Crank OA 12 rad/s, piston B, constrained motion

Assumed layout: $OA$ is vertical, so $v_A = \omega_{OA}\cdot OA$ is horizontal. Rod $AB$ makes $30^\circ$ with the horizontal, and $B$ slides in a vertical guide, so $v_B$ is vertical. Use the instantaneous centre $I$ of rod $AB$.

Locating the ICR

$I$ lies on the perpendicular to $v_A$ through $A$ (the vertical line through $A$) and on the perpendicular to $v_B$ through $B$ (the horizontal line through $B$):

   A *   --> vA (horizontal)
     |\
     | \  AB
   y |  \
     |   \
   I *----* B   (vB vertical)
        x

Horizontal and vertical distances of $B$ from $A$:

$$x = AB\cos30^\circ = 0.6\cos30^\circ = 0.5196\ \text{m}, \qquad y = AB\sin30^\circ = 0.6\sin30^\circ = 0.300\ \text{m}$$

So $IA = y = 0.300$ m and $IB = x = 0.5196$ m.

Velocities

$$v_A = \omega_{OA}\,OA = 12\times0.3 = 3.60\ \text{m/s}$$ $$\omega_{AB} = \frac{v_A}{IA} = \frac{3.60}{0.300} = 12.00\ \text{rad/s}$$ $$v_B = \omega_{AB}\,(IB) = 12.00\times 0.5196 = 6.24\ \text{m/s (vertical)}$$

The midpoint $M$ of $AB$ is at $IM = \sqrt{(x/2)^2 + (y/2)^2} = \sqrt{(0.2598)^2 + (0.150)^2} = 0.300$ m, so

$$v_M = \omega_{AB}\,(IM) = 12.00\times 0.300 = 3.60\ \text{m/s}$$

(perpendicular to $IM$).

Answer: $v_B = 6.24$ m/s, $v_M = 3.60$ m/s, $\omega_{AB} = 12.00$ rad/s.

Constrained motion is motion in which the path or position of a body is restricted by guides, pins, links or contact, so its coordinates are related by geometric equations and it cannot move freely in all directions. The restricted motion lets us find unknown velocities and accelerations from the geometry.

Examples

• A piston in a cylinder: it can move only along the cylinder axis.
• A pendulum bob on a rod: it is forced to move on a circular arc.
• A collar sliding on a fixed rod, or a block on a slotted guide.
• A wheel rolling without slipping: the centre speed and angular speed are related by $v = \omega r$.