---
title: "ICR; angular velocity of bar BD"
subject: "Applied Mechanics (Dynamics)"
chapter: 4
key: "semester_3/applied_mechanics_dynamics:icr-bar-ab-bd-inches"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/icr-bar-ab-bd-inches
site: BCT Engineering
---

# ICR; angular velocity of bar BD

Applied Mechanics (Dynamics) · Chapter 4: Kinematics of Rigid Bodies

- Asked in: 2076 Chaitra, 3+5 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-bar-ab-bd-inches
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-4

## Question

Define instantaneous center of rotation with example. In the position shown, bar AB has an angular velocity of 6 rad/s clockwise. Determine the angular velocity of bar BD. [Figure: bar AB horizontal with A at the right end, B at the left, dimensions 7 in. and 4 in. across the top; bar BD goes down from B (vertical dimensions 6 in. and 3 in.) to D; bar DE continues from D to a slider at E.]

## 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 = 8 rad/s counter-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 (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 $D$:

```text
   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.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. At the instant considered, a body in plane motion behaves as if it were rotating about its ICR. What is true of the ICR's velocity?
   A. It equals the velocity of the mass centre
   B. It is zero
   C. It is perpendicular to the body
   D. It equals omega times the body length
   Answer: B. The ICR is defined as the point of zero velocity at that instant. (Key point: ICR meaning)

2. Here v_B is vertical and v_D (slider end) is horizontal. Where is the ICR of BD?
   A. At the intersection of the horizontal through B and the vertical through D
   B. At the midpoint of BD
   C. At A
   D. At the intersection of the vertical through B and the horizontal through D
   Answer: A. Perpendicular to vertical v_B is the horizontal line through B; perpendicular to horizontal v_D is the vertical line through D; they meet at the ICR. (Key point: Locating ICR of BD)

3. AB = 4 in. and omega_AB = 6 rad/s. What is v_B?
   A. 1.5 in./s
   B. 10 in./s
   C. 24 in./s
   D. 48 in./s
   Answer: C. v_B = 6 x 4 = 24 in./s. (Key point: Speed of B)

4. With IB = 3 in. and v_B = 24 in./s, what is omega_BD?
   A. 8 rad/s
   B. 72 rad/s
   C. 4 rad/s
   D. 0.125 rad/s
   Answer: A. omega_BD = v_B / IB = 24/3 = 8 rad/s, counter-clockwise for the layout. (Key point: omega_BD)

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