---
title: "General plane motion; angular acceleration of arm DE"
subject: "Applied Mechanics (Dynamics)"
chapter: 4
key: "semester_3/applied_mechanics_dynamics:general-plane-motion-arm-de-acceleration"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/general-plane-motion-arm-de-acceleration
site: BCT Engineering
---

# General plane motion; angular acceleration of arm DE

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

- Asked in: 2078 Bhadra, 2+6 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/general-plane-motion-arm-de-acceleration
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-4

## Question

Define General plain motion with suitable example. Knowing that at the instant shown rod AB has zero angular acceleration and an angular velocity of 15 rad/s counter clockwise. Determine a) angular acceleration of arm DE b) the acceleration of Point D. [Figure: four-bar linkage A-B-D-E; horizontal dimensions 40 mm, 50 mm, 50 mm, 40 mm; vertical dimension 30 mm at E.]

## Short answer

General plane motion is motion where all particles remain in parallel planes, combining translation of a reference point and rotation about it; e.g., a rolling wheel. Using relative acceleration equations with given geometry, α_DE = 540 rad/s² clockwise, and a_D = 31.5 m/s² (components -16.2 i + 27.0 j m/s²).

## Answer

**General plane motion (GPM)** is the motion of a rigid body in which every particle stays in a plane parallel to a fixed plane, and the motion is neither pure translation nor pure rotation about a fixed axis. It is the sum of (i) a translation of a chosen reference point and (ii) a rotation of the body about that point.

For two points $A$ and $B$ of the same rigid body:

$$
\vec v_B = \vec v_A + \vec\omega \times \vec r_{B/A}, \qquad \vec a_B = \vec a_A + \vec\alpha \times \vec r_{B/A} - \omega^2 \vec r_{B/A}
$$

Here $v_{B/A} = \omega\, r_{B/A}$ is perpendicular to $AB$, and $a_{B/A}$ has a tangential part $\alpha r$ and a normal part $\omega^2 r$ towards $A$.

**Examples**
- A wheel rolling on the road: the centre translates while the wheel rotates about the centre.
- The connecting rod of an engine: end $A$ (crank pin) moves on a circle, end $B$ (piston) moves in a straight line.
- A ladder sliding down a wall.

#### Given data and assumed geometry
Place $A$ at the origin with $AB$ horizontal (40 mm), $BD$ horizontal (50 mm) and arm $DE$ pivoted at $E$, which is 50 mm to the right of $D$ and 30 mm above it (the figure's dimensions 40, 50, 50 mm and 30 mm):

$$
\vec r_{B/A} = 0.040\hat i,\quad \vec r_{D/B} = 0.050\hat i,\quad \vec r_{D/E} = (-0.050\,\hat i - 0.030\,\hat j)\ \text{m}
$$

$\omega_{AB} = 15$ rad/s ccw, $\alpha_{AB}=0$. A and E are fixed.

#### Velocities
$$
\vec v_B = 15\hat k\times0.04\hat i = 0.6\,\hat j\ \text{m/s}
$$

$$
\vec v_D = \vec v_B + \omega_{BD}\hat k\times\vec r_{D/B} = \omega_{DE}\hat k\times\vec r_{D/E}
$$

$$
\begin{aligned}
\hat i:&\quad 0 = 0.03\,\omega_{DE} \;\Rightarrow\; \omega_{DE} = 0\ \text{rad/s}\\
\hat j:&\quad 0.6 + 0.05\,\omega_{BD} = -0.05\,\omega_{DE} \;\Rightarrow\; \omega_{BD} = -12.00\ \text{rad/s}
\end{aligned}
$$

#### Accelerations of B and D
$$
\vec a_B = -\omega_{AB}^2\vec r_{B/A} = -225(0.04)\hat i = -9\,\hat i\ \text{m/s}^2
$$

From rod $BD$: $\vec a_D = \vec a_B + \alpha_{BD}\hat k\times\vec r_{D/B} - \omega_{BD}^2\vec r_{D/B} = (-9 - 144(0.05))\hat i + 0.05\,\alpha_{BD}\hat j = -16.2\,\hat i + 0.05\,\alpha_{BD}\hat j$

From arm $DE$ (since $\omega_{DE}=0$): $\vec a_D = \alpha_{DE}\hat k\times\vec r_{D/E} = \alpha_{DE}(0.03\,\hat i - 0.05\,\hat j)$

Equating components:

$$
\begin{aligned}
\hat i:&\quad -16.2 = 0.03\,\alpha_{DE} \;\Rightarrow\; \alpha_{DE} = -540.0\ \text{rad/s}^2\\
\hat j:&\quad 0.05\,\alpha_{BD} = -0.05\,\alpha_{DE} \;\Rightarrow\; \alpha_{BD} = 540.0\ \text{rad/s}^2
\end{aligned}
$$

$$
\vec a_D = (-16.2\,\hat i + 27.0\,\hat j)\ \text{m/s}^2, \qquad a_D = 31.5\ \text{m/s}^2
$$

**Answer: $\alpha_{DE} = 540.0$ rad/s² clockwise; $a_D = 31.5$ m/s² (components $-16.2\,\hat i + 27.0\,\hat j$).** The result depends on the assumed layout; the method is the same for the exact figure.

## Check yourself

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

1. Which describes general plane motion?
   A. Translation of a reference point plus rotation of the body about that point
   B. Pure translation
   C. Pure rotation about a fixed axis only
   D. Motion with no rotation
   Answer: A. GPM is the sum of translation and rotation; examples are a rolling wheel or an engine connecting rod. (Key point: Definition of GPM)

2. What is aB of the crank end B (w_AB = 15 rad/s, alpha_AB = 0, AB = 0.04 m)?
   A. Zero
   B. 9 m/s^2 towards A, since w^2 r = 225 x 0.04
   C. 0.6 m/s^2
   D. 15 m/s^2
   Answer: B. With zero angular acceleration only the normal component w^2 r = 9 m/s^2 towards A remains. (Key point: Acceleration of B)

3. What does the velocity analysis give for arm DE and rod BD?
   A. w_DE = 0 and w_BD = 12 rad/s clockwise
   B. w_DE = 15 rad/s
   C. w_DE = 12 rad/s
   D. w_BD = 0 and w_DE = 12 rad/s
   Answer: A. The i equation 0 = 0.03 w_DE gives w_DE = 0; the j equation 0.6 + 0.05 w_BD = 0 gives w_BD = -12 rad/s. (Key point: Velocity analysis)

4. How is alpha_DE found and what is aD?
   A. Equate the i components of aD from rod BD and arm DE, giving alpha_DE = 540 rad/s^2 and aD = 31.5 m/s^2
   B. alpha_DE = alpha_AB = 0
   C. aD = 9 m/s^2
   D. aD = 16.2 m/s^2 along j
   Answer: A. -16.2 = 0.03 alpha_DE gives alpha_DE = -540 rad/s^2 (clockwise), and aD = (-16.2 i + 27.0 j), magnitude 31.5 m/s^2. (Key point: Acceleration of D)

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