DC MECH 242Rigid Body Kinematics I1
2. Rigid Body Kinematics I –
relative motion analysis using
translating axes
(16.116.7 –Hibbeler)
Dr. Daniela Constantinescu
danielac@me.uvic.ca
EOW 543
DC MECH 242Rigid Body Kinematics I2
2.1. Road Map*
Dynamics
Particles
Rigid bodies
Kinematics
Kinetics
I Relative motion –analysis using a
translating frame of reference
II Relative motion –analysis using a
rotating frame of reference
Forcemassacceleration method
Workenergy method
Impulsemomentum method
•Figures numbered 16.* are taken from the course textbook:
Hibbeler, Engineering Mechanics: Dynamics, 11e, Copyrigth2008, Prentice Hall.
DC MECH 242Rigid Body Kinematics I3
2.1. Road Map (ctd.) Notes:
•Two righthanded rectangular
frames will be used throughout
this section:
one
fixed frame Oxyz
shown in
red;
one
translating frame Ax’y’z’with
axes parallel to the axes of the
fixed frame (x’IIx, y’IIy, z’IIz) and
with origin moving with point A on
the rigid
shown in green.
•Absolute motion
(positions,
velocities and accelerations) of
points will be
observed from the
fixed reference frame
.
•Relative motion
(positions,
velocities and accelerations) of
points will be
observed from the
translating reference frame
.
DC MECH 242Rigid Body Kinematics I4
2.2. Objectives:1.To fully describe the planar motion of
rigid bodies.
2.To establish relationships between the
motion of points on the same rigid
body.
3.To establish relationships between the
motion of points on connected rigid
bodies that have nonslipping contacts
to other moving bodies (eg., hinges,
nonslipping cables, meshing teeth,
nonslipping rollers).
DC MECH 242Rigid Body Kinematics I5
2.3. Basic premise: ChaslesTheorem The motion of a rigid body consists of translation with one point on the body
plus rotation about the same point.
Can be shown using:
(1)
(A & P on the same rigid body, i.e., ).
Velocity formula:
(2)
Acceleration formula:
(3)
Note:
•For a rigid body, the angular velocity and angular
acceleration are the sameat all points on the rigid body.
×=
+=
AP
AP
APAP
r
dt
rd
rrr
/
/
/
r
r
r
r
r
r
ω
ctr
AP
=
/
r
AP
A
APPP
P
rv
dtrd
dt
rd
dt
rd
v
/
/
r
r
r
r
r
r
r
×+=+==
ω
()
APAPA
AP
AP
AP
P
rra
dt
rd
r
dt
d
dt
vd
dt
vd
a
//
/
/
r
rr
r
r
r
r
r
r
r
r
r
r
××+×+=
=×+×+==
ωωα
ω
ω
ω
r
α
r
DC MECH 242Rigid Body Kinematics I6
2.4. Translation Definition: A rigid body is translating if and
only if any lineof the body
remains parallel to itself
throughout the motion.
Mathematically:
It follows that:
and the velocity and the acceleration
formulae become:
0
≡
ω
r
0≡=
•
ωα
rr
def
Notes:
•All points on a translating rigid body move
along parallel paths (the kinematic constraints
of the rigid body determine whether the paths are rectilinear orcurvilinear).
•A translating rigid body has 3DOF (translation is fully described by the
displacement of one point, ). If translation is planar, the rigid body
has 2DOF.
≡
≡
AP
AP
aa
vv
rr
r
r
)(trr
AA
r
r
=
DC MECH 242Rigid Body Kinematics I7
2.5. Rotation about a fixed axis Definition: A rigid body is rotating about a fixed
axis if and only iffone lineof the body
remains fixed throughout the motion.
Mathematically:
If A is chosen on the fixed axis of rotation, the
velocity and acceleration formulae become:
Notes:
•All points on a rigid body rotating about a fixed axis move along circles
centered on the axis of rotation.
•A rigid body rotating about a fixed axis has 1DOF (rotation about a fixed axis
is fully described by a single kinematic parameter, the angular displacement
about the axis of rotation ).
)(t
θ
θ
=
APnAPtAPAPP
APAPP
aarra
vrv
/,/,/
2
/
//
rrrr
r
r
r
r
r
r
+=−×=
=×=
ωα
ω
kkkk
dt
d
kk
))))
r
)
)
r
•••
•
====
==
θωα
ω
α
θωω
.
.
DC MECH 242Rigid Body Kinematics I8
2.6. Planar motion Definition: A rigid body is in plane
motion if and only if a plane
of the body remains in a
fixed planethroughout the
motion.
Chasles’Theorem for planar motion
The planar motion of a rigid body
consists of translation with one
pointon the body plus rotation
about an axis through the same
pointand perpendicular on the
plane of the motion at this point.
Notes:
•For a rigid body in planar motion, the angular velocity and the angular
acceleration have constantdirection (perpendicular on the plane of
motion):
APAP
aavv
r
r
r
r
r
r
,,,⊥
αω
ω
r
α
r
DC MECH 242Rigid Body Kinematics I9
•If P and A are both in the plane of motion (i.e., ) then:
and the velocity and the acceleration formulae become:
•Graphically, plane motion can be represented as the superposition of a
translation with point A and a rotation about an axis perpendicular on the
plane of motion at A as follows*:
(
)
APAP
rr
/
2
/
r
r
r
r
ωωω
−=××
APnAPtAAPAPAP
APAAPAP
aaarraa
vvrvv
/,/,/
2
/
//
rrrrr
r
rr
r
r
r
r
r
r
++=−×+=
+=×+=
ωα
ω
AP
r
/
r
r
⊥
ω
* Red vectors represent the absolute motion of points (observed from the fixed frame), while green
vectors represent the relative motion of points (observed from atranslating frame with origin at A).
