9.4 Relate accelerations of points on a planar rigid body | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Introduction to Rigid-Body Dynamics

EN PH 131 Engineering Mechanics: Dynamics — Study topic 9.4

A rigid body may translate and rotate at the same time. Its points share the same angular velocity and angular acceleration, but they do not generally share the same linear acceleration. This lesson develops a direct vector relationship between the accelerations of any two points on a planar rigid body. We will describe motion as seen by an observer fixed to the ground, using fixed horizontal and vertical axes. The body is treated as rigid: distances and directions between its points stay fixed as the body moves.

What you will learn

Explain why two points on the same planar rigid body can have different accelerations.
Apply the acceleration relation between two points on a rigid body.
Resolve tangential and normal acceleration components with consistent signs.
Use known point accelerations and angular motion to find an unknown point acceleration or angular acceleration.

1. Set up the body, observer, and directions

Choose the system as the single rigid body whose point accelerations are being related. The observer is stationary relative to the ground, and the reference frame is an inertial frame fixed to the ground. Use fixed axes:

x

horizontal to the right and

y

vertical upward. In a planar problem, counterclockwise angular motion is positive and clockwise angular motion is negative.

Let

A

and

B

be two points fixed in the body. The vector from

A

to

B

, written

\mathbf r_{B/A}

, points from

A

toward

B

and is measured in the chosen axes. Its length and direction relative to the body remain fixed, even though its direction in the ground frame can change as the body rotates.

Position identifies a location; displacement is a change in position; velocity is the time rate of change of position; and acceleration is the time rate of change of velocity. The relation in this lesson connects the instantaneous accelerations of two body points. It does not say that their acceleration vectors are equal.

\mathbf r_{B/A}=x\mathbf i+y\mathbf j

Use one observer and one fixed coordinate frame for both points.
Set counterclockwise angular velocity and angular acceleration as positive.
Write the relative-position vector from the reference point to the other point.

2. The acceleration relation

The planar rigid-body acceleration relation starts from the fact that the relative position between two body-fixed points rotates with the body. The acceleration of

B

consists of the acceleration carried along from

A

, a tangential contribution due to angular acceleration, and a normal (centripetal) contribution due to angular velocity.

Here,

\boldsymbol\alpha

is the angular-acceleration vector and

\boldsymbol\omega

is the angular-velocity vector. In planar motion, both point perpendicular to the plane. The cross products in the vector equation keep track of directions automatically. The tangential contribution is perpendicular to

\mathbf r_{B/A}

; the normal contribution points from

B

toward

A

.

For components, take

\mathbf r_{B/A}=x\mathbf i+y\mathbf j

, with

x

and

y

in metres, and use signed scalar values

\alpha

and

\omega

for the out-of-plane angular quantities. Then the angular-acceleration contribution has components

(-\alpha y,\alpha x)

, and the centripetal contribution has components

(-\omega^2x,-\omega^2y)

. The sign of

\alpha

matters; the centripetal term uses

\omega^2

and always points inward.

The relationship is useful when the acceleration of one point is known and the rigid body's angular motion is known. It can also be rearranged to find unknown angular acceleration when the accelerations of two points and the angular velocity are known.

\mathbf a_B=\mathbf a_A+\boldsymbol\alpha\times\mathbf r_{B/A}+\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r_{B/A})

Add the reference-point acceleration to the tangential and centripetal contributions.
The centripetal acceleration has magnitude $\omega^2r$ and points toward the reference point.
Use signed angular acceleration in the tangential term.

3. A reliable solution sequence

First identify the body and the two points. State which point is the reference point, and record the known accelerations, angular velocity, angular acceleration, and relative position. If a problem gives the body's orientation, use it to express the relative-position vector in the fixed axes at the instant being studied.

Next choose positive axes and the positive angular direction. Draw the body and mark the vector from

A

to

B

. Resolve all known vector quantities into the same axes. Do not reverse the relative-position vector partway through the calculation: changing it to

\mathbf r_{A/B}

also changes which point's acceleration is on the left side of the relation.

