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KINEMATICS OFKINEMATICS OF
RIGID BODIESRIGID BODIESRIGID BODIESRIGID BODIES
Introduction
In rigid body kinematics, we use the relationships
governing the displacement, velocity and acceleration, but
must also account for the rotational motion of the body.
Description of the motion of rigid bodies is important for
two reasons:
1) To generate, transmit or control motions by using
cams, gears and linkages of various types and analyze the
displacement, velocity and acceleration of the motion to
determine the design geometry of the mechanical parts.
Furthermore, as a result of the motion generated, forces
may be developed which must be accounted for in the
design of the parts.
2) To determine the motion of a rigid body caused by the
forces applied to it. Calculation of the motion of a rocket
under the influence of its thrust and gravitationalunder the influence of its thrust and gravitational
attraction is an example of such a problem.
Rigid Body Assumption
A rigid body is a system of particles for which the
distances between the particles and the angle between
the lines remain unchanged. Thus, if each particle of such
a body is located by a position vector from reference axes
attached to and rotating with the body, there will be no
change in any position vector as measured from these
axes.
Of course this is an idealization since all solid materials
change shape to some extent when forces are applied to
them. Nevertheless, if the movements associated with the
changes in shape are very small compared with the
movements of the body as a whole, then the assumption of
rigidity is usually acceptable.rigidity is usually acceptable.
Plane Motion
All parts of the body move in parallel planes.
The plane motion of a rigid body is divided into several
categories:categories:
1. Translation
2. Rotation
3. General Motion
1. TRANSLATION
It is any motion in which every line in the body remains parallel to itsoriginal position at all times. In translation, there is no rotation of anyline in the body.
1. Rectilinear Translation: All points in the body move in parallelstraight lines.
Rocket test sled
2. Curvilinear Translation: All points move on congruent curves.
In each of the two cases of translation, the motion of the body is
completely specified by the motion of any point in the body, since all
the points have the same motion.
2. Fixed Axis Rotation
Rotation about a fixed axis is the angular motion about the axis. All
particles in a rigid body move in circular paths about the axis of rotation
and all lines in the body which are perpendicular to the axis of rotation
rotate through the same angle at the same time.
A
B
C
A′′′′B′′′′
C′′′′
3. General Plane Motion
It is the combination of translation and rotation.
A
A′′′′
B′′′′
B
ωωωωO
Crank (Krank)
(Rotation)
Connecting rod (General Motion)
Piston
(Translation)
hingehinge
Rotation
The rotation of a rigid body is described byits angular motion. The figure shows a rigidbody which is rotating as it undergoes planemotion in the plane of the figure. The angularpositions of any two lines 1 and 2 attached tothe body are specified by θ1 and θ2 measuredfrom any convenient fixed referencedirection.direction.
12 θθ && = 12 θθ &&&& =Because the angle β is invariant, the relation θ2 = θ1 + β upondifferentiation with respect to time gives andduring a finite interval, ∆ θ2 = ∆ θ1.
All lines on a rigid body in its plane of motion have the same angular displacement, the same angular velocity and the same angular acceleration.
Angular Motion Relations
The angular velocity ω and angular acceleration α of a rigid body in
plane rotation are, respectively, the first and second time derivatives
of the angular position coordinate θ of any line in the plane of motion
of the body. These definitions give
dθ
dθdorαdθ ωdω
dt
dor
dt
d
dt
d
θθθ
θθ
αωω
α
θθ
ω
&&&&
&&&
&
==
====
==
2
2
For rotation with constant angular acceleration, the relationships become
( )020
2
0
2
t
θθαωω
αωω
−+=
+=
200
00
2
1tt αωθθ ++=
Rotation About a Fixed Axis
When a rigid body rotates about a fixed axis, all points other than those
on the axis move in concentric circles about the fixed axis. Thus, for
the rigid body in the figure rotating about a fixed axis normal to the
plane of the figure through O, any point such as A moves in a circle of
radius r. So the velocity and the acceleration of point A can be written
as
α
ωω
ω
ra
vrvra
rv
t
n
=
===
=
/22
These quantities may be expressed using the cross product
relationship of vector notation,
rrvvv&vv
×== ω
kkrrrr
ααωω == ,
( ) ( ){ {
{ ( )43421
rvvrrr
vrv
rvrr
rr
rvrv nt aa
rrdt
rdr
dt
dr
dt
dv
dt
da ××+×=×+×=×== ωωαω
ωω
{ {rvrv n
t aarv ×=ωα
1. The angular velocity of a gear is controlled according to
ω = 12 – 3t2, where ω in rad/s and t is the time in seconds.
