---
title: "Coriolis acceleration; rod AB and BD velocity"
subject: "Applied Mechanics (Dynamics)"
chapter: 4
key: "semester_3/applied_mechanics_dynamics:coriolis-define-rod-ab-bd-velocity"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/coriolis-define-rod-ab-bd-velocity
site: BCT Engineering
---

# Coriolis acceleration; rod AB and BD velocity

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

- Asked in: 2076 Asoj, 2+8 marks
- Video: [Rigid Bodies Relative Motion Analysis: Acceleration Dynamics (step by step)](https://www.youtube.com/watch?v=ee_LrUTVE5o) — Question Solutions
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/coriolis-define-rod-ab-bd-velocity
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-4

## Question

Define Coriolis acceleration of a rigid body in general plane motion. For the figure shown knowing that at the instant shown the velocity of point D is 2.4 m/s upward, determine (a) the angular velocity of rod AB, (b) the velocity of the midpoint of rod BD. [Figure: rod AB vertical with slider A at the top; rod BD inclined from B down to D, D on a horizontal slider E; vertical dimensions 0.2 m and 0.25 m; horizontal dimensions 0.2 m and 0.6 m.]

## Short answer

Coriolis acceleration is the additional acceleration 2ω×v_rel experienced by a point moving relative to a rotating body, e.g., a collar sliding on a rotating arm. From velocity analysis: ω_AB = 3.20 rad/s clockwise; velocity of midpoint of BD is 1.26 m/s.

## Answer

**Coriolis acceleration** is the extra acceleration a point has when it moves relative to a rotating (moving) frame or body. If a point $P$ moves with relative velocity $\vec v_{rel}$ on a body rotating with angular velocity $\vec\omega$, then

$$
\vec a_P = \vec a_{P'} + \vec a_{rel} + 2\vec\omega\times\vec v_{rel}
$$

where $\vec a_{P'}$ is the acceleration of the coincident point of the rotating frame. The term $\vec a_c = 2\vec\omega\times\vec v_{rel}$ is the Coriolis acceleration. Its magnitude is $2\omega v_{rel}$ and its direction is that of $\vec v_{rel}$ rotated through $90^\circ$ in the sense of $\omega$. It exists only when the point slides on the rotating body (for example a collar sliding on a rotating arm, or a slider in a slotted rotating link); it is zero for two points fixed in the same rigid body, such as the crank pin and piston pin of a connecting rod.

#### Velocity analysis
The figure is not fully legible, so I assume: $AB$ is a vertical rod of length $0.25$ m pinned at the top end $A$ to a fixed support; $BD$ runs from $B$ down to $D$ with horizontal projection $0.6$ m and drop $0.2$ m; and $D$ is constrained to move vertically (given $v_D = 2.4$ m/s upward). Take $x$ to the right, $y$ up.

$$
\vec r_{D/B} = (0.6\,\hat i - 0.2\,\hat j)\ \text{m}, \qquad \vec v_D = 2.4\,\hat j\ \text{m/s}
$$

Rod $AB$ rotates about $A$, so $\vec v_B = \omega_{AB}\hat k\times(-0.25\hat j) = 0.25\,\omega_{AB}\,\hat i$ is horizontal.

For rod $BD$:

$$
\vec v_D = \vec v_B + \omega_{BD}\hat k\times\vec r_{D/B} = v_{B}\hat i + \omega_{BD}(0.2\,\hat i + 0.6\,\hat j)
$$

$$
\begin{aligned}
\hat j:&\quad 2.4 = 0.6\,\omega_{BD} \;\Rightarrow\; \omega_{BD} = 4.00\ \text{rad/s (ccw)}\\
\hat i:&\quad 0 = v_B + 0.2(4.00) \;\Rightarrow\; v_B = -0.80\ \text{m/s}
\end{aligned}
$$

#### (a) Angular velocity of AB
$$
\omega_{AB} = \frac{v_B}{0.25} = \frac{-0.80}{0.25} = -3.20\ \text{rad/s}
$$

so $|\omega_{AB}| = 3.20$ rad/s, clockwise (the sign shows that $B$ moves to the left).

#### (b) Velocity of the midpoint of BD
$$
\vec v_M = \vec v_B + \omega_{BD}\hat k\times\left(\tfrac12\vec r_{D/B}\right) = (-0.80\,\hat i) + 4.00(0.1\,\hat i + 0.3\,\hat j) = (-0.40\,\hat i + 1.20\,\hat j)\ \text{m/s}
$$

$$
v_M = 1.26\ \text{m/s}
$$

**Answer: $\omega_{AB} = 3.20$ rad/s clockwise; $v_M = 1.26$ m/s.**

## Check yourself

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

1. Coriolis acceleration equals
   A. 2 w x v_rel
   B. w^2 r
   C. alpha x r
   D. w x r
   Answer: A. It is 2 w x v_rel, of magnitude 2 w v_rel, present when a point slides relative to a rotating body. (Key point: Coriolis acceleration)

2. Why is vB horizontal?
   A. Rod AB rotates about the fixed top end A, so vB is perpendicular to the vertical rod AB
   B. B moves along a vertical guide
   C. D moves horizontally
   D. AB is at rest
   Answer: A. A is fixed and AB is vertical, so B moves perpendicular to it: horizontally. (Key point: Direction of vB)

3. How is w_BD obtained and what is it?
   A. Using the j component of vD = 2.4 m/s = 0.6 w_BD, giving 4 rad/s
   B. Using the i component, giving 12 rad/s
   C. w_BD = 2.4/0.2 = 12 rad/s
   D. w_BD = 2.4 rad/s
   Answer: A. vD = vB i + w_BD (0.2 i + 0.6 j) with vD vertical gives 2.4 = 0.6 w_BD, so w_BD = 4 rad/s ccw. (Key point: Angular velocity of BD)

4. Which results follow for w_AB and the speed of the midpoint of BD?
   A. w_AB = 3.20 rad/s clockwise, v_M = 1.26 m/s
   B. w_AB = 4 rad/s, v_M = 2.4 m/s
   C. w_AB = 0.8 rad/s, v_M = 1.2 m/s
   D. w_AB = 3.20 rad/s, v_M = 0.80 m/s
   Answer: A. vB = -0.80 m/s so w_AB = 0.80/0.25 = 3.20 rad/s; vM = (-0.40 i + 1.20 j) so 1.26 m/s. (Key point: Results)

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