Apply the vector relation, or its component form. Solve for unknowns symbolically before inserting numerical values when practical. Keep acceleration units in

\mathrm{m/s^2}

, angular acceleration in

\mathrm{rad/s^2}

, angular velocity in

\mathrm{rad/s}

, and distances in metres. Radians are dimensionless in the unit check.

Finally, interpret the component signs. A negative

x

component means acceleration toward the left, and a negative

y

component means downward. Check that angular contributions vanish in the appropriate limits: with no rotation, points have the same acceleration; with constant angular velocity, the tangential term vanishes but the centripetal term remains.

\mathbf a_B=\mathbf a_A+(-\alpha y-\omega^2x)\mathbf i+(\alpha x-\omega^2y)\mathbf j

Draw and label the relative-position vector before substituting.
Use a common coordinate frame for every vector.
Check components, units, and limiting cases.

4. What the relation does—and does not—say

The equation applies to points on one rigid body at the same instant. It describes how their accelerations differ because of the body's rotation. It does not require the body to be pinned or to rotate about a fixed point; the reference point may itself accelerate.

A useful special case is pure translation: when the body has zero angular velocity and zero angular acceleration, every point has the same acceleration. Another useful case is rotation at constant angular velocity: the tangential contribution is zero, but points away from the chosen reference point can still have centripetal acceleration relative to it.

The relation is a kinematic relationship, meaning it describes motion. It does not by itself determine the forces that cause the motion. Choose it when the unknowns are accelerations or angular acceleration; a force analysis may be needed separately if the problem asks what forces produce the motion.

\omega=0,\ \alpha=0\ \Longrightarrow\ \mathbf a_B=\mathbf a_A

The points must belong to the same rigid body.
The reference point may translate and accelerate.
This relation connects motion quantities; it is not a force equation.

Worked example

A rotating body with a fixed reference point

A rigid plate rotates counterclockwise in the ground-fixed

xy

frame. At one instant, point

A

is fixed, and point

B

is at

\mathbf r_{B/A}=(0.60\mathbf i+0.80\mathbf j)\,\mathrm m

. The plate has

\omega=3.0\,\mathrm{rad/s}

and

\alpha=2.0\,\mathrm{rad/s^2}

. Find

\mathbf a_B

.

Rotating plate at one instant

Schematic only; the relative-position components are stated in the problem.

Set the frame and knowns
The system is the plate, observed from the ground-fixed frame. Take right and up as positive, and counterclockwise as positive. Point $A$ is fixed, so its acceleration is zero. The relative-position vector points from $A$ to $B$ .
$\mathbf a_A=\mathbf 0,\quad \mathbf r_{B/A}=(0.60\mathbf i+0.80\mathbf j)\,\mathrm m$
Find the rotational contributions
Use the component form. The tangential contribution is perpendicular to the radius vector; the centripetal contribution points back toward $A$ . With the given positive angular values, their components follow directly from the signed component equations.
$\mathbf a_{B/A}=(-2.0(0.80)-3.0^2(0.60))\mathbf i+(2.0(0.60)-3.0^2(0.80))\mathbf j$
Evaluate and interpret
The result points left and downward. The tangential and centripetal contributions each have acceleration units: angular acceleration times distance, and angular speed squared times distance.
$\mathbf a_B=(-7.0\mathbf i-6.0\mathbf j)\,\mathrm{m/s^2}$

Answer: The acceleration of

B

is

7.0\,\mathrm{m/s^2}

left and

6.0\,\mathrm{m/s^2}

down, or

\mathbf a_B=(-7.0\mathbf i-6.0\mathbf j)\,\mathrm{m/s^2}

.

Check: The tangential term is

(-1.6\mathbf i+1.2\mathbf j)\,\mathrm{m/s^2}

and the centripetal term is

(-5.4\mathbf i-7.2\mathbf j)\,\mathrm{m/s^2}

. Their sum matches the result. Both terms have units of acceleration.