Find the net angular displacement ∆θ from the time t = 0 to
t = 3 s. Also find the total number of revolutions N through
which the gear turns during the three seconds.
PROBLEMS
( ) ( )
rad
radttdttd
dtddt
d
9
933123
312 , 312
33
0
33
0
2
0
=∆
=−=−=−=
=⇒=
∫∫θ
θθ
ωθωθ
θ
SOLUTION
( ) ( ) radttdttd
statstopsitsttt
1622123
312 312
) 2 ( 2 312 0312
32
0
31
2
0
2
0
22
1
=−=−=⇒−=
====−=
∫∫ θθ
ω
θ
Does the gear stop between t = 0 and t = 3 seconds?
SOLUTION
( )
srevolutionNradsrevolution N
radrevolution
rad
radttdttd
66.3 23
2 1
23716
73
312 312
3
2
32
3
2
2
0
2
=⇒
=−+
−=−=⇒−= ∫∫
π
θθθ
2. The belt-driven pulley and attached disk are rotating with increasing
angular velocity. At a certain instant the speed v of the belt is 1.5 m/s,
and the total acceleration of point A is 75 m/s2. For this instant
determine (a) the angular acceleration α of the pulley and disk, (b) the
total acceleration of point B, and (c) the acceleration of point C on the
belt.
PROBLEMS
belt.
222
222
CB2
45
/456075
/6015.020
/ 20075.0
5.1
?ac)?ab)?a) / 75 / 5.1
a
sma
smRa
sradr
v
smasmv
A
A
C
AC
t
n
=−=
===
===
=====
ω
ω
α
SOLUTION
2
222
222
2
2
/5.22075.0300
/5.37305.22/30075.020
/5.22075.0300
/30015.0
45
smra
smasmra
smra
sradR
a
C
B
B
B
A
n
t
t
=⋅=⋅=
=+=
=⋅=⋅=
=⋅=⋅=
===
α
ω
α
α
3. The design characteristics of a gear-reduction unit are under
review. Gear B is rotating clockwise (cw) with a speed of 300 rev/min
when a torque is applied to gear A at time t=2 s to give gear A a
counterclockwise (ccw) acceleration α which varies with time for a
duration of 4 seconds as shown. Determine the speed NB of gear B when
t=6 s.
PROBLEMS
t=6 s.
SOLUTION
srad
revNst
B
B
/1060
2300
min/3002
ππ
ω =⋅=
=⇒=
The velocities of gears A and B are same at the contact point.
( ) ( ) /202 sradbbvv ABABA =⇒=⇒= πωωω
( )
( ) ( )min/59.414
/415.4326
)6(/83.8622
20
22
6
2
2
6
220
revN
sradbbst
statsradtt
dttddt
dt
B
BBA
AA
t
AA
AA
A
=
=⇒=⇒=
==⇒+=−
+=⇒=⇒+= ∫∫=
ωωω
ωπω
ωω
ααω
π
Absolute Motion
In this approach, we make use of the geometric relations
which define the configuration of the body involved and
then proceed to take the time derivatives of the definingthen proceed to take the time derivatives of the defining
geometric relations to obtain velocities and accelerations.
A wheel of radius r rolls on a flat surface without slipping. Determine
the angular motion of the wheel in terms of the linear motion of its
center O. Also determine the acceleration of a point on the rim of the
wheel as the point comes into contact with the surface on which the
wheel rolls.
PROBLEM
Relative Motion
The second approach to rigid body kinematics uses the principles of
relative motion. In kinematics of particles for motion relative to
translating axes, we applied the relative velocity equation
to the motions of two particles A and B.
BABA vvv /vvv
+=
to the motions of two particles A and B.
We now choose two points on the same rigid body for our two particles.
The consequence of this choice is that the motion of one point as seen by
an observer translating with the other point must be circular since the
radial distance to the observed point from the reference point does not
change.
The figure shows a rigid body moving in the plane of the figure from
position AB to A´B´ during time ∆t. This movement may be visualized
as occurring in two parts. First, the body translates to the parallel
position A´´B´ with the displacement . Second, the body rotates
about B´ through the angle ∆θ, from the nonrotating reference axes
x´-y´ attached to the reference point B´, giving rise to the
displacement of A with respect to B.BAr /
v∆
Brv
∆
BAr /v
∆
With B as the reference point, the total displacement of A is
BABA rrr /vvv
∆∆∆ +=
Where has the magnitude r∆θ as ∆θ approaches zero.
