ICR; angular velocity of bar BD

Instantaneous centre of rotation (ICR) is the point in the plane of motion (on the body or on its imaginary extension) whose velocity is zero at that instant. At that instant the body behaves as if it were rotating about this point, so every point has speed $v = \omega\,r$, where $r$ is its distance from the ICR, and its velocity is perpendicular to the line joining it to the ICR.

Location: draw perpendiculars to the velocity directions of two points of the body; they meet at the ICR. For a body rolling without slipping, the ICR is the contact point.

Examples

• Rolling wheel: the ICR is the point of contact with the ground, so $v_{centre} = \omega r$ and the top point moves at $2\omega r$.
• Ladder sliding on a wall and floor: the ICR is the intersection of the vertical through the foot and the horizontal through the top end.
• Connecting rod of an engine: the ICR lies where the perpendicular to the piston path through the piston meets the perpendicular to the crank-pin velocity (the crank line extended).

Solution

Assumed layout (from the figure's dimensions): $A$ is a fixed pin at the right end of the horizontal bar $AB$ ($AB = 4$ in.). $D$ is 7 in. to the left of $A$ and 6 in. below it, so $BD$ runs from $B$ down and to the left (3 in. horizontally, 6 in. vertically). Bar $DE$ is vertical (3 in.) and its end $E$ is a slider in a horizontal guide; since $\vec v_E$ is horizontal and $DE$ is vertical, $\vec v_D$ is also horizontal.

$$v_B = \omega_{AB}\,AB = 6\times4 = 24\ \text{in./s}$$

$B$ is to the left of $A$ and the bar turns clockwise, so $\vec v_B$ is vertical, upward.

ICR of BD: the perpendicular to $\vec v_B$ at $B$ is the horizontal line through $B$; the perpendicular to the horizontal $\vec v_D$ at $D$ is the vertical line through $D$. They meet at $I$, which is 3 in. to the left of $B$ and 6 in. above $D$:

   I *-----* B -------* A
     |    /   (v_B up)
     |   /
     |  /
     | /
     * D   (v_D horizontal)

$$IB = 3\ \text{in.}, \qquad ID = 6\ \text{in.}$$ $$\omega_{BD} = \frac{v_B}{IB} = \frac{24}{3} = 8\ \text{rad/s}$$

$B$ is to the right of $I$ and moves up, so the rotation is counter-clockwise.

$$v_D = \omega_{BD}\,(ID) = 8\times6 = 48\ \text{in./s}$$

Answer: $\omega_{BD} = 8$ rad/s counter-clockwise. (The numerical result depends on the assumed layout above; the ICR method is the same for the actual figure.)