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1. Acceleration Analysis
Accelerationanalysisinvolvesdeterminingthemannerinwhichcertainpointsonthelinksofa
mechanismareeither“speedingup”or“slowingdown.”Accelerationisacriticalproperty
becauseoftheinertialforcesassociatedwithit.Inthestudyofforces,SirIsaacNewton
discoveredthataninertialforceisproportionaltotheaccelerationimposedonabody.This
phenomenoniswitnessedanytimeyoulungeforwardasthebrakesareforcefullyappliedon
yourcar.
Of course, an important part of mechanism design is to ensure that the strength of the links and
joints is sufficient to withstand the forces imposed on them. Understanding all forces, especially
inertia, is important. Force analysis is introduced in Chapters 13 and 14.However, as a
preliminary step, acceleration analysis of a mechanism’s links must be performed.
The determination of accelerations in a linkage is the purpose of this chapter. The primary
procedure used in this analysis is the relative acceleration method, which utilizes the results of
the relative velocity method introduced in Chapter 6. Consistent with other chapters in this
book, both graphical and analytical techniques are utilized.

3. Ex. The crank and connecting rod of a theoretical steam engine are 0.5 m and 2 m long respectively. The crank makes 180 r.p.m. in
the clockwise direction. When it has turned 45°from the inner dead center position, determine : 1. velocity of piston, 2. angular
velocity of connecting rod, 3. velocity of point E on the connecting rod 1.5 m from the gudgeon pin, 4. velocities of rubbing at the
pins of the crank shaft, crank and crosshead when the diameters of their pins are 50 mm, 60 mm and 30 mm respectively, 5.
position and linear velocity of any point G on the connecting rod which has the least velocity relative to crank shaft.

6. Forces Acting in a Mechanism
Consider a mechanism of a four barchain, Let force FA newton is acting at the joint A in the direction of the velocity of A (vAm/s)
which is perpendicular to the link DA. Suppose a force FB newton is transmitted to the joint B in the direction of the velocity of B
(i.e.vBm/s) which is perpendicular to the link CB. If we neglect the effect of friction and the change of kinetic energy of the link
(i.e.), assuming the efficiency of transmission as 100%), then by the principle of conservation of energy,

7. Mechanical Advantage
It is defined as the ratio of the load to the effort. In a four barmechanism, as shown in Fig. the link DA is called the driving
link and the link CB as the driven link. The force FA acting at A is the effort and the force FB at B will be the load or the
resistance to overcome. We know from the principle of conservation of energy, neglecting effect of friction,
If we consider the effect of friction, less resistance will be
overcome with the given effort. Thereforethe actual mechanical
advantage will be less.

11. LINEAR ACCELERATION
Linear acceleration, A, of a point is the change of linear velocity of that point per unit of time.
Chapter 6 was dedicated to velocity analysis. Velocity is a vector quantity, which is defined
with both a magnitude and a direction. Therefore, a change in either the magnitude or
direction of velocity produces an acceleration. The magnitude of the acceleration vector is
designateda = ??????
Linear Acceleration of Rectilinear Points
Consider the case of a point having straight line, or rectilinear, motion. Such a point is most commonly found on a
Link that is attached to the frame with a sliding joint. For this case, only the magnitude of the velocity vector can
change. The acceleration can be mathematically described as
Because velocity is a vector, equation (7.1) states that
acceleration is also a vector. The direction of linear
acceleration is in the direction of linear movement when
the link accelerates. Conversely, when the link decelerates,
the direction of linear acceleration is opposite to the
direction of linear movement.

12. Linear acceleration is expressed in the units of velocity (length per time) divided by time, or
length per squared time. In the U.S. Customary System, the common units used are feet per
squared second (ft/s
2
) or inches per squared second (in./s
2
). In the International System, the
common units used are meters per squared second (m/s
2
) or millimeters per squared second
(mm/s
2
).
Constant Rectilinear Acceleration
Rewriting equation (7.3), the velocity change that occurs during a period of constant
acceleration is expressed as
Additionally, the corresponding displacement that occurs during a period of constant acceleration can be written
as

