B = \mathbf{v}_A + \mathbf{v}_{B/A}$, with $\mathbf{v}_B$ horizontal and $\mathbf{v}_{B/A}$ perpendicular to AB (magnitude $\omega_{AB}\times l$).

Vertical components (B has none): $v_{Ay} = \omega_{AB}\,d$

$$\omega_{AB} = \frac{3.142}{0.7810} = 4.022\ \text{rad/s (clockwise)}$$

Horizontal components:

$$v_B = v_{Ax} - \omega_{AB}(AB\text{ vertical drop}) = -5.441 - (4.022)(0.1732) = -6.138\ \text{m/s}$$

The negative sign means the slider moves toward O (to the left).

Verification by instantaneous centre

I is the intersection of the perpendicular to $v_A$ at A (line OA extended) and the perpendicular to $v_B$ at B (vertical through B).

• $x_B = 0.2\cos60^\circ + 0.7810 = 0.8810$ m
• I lies on line OA at that x: $y_I = x_B\tan60^\circ = 1.5260$ m, so $IB = 1.526$ m
• $IA = \sqrt{(x_B - 0.1)^2 + (y_I - 0.1732)^2} = 1.562$ m

$$\omega_{AB} = \frac{v_A}{IA} = \frac{6.283}{1.562} = 4.022\ \text{rad/s}$$ $$v_B = \omega_{AB}\times IB = 4.022 \times 1.526 = 6.138\ \text{m/s}$$

The two methods agree.

Answer: Slider velocity = 6.14 m/s toward the crank centre; ω of connecting rod = 4.02 rad/s clockwise.