ICR; angular velocities of BD and DE

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: $A$ and $E$ are fixed pins; $AB$ hangs vertically from $A$ (0.250 m); $BD$ is horizontal (0.200 m) from $B$ to $D$; $E$ is 0.150 m above $D$ and 0.075 m to the right of it, so $D$ moves on a circle about $E$.

$$v_B = \omega_{AB}\,AB = 6\times0.250 = 1.50\ \text{m/s}$$

$B$ is below $A$ and the bar turns clockwise, so $\vec v_B$ points to the left.

ICR of bar BD: the perpendicular to $\vec v_B$ at $B$ is the vertical line through $B$. Since $\vec v_D \perp DE$, the perpendicular at $D$ is the line $ED$ extended. The ICR of $BD$ is where these two lines meet. Line $ED$ has slope $0.150/0.075 = 2$, so going $0.200$ m to the left of $D$ along it takes the line $0.400$ m below $D$. Hence the ICR $I$ is $0.400$ m below $BD$, directly below $B$ and $0.200$ m from $D$ horizontally:

$$IB = 0.400\ \text{m}, \qquad ID = \sqrt{0.200^2 + 0.400^2} = 0.4472\ \text{m}$$ $$\omega_{BD} = \frac{v_B}{IB} = \frac{1.50}{0.400} = 3.75\ \text{rad/s}$$ $$v_D = \omega_{BD}\,(ID) = 3.75\times0.4472 = 1.68\ \text{m/s}$$ $$\omega_{DE} = \frac{v_D}{DE} = \frac{1.68}{\sqrt{0.075^2+0.150^2}} = \frac{1.68}{0.1677} = 10.00\ \text{rad/s}$$

Answer: $\omega_{BD} = 3.75$ rad/s (counter-clockwise), $\omega_{DE} = 10.00$ rad/s (clockwise).

(Check by vectors: $\vec v_D = \vec v_B + \omega_{BD}\hat k\times\vec r_{D/B}$ gives $v_{Dx} = -1.5$ m/s and $v_{Dy} = 0.2\,\omega_{BD}= 0.75$ m/s, and $\omega_{DE} = 1.5/0.15 = 10$ rad/s.)