ME 3507: Theory of Machines

Acceleration Analysis

Dr. Faraz Junejo

Objective

• Determine the accelerations of links

and points on mechanisms

Introduction

Introduction (contd.)

It is important to determine the acceleration

of links because acceleration produces inertia

forces in the link which stress the component

parts of the mechanism.

Accelerations may be relative or absolute in

the same way as described for velocity.

Acceleration

Acceleration is rate of change of velocity with

respect to time

Acceleration can be linear “A” (rate of change in

linear velocity) or angular “” (rate of change in

angular velocity).

Linear acceleration

Angular acceleration

VRA

Acceleration of a point Figure 7-1 shows a

link PA in pure rotation, pivoted at point A in the xy plane.

We are interested in the acceleration of point P when the link is subjected to an angular velocity (ω) and an angular acceleration , which need not have the same sense.

Acceleration of a point (contd.)

As a result of the differentiation, the tangential component is multiplied by the (constant j) complex operator. This causes a rotation of this acceleration vector through 90 ° with respect to the original position vector.

However, the tangential component is also multiplied by , which may be either positive or negative. As a result, the tangential component of acceleration will be rotated 90° from the angle θ of the position vector in a direction dictated by the sign of .

The normal, or centripetal, acceleration component is multiplied by j2, or -1.

This directs the centripetal component at 180° to the angle θ of the original position vector, i.e., toward the center (centripetal means toward the center).

Substituting the Euler’s identity in above expression yields:

APA is absolute acceleration in this case i.e. APA = AP

Tangential Acceleration

Recall that the instantaneous velocity of a point on a

rotating link is perpendicular to a line that connects

that point to the center of rotation.

Any change in the magnitude of this velocity creates

tangential acceleration, which is also perpendicular

to the line that connects the point with the center of

rotation.

Tangential Acceleration

The magnitude of the tangential acceleration of

point A on a rotating link 2 can be expressed as:

It is important to remember that the angular

acceleration, α, in equation must be expressed as units

of radians per squared time.

Tangential acceleration acts in the direction of

motion when the velocity increases or the

point accelerates.

Conversely, tangential acceleration acts in the

opposite direction of motion when the velocity

decreases or the point decelerates.

Tangential Acceleration (contd.)

Normal acceleration Any change in velocity direction creates normal acceleration,

which is always directed toward the center of rotation.

Because acceleration is defined as the time rate of velocity change, both sides should be divided by time:

Using equation v = r ω, the relationships between the magnitude of the linear velocity and angular velocity, the following equations for the magnitude of the normal acceleration of a point can be derived:

Magnitudes of acceleration components

A

PAt

PA

AnPA

APA

,

Magnitude of tangential component = p, magnitude of normal component = p 2

Length of link: p2 i i

PAA pe i pe

PA

tAPA

nA

Example: 1• The mechanism shown in Figure is used in a distribution

center to push boxes along a platform and to a loading area. The input link is driven by an electric motor, which, at the instant shown, has a velocity of 25 rad/s and accelerates at a rate of 500 rad/s2. Knowing that the input link has a length of 250 mm, determine the instantaneous acceleration of the end of the input link in the position shown.

1. Draw a Kinematic Diagram

• The kinematic diagram for the transfer mechanism is shown below. Notice that it is the familiar four-bar mechanism.

2. Determine the Tangential Acceleration of Point A

• Because the input link (link 2) is in pure rotation, the acceleration components of the end of the link can be readily obtained. the magnitude of the tangential acceleration is given by:

Because the link is accelerating, the direction of the vector is in the direction of the motion at the end of the link, which is perpendicular to the link itself. Thus, the tangential acceleration is

3. Determine the Normal Acceleration of Point A

Magnitude of the normal acceleration is given by:

Normal acceleration always occurs toward the center

of rotation. Thus, normal acceleration is calculated as:

4. Determine the Total Acceleration of Point A The magnitude of the total acceleration is computed as:

The angle of the total acceleration vector from the normal component can be calculated as:

Therefore, direction of the total acceleration vector from the horizontal axis is

Acceleration Difference / Relative Acceleration• Next consider point A on a sliding block with link 2 in

pure rotation about point A

AP = AA + APA

AP = APA + AA

i.e. Absolute acceleration of point P is equal to acceleration of P relative to A (acceleration difference) plus acceleration of A relative to ground

Acceleration Difference / Relative Acceleration

Please note vector AA has zero normal component in this case, as it is in pure translation

Now consider the relative acceleration of two independent bodies

If their independent accelerations i.e. AP and AA are known , then their relative acceleration can be found using

APA = AP - AA

Just as with velocity analysis, we will have two cases:

Acceleration Difference / Relative Acceleration (contd.)

Case 1 – Two points in the same body• acceleration difference

Case 2 – Two points in the different bodies• relative acceleration

Example: 2• Figure shows a power hacksaw. At this instant, the electric motor

rotates counterclockwise and drives the free end of the motor crank (point B) at a velocity of 12 in./s. Additionally, the crank is accelerating at a rate of 37 rad/s2.

• The top portion of the hacksaw is moving toward the left with a velocity of 9.8 in./s and is accelerating at a rate of 82 in./s2. Determine the magnitude of relative acceleration of point C with respect to point B.

1. Draw a Kinematic Diagram

• The kinematic diagram for the power hacksaw is shown below. Notice that it is the familiar slider-crank mechanism.

