--- title: "ICR; angular velocities of BD and DE" subject: "Applied Mechanics (Dynamics)" chapter: 4 key: "semester_3/applied_mechanics_dynamics:icr-bars-ab-bd-de-6rads" source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/icr-bars-ab-bd-de-6rads site: BCT Engineering --- # ICR; angular velocities of BD and DE Applied Mechanics (Dynamics) · Chapter 4: Kinematics of Rigid Bodies - Asked in: 2079 Baishakh, 2+6 marks - Video: [Dynamics - Instantaneous Center of Zero Velocity example 1](https://www.youtube.com/watch?v=wydw61lmYHw) — Engineering Deciphered - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/icr-bars-ab-bd-de-6rads - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-4 ## Question Define instantaneous center of rotation with an example. In the position shown, bar AB has an angular velocity of 6 rad/s clockwise. Determine the angular velocity of bars BD and DE. [Figure: bar AB vertical, 250 mm, with A at the top; horizontal bar BD, 200 mm, from B to D; bar DE inclined, 75 mm horizontal offset, with E at a slider 150 mm above D.] ## Short answer Instantaneous center of rotation is the point on a body with zero velocity at an instant, e.g., the contact point of a rolling wheel. Using ICR method: ω_BD = 3.75 rad/s counter-clockwise, ω_DE = 10.00 rad/s clockwise. ## Answer **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.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the instantaneous centre of rotation of a body in plane motion? A. The mass centre of the body B. The point, on the body or its extension, whose velocity is zero at that instant C. The point with the maximum velocity D. The point about which the body always rotates Answer: B. The ICR is the point with zero velocity at that instant; the body then behaves as if rotating about it. (Key point: Definition of ICR) 2. How is the ICR of a body located from the velocity directions of two of its points? A. Draw lines parallel to the two velocities B. Join the two points and take the midpoint C. Draw perpendiculars to the two velocity directions; they meet at the ICR D. Take the intersection of the two velocity vectors Answer: C. Velocity of each point is perpendicular to its line to the ICR, so perpendiculars to the velocities intersect at the ICR. (Key point: Locating the ICR) 3. With AB = 0.250 m and omega_AB = 6 rad/s, what is v_B? A. 24 m/s B. 1.50 m/s C. 0.042 m/s D. 6.00 m/s Answer: B. v_B = omega_AB x AB = 6 x 0.250 = 1.50 m/s. (Key point: Speed of B from AB) 4. If the ICR of BD is 0.400 m from B and v_B = 1.50 m/s, what is omega_BD? A. 0.60 rad/s B. 6.00 rad/s C. 3.75 rad/s D. 1.50 rad/s Answer: C. omega_BD = v_B / IB = 1.50 / 0.400 = 3.75 rad/s. 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