Kinematic Analysis of Mechanisms(3)

The acceleration analysis plays an important role in the development of machines and mechanisms.

Acceleration diagram for link

Fig: Acceleration for a link

Consider the two points A and B on a rigid link as shown in fig (a). Let the point B moves with respect to A with angular velocity of ω rad/s and let α rad/s2 be the angular acceleration of the link AB. The acceleration of a particle whose velocity changes both in magnitude and direction at any instant has the following two components.

• The radial or centripetal component which is perpendicular to the velocity of the particle at the given instant.
• The tangential component which is parallel to the velocity of the particle at the given instant.

Thus for given link AB the velocity of point B with respect to A i.e. v_BA is perpendicular to the link AB as shown in fig (a). As the point B moves with respect to A with angular velocity of ω rad/s. Therefore centripetal or radial component of the acceleration of B with respect to A,

a^r_BA = ω² × Length of link AB = ω² × AB = v²_BA / AB

This radial component of acceleration acts normal to velocity v_BA. In other words it acts parallel to link AB.

We know that tangential component of the acceleration of B with respect to A,

a^r_BA = α × Length of the link AB = α × AB

This tangential component of acceleration acts as parallel to velocity v_BA. In other words it acts perpendicular to the link AB. In order to draw acceleration diagram for a link AB which is shown in fig (b) from any point b', draw vector b'x parallel to BA which represent the radial component of acceleration of B with respect to A i.e. a^r_BA and from point x draw vector xa' perpendicular to the BA to represent tangential component of acceleration of B with respect to A i.e. a^t_BA. Join b' a'. The vector b' a' known as acceleration image of link AB represents the total acceleration of B with respect to A i.e. a_BA and it is vector sum of radial component (a^r_BA) and tangential component (a^t_BA) of acceleration.

Acceleration of a Point on a Link

Fig: Acceleration of a point on a link

Consider two points A and B on rigid link which is shown in fig (a). Let the acceleration of the point A i.e. a_A is known in magnitude and direction of path of B is given. The acceleration of point B is determined in magnitude and direction by drawing acceleration diagram as given below.

1. From any point o' draw vector o'a' parallel to the direction of absolute acceleration at point A i.e. a_A , to some suitable scale as shown in fig (b).
2. We know that the acceleration of B with respect to A i.e. a_BA has two components.

• Radial component of acceleration of B with respect to A i.e. a^r_BA , and
• Tangential component of the acceleration B with respect to A i.e. a^t_BA. These two components are mutually perpendicular.

3. Draw vector a'x parallel to link AB because radial component of acceleration of B with respect to A will pass through AB, such that

Vector a′x = a^r_BA = v²_BA/AB

Where v_BA = Velocity of B with respect to A.

4. From the point x draw vector xb' perpendicular to AB or vector a'x because tangential component of B with respect to A i.e. a^t_BA, is perpendicular to radial component a^r_BA and through o' draw a line parallel to the path of B to represent absolute acceleration of B is a_B. The vectors xb' and o' b' intersect at b'. Now the values of a_B and a^t_BA at may be measured to the scale.
5. By joining points a' and b' we determine total acceleration of B with respect to A i.e. a_BA. The vector a'b' is known as acceleration image of link AB.
6. For any other point C on link draw triangle a'b'c' similar to triangle ABC. Now vector b'c' represents acceleration of C with respect to B i.e. a_CB and vector a' c' represents the acceleration of C with respect to A is a_CA. As a_CB and a_CA will each have two components as follows.

• a_CB has two components; a^r_CB and a^t_CB as shown by triangle b' zc' in fig (b) , in which b' z is parallel to BC and zc' is perpendicular to b'z or BC.
• a_CA has two components a^r_CA and a^t_CA as shown by triangle a' yc' in Fig (b) in which a'y is parallel to AC and yc' is perpendicular to a'y or AC.

7. The angular acceleration of link AB is obtained by dividing tangential components of acceleration of B with respect to A (a^t_BA ) to the length of the link. Mathematically, angular acceleration of the link AB,

α_AB = a^t_BA/ AB

Coriolis Component of Acceleration

When a point on one link is sliding along another rotating link such as in quick return motion mechanism, then the Coriolis component of the acceleration must be calculated. Consider a link OA and a slider B as shown in fig (a). The slider B moves along the link OA. The point C is the coincident point on the link OA.

Let

ω = Angular velocity of the link OA at time t seconds.

v = Velocity of the slider B along the link OA at time t seconds.

