Slider crank velocity by instantaneous centre method

Configuration

Take O as origin and the line of stroke along the $x$-axis.

$$\omega_{OA} = \frac{2\pi \times 300}{60} = 31.416\ \text{rad/s}, \qquad v_A = \omega_{OA}\cdot OA = 31.416 \times 0.150 = 4.712\ \text{m/s}$$

$v_A$ is perpendicular to OA. Obliquity of the rod: $\sin\phi = \dfrac{OA\sin 40^\circ}{AB} = 0.1607 \Rightarrow \phi = 9.25^\circ$.

Coordinates: A $= (114.9,\ 96.4)$ mm, B $= (707.1,\ 0)$ mm.

Locating the instantaneous centre $I_{13}$ (connecting rod)

• The velocity of A is perpendicular to OA, so $I$ lies on the line OA produced.
• The velocity of B is along the line of stroke, so $I$ lies on the perpendicular to the line of stroke at B.

The intersection gives $I$ at $(707.1,\ 593.3)$ mm.

           I  (instantaneous centre)
          /|
        /  |
      A/   |        IA = 773.1 mm
     /\    |        IB = 593.3 mm
   O/__\___|__B ->

$$IA = \sqrt{(707.1 - 114.9)^2 + (593.3 - 96.4)^2} = 773.1\ \text{mm}, \qquad IB = 593.3\ \text{mm}$$

(b) Angular velocity of the rod

The rod rotates about $I$, so $v_A = \omega_{AB}\cdot IA$:

$$\omega_{AB} = \frac{v_A}{IA} = \frac{4.712}{773.1/1000} = 6.096\ \text{rad/s}$$

(clockwise: A lies to the lower left of $I$ and moves to the upper left.)

(a) Velocity of slider

$$v_B = \omega_{AB}\cdot IB = 6.096 \times \frac{593.3}{1000} = 3.617\ \text{m/s}$$

(towards the crank end, towards O.)

Check with the analytic result $v_B = \omega r\left[\sin\theta + \dfrac{r\sin2\theta}{2\sqrt{l^2 - r^2\sin^2\theta}}\right] = 3.617$ m/s. It agrees.

(c) Velocity of midpoint G

Distance from I to G (G is the midpoint of AB): $IG = 620.4$ mm.

$$v_G = \omega_{AB}\cdot IG = 6.096 \times \frac{620.4}{1000} = 3.781\ \text{m/s}$$

perpendicular to IG.

Answer: $v_B = 3.617$ m/s, $\omega_{AB} = 6.096$ rad/s (clockwise), $v_G = 3.781$ m/s.