---
title: "Link CD 6 rad/s, point E and link BC"
subject: "Applied Mechanics (Dynamics)"
chapter: 4
key: "semester_3/applied_mechanics_dynamics:link-cd-6rads-velocity-e-link-bc"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/link-cd-6rads-velocity-e-link-bc
site: BCT Engineering
---

# Link CD 6 rad/s, point E and link BC

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

- Asked in: 2081 Bhadra, 2+6 marks
- Video: [Rigid Bodies Relative Motion Analysis: Velocity Dynamics (Learn to solve any question step by step)](https://www.youtube.com/watch?v=qInzrJ3-X-c) — Question Solutions
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/link-cd-6rads-velocity-e-link-bc
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-4

## Question

What is meant by coriolis's acceleration in plane motion of a rigid body? If link CD has an angular velocity of $\omega_{CD} = 6$ rad/sec, determine the velocity of point E on link BC and the angular velocity of link BC at the instant shown. [Figure: link AB inclined at $30^\circ$ pinned at A; link BC horizontal with midpoint E (0.3 m on each side); link CD vertical, 0.6 m long, pinned at D; $\omega_{CD} = 6$ rad/s.]

## Short answer

Using relative velocity equations, v_E = 4.76 m/s at 40.9° above horizontal towards left, ω_BC = 10.39 rad/s clockwise. Coriolis acceleration is 2ω × v_rel, the extra acceleration of a point sliding on a rotating body.

## 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 body, $\vec a_{rel}$ is the acceleration of $P$ relative to the body, and $2\vec\omega\times\vec v_{rel}$ is the Coriolis acceleration (magnitude $2\omega v_{rel}$, perpendicular to $\vec v_{rel}$). The velocity problem below involves only points of rigid links joined by pins, so no Coriolis term appears.

Assumed layout (from the figure): $D$ and $A$ are fixed pins at the same level; $CD$ is vertical (0.6 m) with $C$ at the top; $BC$ is horizontal (0.6 m) with $B$ to the left of $C$ and $E$ its midpoint; $AB$ rises at $30^\circ$ from $A$ to $B$, so $AB = 0.6/\sin30^\circ = 1.2$ m. Take $\omega_{CD}=6$ rad/s counter-clockwise, $x$ to the right and $y$ up (reverse all senses if $CD$ turns the other way).

#### Step 1: velocity of C
$$
\vec v_C = \omega_{CD}\hat k\times\vec r_{C/D} = 6\hat k\times0.6\hat j = -3.6\,\hat i\ \text{m/s}
$$

#### Step 2: direction of v_B
Link $AB$ rotates about $A$, so $\vec v_B \perp AB$. With $\vec v_B = (v_{Bx}, v_{By})$ and $AB$ along $(\cos30^\circ, \sin30^\circ)$:

$$
v_{Bx}\cos30^\circ + v_{By}\sin30^\circ = 0
$$

Link $BC$ is horizontal, so $\omega_{BC}\hat k\times\vec r_{B/C}$ is vertical and $v_{Bx} = v_{Cx} = -3.6$ m/s. Hence

$$
v_{By} = \frac{3.6\cos30^\circ}{\sin30^\circ} = 6.24\ \text{m/s}, \qquad v_B = \sqrt{3.6^2 + 6.24^2} = 7.20\ \text{m/s}
$$

#### Step 3: link BC
$$
\vec v_B = \vec v_C + \omega_{BC}\hat k\times(-0.6\hat i) \;\Rightarrow\; v_{By} = -0.6\,\omega_{BC} \;\Rightarrow\; \omega_{BC} = -10.39\ \text{rad/s}\ \text{(clockwise)}
$$

#### Step 4: point E
$$
\vec v_E = \vec v_C + \omega_{BC}\hat k\times(-0.3\hat i) = -3.6\hat i + (-10.39)(-0.3)\hat j = (-3.60\,\hat i + 3.12\,\hat j)\ \text{m/s}
$$

$v_E = 4.76$ m/s, directed 40.9° above the horizontal, towards the left.

**Answer: $v_E = 4.76$ m/s (40.9° above horizontal, towards the left); $\omega_{BC} = 10.39$ rad/s clockwise.**

## Check yourself

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

1. What is the velocity of C on link CD (w = 6 rad/s, CD = 0.6 m)?
   A. 3.6 m/s, perpendicular to CD (horizontal)
   B. 3.6 m/s along CD
   C. 6 m/s horizontal
   D. 0.1 m/s
   Answer: A. vC = w x r = 6 x 0.6 = 3.6 m/s, perpendicular to the vertical link CD. (Key point: Velocity of C)

2. How is the x component of vB found?
   A. Since BC is horizontal, w_BC x r is vertical, so vBx = vCx = -3.6 m/s
   B. vBx = 0 since AB is fixed
   C. vBx = vBy
   D. vBx = 6.24 m/s
   Answer: A. Relative velocity along a horizontal link is vertical, so horizontal components of B and C are equal. (Key point: Direction of vB)

3. What is w_BC and how is it found?
   A. 10.39 rad/s clockwise from vBy = -0.6 w_BC with vBy = 6.24 m/s
   B. 6 rad/s
   C. 7.2 rad/s
   D. 3.6 rad/s
   Answer: A. vB is perpendicular to AB, giving vBy = 3.6 cos30/sin30 = 6.24 m/s; then w_BC = 6.24/0.6 = 10.39 rad/s clockwise. (Key point: Angular velocity of BC)

4. What is the speed of the midpoint E of BC?
   A. 4.76 m/s
   B. 3.60 m/s
   C. 7.20 m/s
   D. 3.12 m/s
   Answer: A. vE = (-3.6 i + 3.12 j) m/s so vE = sqrt(3.6^2 + 3.12^2) = 4.76 m/s. (Key point: Velocity of E)

## Close variants set in other exams

- [Link CD 6 rad/s, point E and link AB](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Alink-cd-6rads-velocity-e-link-ab-gpm) — Applied Mechanics (Dynamics), chapter 4, asked 2×

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