Dividing the time interval ∆t and passing to the limit, we obtain the
relative velocity equation
BAr /v
∆
vvvvvv
+= BABA vvv /vvv
+=
The distance r between A
and B remains constant.
The magnitude of the relative velocity is thus seen to be
which, with becomes
Using to represent the vector , we may write the
dt
dr
t
r
t
rv
t
BA
tBA
θθ=
∆∆
=
∆
∆=
→∆→∆ 0
/
0/ limlim
v
θω &=
ωrv BA =/
BAr /v
rv
rv BAvvv
×=ω/
Using to represent the vector , we may write the relative velocity as the vector
BAr /r
Therefore, the relative velocity equation becomes
rvv BAvvvv
×+= ω
Here, is the angular velocity vector normal to the plane of the
motion in the sense determined by the right hand rule.
It should be noted that the direction of the relative velocity will
always be perpendicular to the line joining the points A and B.
Interpretation of the Relative Velocity Equation
We can better understand the relative velocity equation by visualizing
ωv
We can better understand the relative velocity equation by visualizing
the translation and rotation components separately.
Translation Fixed axid rotation
In the figure, point B is chosen as the reference point and theIn the figure, point B is chosen as the reference point and the
velocity of A is the vector sum of the translational portion , plus
the rotational portion , which has the magnitude
vA/B=rω, where , the absolute angular velocity of AB . The
relative linear velocity is always perpendicular to the line joining the
two points A and B.
rv BAvvv
×=ω/
Bvv
θω &v=
Relative Acceleration
Equation of relative velocity is
By differentiating the equation with respect to time, we obtain the
relative acceleration equation, which is
BABA vvv /vvv
+=
BABA vvv /&v&v&v +=
or
This equation states that the acceleration of point A equals the vector
sum of the acceleration of point B and the acceleration which A appears
to have to a nonrotating observer moving with B.
BABA vvv /&&& +=
BABA aaa /vvv
+=
If points A and B are located on the same rigid body, the distance r
between them remains constant. Because the relative motion is
circular, the relative acceleration term will have both a normal
component directed from A toward B due to the change of direction
of and a tangential component perpendicular to AB due to the
change in magnitude of . Thus, we may write,BAv /v
BAv /v
Where the magnitudes of the relative acceleration components are
( ) ( )tBAnBABA aaaa //
vvvv++=
( )( ) α
ω
rva
rrva
BAtBA
BAnBA
==
==
//
22// /
&
In vector notation the acceleration components are
( ) ( )( ) ra
ra
tBA
nBAvvv
vvvv
×=
××=
α
ωω
/
/
The relative acceleration equation, thus, becomes
( ) rraa BAvvvvvvv
×+××+= αωω ( ) rraa BAvvvv
×+××+= αωω
The figure shows the acceleration of A to be composed of two parts:
the acceleration of B and the acceleration of A with respect to B.
Solution of the Relative Acceleration Equation
As in the case of the relative velocity equation, the
relative acceleration equation may be carried out by
scalar or vector algebra or by graphical construction.
Because the normal acceleration components depend on
velocities, it is generally necessary to solve for the
velocities before the acceleration calculations can be
made.
1. The center O of the disk has the velocity and acceleration shown.
If the disk rolls without slipping on the horizontal surface, determine
the velocity of A and the acceleration of B for the instant
represented.
PROBLEMS
2. The triangular plate has a constant clockwise angular velocity of
3 rad/s. Determine the angular velocities and angular accelerations
of link BC for this instant.
PROBLEMS
3. If the velocity of point A is 3 m/s to the right and is constant for
an interval including the position shown, determine the tangential
acceleration of point B along its path and the angular acceleration of
the bar AB.
PROBLEMS
4. The flexible band F attached to the sector at E is given a
constant velocity of 4 m/s as shown. For the instant when BD is
perpendicular to OA, determine the angular acceleration of BD.
PROBLEMS
5. At a given instant, the gear has the angular motion shown.
Determine the acceleration of points A and B on the link and the
link’s angular accelartion at this instant.
PROBLEMS
6. The center O of the disk rolling without slipping on the horizontal
surface has the velocity and acceleration shown. Radius of the disk is
4.5 cm. Calculate the velocity and acceleration of point B.
PROBLEMS
v =45 cm/s a =90 cm/s
37o
A
O4 cm
4.5 cm
vo=45 cm/s ao=90 cm/s2
10 cm
6 cm
B
x
y
x=2 cm2
4
1xy =