13. Since rectilinear motion is along a straight line, the direction of the displacement, velocity, and
acceleration (r, v, a) can be specified with an algebraic sign along a coordinate axis.
Thus, equations (7.4), (7.5), and (7.6) can be written in terms of the vector magnitudes (r, v, a).
An express elevator used in tall buildings can reach a full speed of 15 mph in 3 s. Assuming thatthe elevator
experiences constant acceleration, determine the acceleration and the displacement during the 3 s.
EXAMPLE
Because the elevator starts at rest, the velocity change is calculated as
Normalize the Acceleration with Respect to Gravity
When people accelerate in an elevator, the acceleration is often “normalized” relative to the acceleration due to
gravity. The standard acceleration due to gravity (g) on earth is 32.17 ft/s2 or 9.81 m/s2. Therefore, the acceleration
of the elevator can be expressed as
Calculate the Displacement during the 3-Second Interval
The displacement can be determined from equation

14. EXAMPLE
An automated assembly operation requires linear motion from a servo actuator. The total displacement must be
10 in. For design reasons, the maximum velocity must be limited to 2 in./s, and the maximum acceleration or
deceleration should not exceed 4 in./s2. Plot the velocity profile for this application.
SOLUTION: 1. Determine the Motion Parameters during Speed-Up
Determine the Motion Parameters during Slow-Down
Determine the Motion Parameters during Steady-State
Determine the Motion Parameters during Steady-State

23. CORIOLISACCELERATION
Throughouttheprecedinganalyses,twocomponentsofan
accelerationvector(i.e.,normalandtangential)were
thoroughlyexamined.Incertainconditions,athird
componentofaccelerationisencountered.Thisadditional
componentisknownastheCorioliscomponentof
accelerationandispresentincaseswhereslidingcontact
occursbetweentworotatinglinks.
Mechanismsusedinmachineshavebeenknowntofaildue
tothemistakenomissionofthisCorioliscomponent.
OmittingtheCorioliscomponentunderstatesthe
accelerationofalinkandtheassociatedinertialforces.The
actualstressesinthemachinecomponentscanbegreater
thanthedesignallows,andfailuremayoccur.Therefore,
everysituationmustbestudiedtodeterminewhethera
Coriolisaccelerationcomponentexists.
Specifically, the Coriolis component is encountered in the relative
acceleration of two points when all of the following three
conditions are simultaneously present:
1. The two points are coincident, but on different links;
2. The point on one link traces a path on the other link; and
3. The link that contains the path rotates.
❑The points are coincident, but not on the same link
(condition 1);
❑Point B2 slides along and traces a path on link 4
(condition 2); and
❑The link that contains the path, link 4, rotates
(condition 3).

25. 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 corioliscomponent of the acceleration must be calculated. Consider a link OA and a slider B as shown in Fig.
The slider B moves along the link OA. The point C is the coincident point on the link OA.
Let = Angular velocity of the link OA at time t seconds.
v = Velocity of the slider B along the link OA at time t seconds.
.r = Velocity of the slider B with respect to O (perpendicular to the link OA) at time t seconds, and
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 the link OA.

26. the ratio of the time of cutting to the time of return stroke
Since the tool travels a distance of R1 R2 during cutting and
return stroke, therefore travel of the tool or length of stroke

27. Ex. A mechanism of a crank and slotted lever quick return motion is shown in Fig. 8.28. If the
crank rotates counter clockwise at 120 r.p.m., determine for the configuration shown, the
velocity and acceleration of the ram D. Also determine the angular acceleration of the slotted
lever. Crank, AB = 150 mm ; Slotted arm, OC = 700 mm and link CD = 200 mm.
Solution.
v
D
V
B
V
BB’

28. a
B’
a
D
a
t
OB’
B’’

29. Ex. The driving crank AB of the quick-return mechanism, as shown in Fig.,
revolves at a uniform speed of 200 r.p.m. Find the velocity and acceleration
of the tool-box R, in the position shown, when the crank makes an angle of
60°with the vertical line of centresPA. What is the acceleration of sliding
of the block at B along the slotted lever PQ ?
VR
VB
B
O,P
VBB
VQ
B’
VRQ

31. Figure illustrates handheld grass shears, used for trimming areas that are
hard to reach with mowers. The drive wheel rotates counterclockwise at
400 rpm. Determine the angular acceleration of the oscillating blades at
the instant shown.