ω.r = Velocity of the slider B with respect to O (perpendicular to the link OA)

At time t seconds and

(ω + δω), (v + δv) and (ω + δω) (r + δr) = Corresponding values at time (t + δt) seconds.

Fig: Coriolis component of acceleration.

Let us now find out the acceleration of the slider B with respect to O and with respect to its coincident point C lying on link OA.

∴ Radial component of the acceleration of the slider B with respect to O on the link OA, acting radially outwards from O to A,

$$a_{_{BO}}^{^{{r_{}}}} = Lt\frac{{\delta v - \omega .r.\delta \theta }}{{\delta t}} = \frac{{dv}}{{dt}} - \omega .r\frac{{d\theta }}{{dt}} = \frac{{dv}}{{dt}} - {\omega ^2}r......................(i)$$

Also, the total component of change of velocity along tangential direction

= xb1 + by = v. δ θ + (ω.δr + r.δω)……………………………………..... (Perpendicular to OA and towards left)

∴ Tangential component of acceleration of slider B with respect to O on the link OA acting perpendicular to OA and towards left,

$$\begin{array}{l} a_{_{BO}}^t = Lt\frac{{v\delta \theta + (\omega .\delta r + r\delta \omega )}}{{\delta t}} = v\frac{{d\theta }}{{dt}} + \omega .\frac{{dr}}{{dt}} + r.\frac{{d\omega }}{{dt}}\\ \;\;\;\;\;\;\; = v\omega + \omega v + r\alpha \;\;\;\;\;\;\;......................(ii) \end{array}$$

Now radial component of acceleration of the coincident point C with respect to O, acting in a direction from C to O,

a^r_CO = ω².r ↑ ...(iii)

And tangential component of acceleraiton of the coincident point C with respect to O acting in a direction perpendicular to CO and towards left,

A^t_CO = α.r← ...(iv)

Radial component of the slider B with respect to the coincident point C on the link OA, acting radially outwards,

a^r_BC = a^r_BO − a^r_CO = dv/dt↑

And tangential component of the slider B with respect to coincident point C on link OA acting in a direction perpendicular to OA and towards left,

a^t_BC = a^t_BO − a^t_CO = (2ω.v + α.r) − α.r = 2ω.v←

Fig: Direction of Coriolis component of acceleration

This tangential component of acceleration of slider B with respect to the coincident point C on link is known as Coriolis component of acceleration and is always perpendicular to link.

∴ Coriolis component of acceleration of B with respect of C,

a^c_BC = a^t_BC = 2ω.v

Where ω = Angular velocity of the link OA

v = Velocity of slider B with respect to coincident point C.

The anticlockwise direction for ω and the radially outward direction for v are taken as positive. It may be noted that the direction of Coriolis component of acceleration changes sign, if either ω or v is reversed in direction. But the direction of Coriolis component of acceleration will not changed in sign if both ω and v are reversed in direction. It is concluded that given direction of Coriolis component of acceleration is obtained by rotating v at 90° about its origin in the same direction as that of ω. The direction of Coriolis component of acceleration (2 ω.v) for all four possible cases is shown in fig. The directions of ω and v are given.

Kinematic Analysis by Complex Number

This is done by loop closure equation which is the vector equation that represent the closed polygon which encompasses all active joints in the mechanism.

Consider the four bar linkage

Fig: four bar linkage

Now the loop closure equation is given by

$$\overrightarrow {{r_1}} \; + \overrightarrow {{r_2}} \; + \overrightarrow {{r_3}} \; + \overrightarrow {{r_4}} \; = 0........................(a)$$

This can be written in complex number notation as

r₁е^jθ1+ r₂ е^jθ2+ r₃ е^jθ3+ r₄ е^jθ4 = 0………………….(b)

Now the Eq (b) can be solved for the position analysis and for the velocity and acceleration again differentiation twice and thrice respectively.

References:
1. H.H. Mabie and C. F. Reinholtz, “Mechanism and Dynamics of Machinery”, Wiley.
2. J.S. Rao & R.V. Dukkipati Mechanisms and Machine Theory, New Age International (P) Limited..
3. J.E. Shigley and J.J. Uicker, Jr., “ Theory of Machines and Mechanisms”, McGraw Hill.
4. B. Paul, “Kinematics and Dynamics of Planar Machinery”, Prentice Hall.
5. C. E. Wilson, J.P. Sadler and W.J. Michels, “Kinematics and Dynamics of Machinery”, Harper Row.