Worked example

A translating and rotating body

At an instant, point

A

on a rigid body has acceleration

\mathbf a_A=(1.0\mathbf i+2.0\mathbf j)\,\mathrm{m/s^2}

. Point

B

is located at

\mathbf r_{B/A}=(0.40\mathbf i-0.30\mathbf j)\,\mathrm m

. The body rotates counterclockwise at

4.0\,\mathrm{rad/s}

while its angular acceleration is

3.0\,\mathrm{rad/s^2}

clockwise. Find

\mathbf a_B

.

Translation and rotation

Acceleration arrows indicate component directions only; magnitudes are calculated below.

Define signed quantities
Use the ground-fixed frame with right and up positive and counterclockwise positive. The body is the system. The clockwise angular acceleration must therefore be entered as a negative scalar, while the counterclockwise angular velocity is positive.
$\alpha=-3.0\,\mathrm{rad/s^2},\quad \omega=4.0\,\mathrm{rad/s}$
Apply the component relation
The known acceleration of $A$ is included because the body translates as well as rotates. Substitute the signed values and the components of the vector from $A$ to $B$ .
$\mathbf a_B=(1.0-\alpha y-\omega^2x)\mathbf i+(2.0+\alpha x-\omega^2y)\mathbf j$
Calculate and check direction
The horizontal component is negative and the vertical component is positive, so point $B$ accelerates left and up. The signs follow from the chosen fixed axes.
$\mathbf a_B=(-6.3\mathbf i+5.6\mathbf j)\,\mathrm{m/s^2}$

Answer: The acceleration is

\mathbf a_B=(-6.3\mathbf i+5.6\mathbf j)\,\mathrm{m/s^2}

, directed left and upward.

Check: The tangential term is

(-0.9\mathbf i-1.2\mathbf j)\,\mathrm{m/s^2}

, and the centripetal term is

(-6.4\mathbf i+4.8\mathbf j)\,\mathrm{m/s^2}

. Adding these to

\mathbf a_A

gives the stated result.

Worked example

Find angular acceleration from two point accelerations

Two points

A

and

B

lie on a rigid body. At an instant,

\mathbf r_{B/A}=(0.50\mathbf i)\,\mathrm m

,

\omega=2.0\,\mathrm{rad/s}

counterclockwise,

\mathbf a_A=(0\mathbf i+1.0\mathbf j)\,\mathrm{m/s^2}

, and

\mathbf a_B=(-2.0\mathbf i+3.0\mathbf j)\,\mathrm{m/s^2}

. Find the signed angular acceleration.

Two known point accelerations

The stated relative-position vector lies along the positive x-axis.

Subtract the reference acceleration
Work in the ground-fixed frame with right and up positive. Subtracting $\mathbf a_A$ isolates the angular contributions between the points. Since the relative-position vector lies along positive $x$ , the tangential contribution is vertical and the centripetal contribution is horizontal.
$\mathbf a_B-\mathbf a_A=(-2.0\mathbf i+2.0\mathbf j)\,\mathrm{m/s^2}$
Match components
The centripetal contribution is $-\omega^2(0.50)\mathbf i=-2.0\mathbf i\,\mathrm{m/s^2}$ , which matches the horizontal difference. The vertical difference is $\alpha(0.50)\mathbf j$ , so it determines the signed angular acceleration.
$2.0=0.50\alpha$
Solve and interpret
A positive answer means the angular acceleration is counterclockwise under the selected convention. Its units are inverse seconds squared, conventionally written as radians per second squared.
$\alpha=4.0\,\mathrm{rad/s^2}$

Answer: The angular acceleration is

4.0\,\mathrm{rad/s^2}

counterclockwise.

Check: Substituting this value gives tangential acceleration

2.0\mathbf j\,\mathrm{m/s^2}

and centripetal acceleration

-2.0\mathbf i\,\mathrm{m/s^2}

. Adding both to

\mathbf a_A

recovers

\mathbf a_B=(-2.0\mathbf i+3.0\mathbf j)\,\mathrm{m/s^2}

.

