In kinematic analysis of mechanisms, acceleration analysis is usually performed following a velocity analysis; i.e., the positions and orientations, and the velocities of all the links in a mechanism are assumed known.
Acceleration of a point - The acceleration of a point is the relationship between the change of its velocity vector and time.
Relative Acceleration of Two Points in the Same Rigid Body
As the distance between two points of a rigid body cannot change, relative motion between them is a rotation of one point about the other. Let point B rotates about point A, both being part of a link that moves with angular velocity ω and angular acceleration α. The relative acceleration vector of point B with respect to point A can be broken into two components:
![](/CUPS/AccelerationAndVelocity_files/img17.jpg)
Acceleration Analysis (Fig.- 6.14)
The normal component, anBA, is always perpendicular to the relative velocity vector and it points towards the center of curvature of the trajectory. In this case, it points towards point A.
The tangential component, atBA, has the same direction as the relative velocity vector. In the example above as the direction of angular acceleration α opposes the direction of angular velocity ω, the tangential component points in the opposite direction to relative velocity vBA, which means that the magnitude of this velocity is decreasing. These normal and tangential components of the relative acceleration of point B with respect to point A can be obtained by the below equations
anBA=ω2.AB−−− (Eqn.-6.9)
atBA=AB.α−−− (Eqn.-6.10)
Where,
ω is the angular velocity of the link.
α is the angular acceleration of the link.
AB is the length of the link.
The magnitude of the acceleration can be determined by the magnitudes of its normal and tangential components.
aA=anA2+(atA)2−−−−−−−−−−√−−− (Eqn.-6.11)
The angle formed by the acceleration vector and the normal direction to the trajectory is given by
ϕ=tan−1atAanA=tan−1αω2−−− (Eqn.-6.12)
The above equations are valid only when the radius is constant.
Acceleration in four bar mechanism -Polygon method
![](/CUPS/AccelerationAndVelocity_files/img18.jpg)
Acceleration 4Bar
For a known four-bar mechanism, in a given configuration and known velocities, and a given angular acceleration of the crank, α2 (say CCW), construct the acceleration polygon. Determine α3 and α4 .
Secondary equation :
We can use the polygon method to determine the acceleration of a coupler point, such as P. It is assumed that all the angular velocities and acceleration have already been determined.
For the position vector
RPO2=RAO2+RPA−−− (Eqn.-6.13)
acceleration expression becomes
AP=AA+APA=AnA+AtA+AnPA+AtPA−−− (Eqn.-6.14)
−ω22RAO2+α2RAO2−ω23RPA+α3RPA−−− (Eqn.-6.15)
All four vector can be constructed graphically. The vector sum is the acceleration of P.
![](/CUPS/AccelerationAndVelocity_files/img34.jpg)
Acceleration Secondary