2. Determine the Tangential Acceleration of Point B• From the kinematic diagram, it should be apparent that point

B travels up and to the left as link 2 rotates counterclockwise.• Because the motor crank (link 2) is in pure rotation, the

components of the acceleration at the end of the link can be readily obtained. The magnitude of the tangential acceleration is given by:

Because the link is accelerating, the direction of the vector is in the direction of the motion at the end of the link, which is perpendicular to the link itself. Thus, the tangential acceleration is

3. Determine the Normal Acceleration of Point B Magnitude of the normal acceleration is given by:

Normal acceleration always occurs toward the center of rotation. Thus, normal acceleration is calculated as:

4. Specify the Acceleration of Point C Point C is constrained to linear motion. Therefore, point C does

not experience a normal acceleration. The total acceleration is given in the problem statement as:

5. Construct the Acceleration Polygon for the Acceleration of C Relative to B

6. Solve for the Unknown Vector Magnitudes• the acceleration AC/B can be found by separating the

vectors into horizontal and vertical components.

• Separate algebraic equations can be written for the horizontal and vertical components as follows:

The magnitude of the acceleration can be found by

Example: 2 (contd.)• The direction of the vector can be determined

by

• Finally, the relative acceleration of C with respect to B is

Graphical Acceleration analysis

Graphical acceleration analysis for one position of a fourbar linkage

Given θ2, θ3, θ4, ω2 ,ω3, ω4,α2 find:

– angular acceleration (α3 , α4) and

– linear accelerations (AA, AB and AC )

– Velocity analysis already performed

1. Start at the end of the linkage about which you have the most information. Calculate the magnitude of the acceleration of point A

2. Draw the acceleration AA

3. Move next to a point which you have some information, point B. Draw the construction line pp through B perpendicular to BO4

4. Write the acceleration difference equation for point B vs. A i.e.

AB = AA + ABA

Now substituting the normal and tangential components of each term in above expression:

We will use point A as the reference point to find AB because A is in the same link as B and we have already solved for At

A and AnA .

Knowing ω4 we can solve for AnB

5. Draw construction line qq through point B and perpendicular to BA to represent the direction of At

BA

The term ABA represents the acceleration difference of B with respect to A. This has two components. The normal component is directed along the line BA because we are using point A as the reference center of rotation for the free vector ω3, and its magnitude can be calculated from equation:

6. The vector equation

can be solve graphically by drawing the following vector diagram.

i. First draw acceleration vectors AtA and An

A tip to tail, carefully to some scale, maintaining their directions. Note that the sum of these two components is the vector AA.

ii. The acceleration difference equation in step 4 says to add ABA to AA. We know An

BA (i.e. both magnitude & direction), so we can draw that component at the end of AA.

iii. We also know AnB, but this component is on the left side of

equation, so we must subtract it. Draw the negative (opposite sense) of An

B at the end of AnBA.

iv. This finishes components for which we know both magnitude and direction. Our two remaining knowns are the directions of At

B and AtBA which lie along the lines pp and qq, respectively.

v. Draw a line parallel to line qq (direction of AtBA )

across the tip of the vector representing minus AnB .

vi. The resultant, or left side of the equation, must close the vector diagram, from the tail of the first vector drawn (AA) to the tip of the last, so draw a line parallel to pp (direction of At

B )across the tail of AA.

vii. The intersection of these lines parallel to pp and qq defines the lengths of At

B and AtBA.

viii.Vector AA was added to ABA, so their components must be arranged tip to tail. Vector AB is the resultant, so its component At

B must be from the tail of the first to the tip of the last.

7. The angular accelerations of link 3 and 4 can now be calculated

8. Solve for AC

• Since, magnitudes and directions of both AA and ACA are known,

the vector diagram using acceleration difference equation (AC =

AA + ACA)can directly be drawn as shown below.

• Following figure shows the calculated acceleration vectors on the fourbar linkage diagram.

Example 1: Solve using a graphical approach

Example: 1 (contd.)

Example: 1 (contd.)

From Vel. Analysis, Ex. 1 VA = 300 mm/sec

222

22 /3000)/10)(30()( smmsradmmAOaOrn

A

Example: 1 (contd.)

Example: 1 (contd.)

Example: 1 (contd.)

222

3//1325)/64.3)(100()( smmsradmmBAaOr

n

AB

From Vel. Analysis, Ex. 1 VB/A = 364 mm/sec

Example: 1 (contd.)

Example: 1 (contd.)

From Vel. Analysis, Ex. 1 VB = 184 mm/sec

222

44 /375)/04.2)(90()( smmsradmmBOaOrn

B

Example: 1 (contd.)

Example: 1 (contd.)

Example: 1 (contd.)

Example: 1 (contd.)

Example: 1 (contd.)

Example: 1 (contd.)

Example: 1 (contd.)

Example: 1 (contd.)

Example: 1 (contd.)

From Vel. Analysis, Ex. 1 VC/A = (33.1 mm) * ((5mm/s)/mm)

Scale used in Vel. Analysis

2223/

/2.604)/64.3)(6.45()( smmsradmmACaOrn

AC

Example: 1 (contd.)

Exercise: 1• Link 2 is isolated from a

kinematic diagram and shown in Figure. The link is rotating counterclockwise at a constant rate of 300 rpm. Determine the total linear acceleration of points A and B. Use y = 50° and β = 60°.

Exercise: 2• Link 2 is isolated from a

kinematic diagram and shown in Figure. The link is rotating counterclockwise at a rate of 200 rpm, and accelerating at 400 rad/s2. Determine the total linear acceleration of points A and B. Use y = 50° and β = 60°.