Common mistakes and how to avoid them

Using the vector from

B

to

A

while still calculating

\mathbf a_B

from

\mathbf a_A

.

Correction: Keep the relative-position vector directed from the reference point to the point whose acceleration is sought: use

\mathbf r_{B/A}

with

\mathbf a_B

on the left.

Giving the centripetal term the same direction as the relative-position vector.

Correction: The centripetal contribution is inward, opposite to the relative-position vector in the double-cross-product term.

Treating clockwise angular acceleration as positive after choosing counterclockwise positive.

Correction: Use a negative signed value for clockwise angular acceleration; this sign controls the tangential contribution.

Assuming every point has the same acceleration because the body is rigid.

Correction: Rigidity fixes distances between points. Rotation can still give different point accelerations.

Omitting the reference point's acceleration.

Correction: Include

\mathbf a_A

unless the point is known to have zero acceleration.

Lesson summary

For points $A$ and $B$ on one planar rigid body, use $\mathbf a_B=\mathbf a_A+\boldsymbol\alpha\times\mathbf r_{B/A}+\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r_{B/A})$ .
The angular-acceleration term is tangential; the angular-velocity term is centripetal and points inward.
Use a consistent observer, fixed axes, signed angular quantities, and a relative-position vector from the reference point to the point of interest.
Check units and signs, and verify the result by substituting it back into the component relations.

Check your understanding

Question 1

A body translates without rotating at an instant. What is the acceleration relationship between any two of its points at that instant?

Their accelerations are equal.
Their accelerations are opposite.
The point farther from the centre must have greater acceleration.
Their acceleration difference is always directed inward.

Show answer and explanation

Their accelerations are equal.

With zero angular velocity and zero angular acceleration, both rotational contributions vanish, leaving equal point accelerations.

Question 2

A point lies

0.40\,\mathrm m

to the right of a reference point. The body has angular velocity

3.0\,\mathrm{rad/s}

counterclockwise. What is the centripetal contribution to the point's acceleration?

$3.6\,\mathrm{m/s^2}$ to the left
$3.6\,\mathrm{m/s^2}$ to the right
$1.2\,\mathrm{m/s^2}$ upward
Zero, because angular velocity is constant.

Show answer and explanation

3.6\,\mathrm{m/s^2}

to the left

The centripetal contribution has magnitude

\omega^2r=(3.0)^2(0.40)=3.6\,\mathrm{m/s^2}

and points toward the reference point, hence left.

Question 3

With counterclockwise defined as positive, what sign should be used for a clockwise angular acceleration of

5.0\,\mathrm{rad/s^2}

?

$-5.0\,\mathrm{rad/s^2}$
$+5.0\,\mathrm{rad/s^2}$
$0\,\mathrm{rad/s^2}$
Its sign depends on the body's mass.

Show answer and explanation

-5.0\,\mathrm{rad/s^2}

Clockwise is opposite to the chosen positive angular direction, so the signed angular acceleration is negative.

Key terms

Rigid body: A body whose points keep the same distances from one another during the motion considered.
Relative-position vector: A vector from one chosen point on a body to another, expressed in the selected coordinate frame.
Tangential acceleration: The acceleration contribution associated with angular acceleration; it is perpendicular to the relative-position vector.
Centripetal acceleration: The inward acceleration contribution associated with angular velocity; its magnitude is angular speed squared times distance from the reference point.

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9.4 · Relate accelerations of points on a planar rigid body

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

EN PH 131 Engineering Mechanics: Dynamics — Study topic 9.4

What you will learn

1. Set up the body, observer, and directions

2. The acceleration relation

3. A reliable solution sequence

4. What the relation does—and does not—say

A rotating body with a fixed reference point

A translating and rotating body

Find angular acceleration from two point accelerations

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through EN PH 131

About this lesson and its review