DC MECH 242Rigid Body Kinematics I10
•At each moment in time, there exists anaxis at reston a rigid body in plane
motion. This axis is called the instantaneous axis of rotationand is
perpendicular on the plane of motion. The point on the instantaneous axis of
rotation in the plane of motion is called instantaneous center of zero
velocity (ICZV).
•A rigid body in planar motion has 3DOF (i.e., planar motion is fully
described by the displacement of one point plus the angular
displacement of the rigid body about an axis perpendicular on the plane of
motion ).
Example mechanism: a system of rigid bodies connected to each other, in
translation, in rotation about a fixed axis, and in general plane motion (i.e.,
translation plus rotation).
)(trr
AA
r
r
=
)(t
θ
θ
=
DC MECH 242Rigid Body Kinematics I11
2.6.1. Plane motion of a rigid body properties of the ICZV
1.The ICZV is on a line passing through a point A on the rigid body and
perpendicular on the velocity of point A.
The ICZV can be found if the direction of the velocity of two points on
the rigid body is known.
DC MECH 242Rigid Body Kinematics I12
2.The ICZV is not necessarilyon
the rigid body.
3.The ICZV is an instantaneouscentre of rotation,
i.e., it has zero velocity, but its acceleration is
always different than zero.
4.If two points on a rigid body:
1.have the same velocity ;
or
2.have parallel velocities that are not
perpendicular on the line between the two
points ,
then the ICZV of the body is at infinityand the body
is instantaneously translating.
≠
=
0
0
ICZVICZV
a
v
r
r
0
=
⇒
ω
r
AB
vv
r
r
=
ABvv
AB
⊥
r
r
DC MECH 242Rigid Body Kinematics I13
Note the major difference between an instantaneous axis of rotation
(instantaneous centre of zero velocity) and a fixed axis of rotation (fixed
centre of zero velocity):
The instantaneous centre of zero velocity has zero velocityand
acceleration different than zero, i.e., it is at rest instantaneously (it will
start moving at the next moment in time).
The fixed centre of zero velocity has zero velocityand zero
acceleration, i.e., it is at rest throughout the motion (it will remain at rest
at next moment in time).
≠
=
0
0
ICZVICZV
a
v
r
r
≡
≡
0
0
A
A
a
v
r
r
, A on fixed axis of rotation.
DC MECH 242Rigid Body Kinematics I14
2.7. Problem solving technique
1.Number the rigid bodies, starting from the rigid body with knownmotion.
2.Determine the connections (kinematic constraints) between the rigid
bodies, and between the rigid bodies and the ground.
3.Determine the motion of each rigid body: translation, or rotation about a
fixed axis, or plane motion.
4.Choose a coordinate frame in which to represent all vectors.
5.Analyze of motion of the mechanism using the following steps:
1.Start from the rigid body with known motion.
2.Use to determine:
1.the velocity of point P connecting the body with known motion ( )
to the next body in the mechanism; and
2.the angular velocity of the next body in the mechanism.
3.Repeat step 2 until the angular velocities of all rigid bodies in the
mechanism have been determined.
4.Start from the rigid body with know motion.
5.Use to determine:
1.the acceleration of point P connecting the body with known motion
( ) to the next body in the mechanism; and
2.the angular acceleration of the next body in the mechanism.
6.Repeat step 5 until the angular accelerations of all rigid bodies in the
mechanism have been determined.
APAPAP
rvv
/
r
r
r
r
×
+
=
ω
AP
ω
)(
//APAPAPAPAPAP
rraa
r
r
r
r
r
r
r
×
×
+
×
+
=
ω
ω
α
APAP
αω
r
r
,
DC MECH 242Rigid Body Kinematics I15
2.7. Problem solving technique (ctd)
Notes:
•The velocity analysis should typically be completed before starting the
acceleration analysis.
•In Equations
and
A and P are points on the same rigid body chosen such that:
1.the motion of A is known, i.e., and are known.
2.the path of P is known locally. This means that:
1.during the velocity analysis, the direction of is known
(tangent to the path);
2.during the acceleration analysis:
1.the direction of its tangential acceleration is known
(tangent to the path);
2.its normal acceleration is known: it has direction
from P to the centre of curvature of its path, and it has
magnitude
.
ρ
/
2
,PPn
va=
A
v
r
A
a
r
P
v
r
Pt
a
,
r
Pn
a
,
r
APAPAP
rvv
/
r
r
r
r
×
+
=
ω
)(
//APAPAPAPAPAP
rraa
r
r
r
r
r
r
r
×
×
+
×
+
=
ω
ω
α
DC MECH 242Rigid Body Kinematics I16
Example
1.3 rigid bodies: slider P, link BP, link AB.
2.Body connections (kinematic constraints):
1.Slider P: connected to the ground.
2.Link BP: pinned to link AB at B, pinned to the slider at P.
3.Link AB: pinned to link BP at B , pinned to the ground at A.
3.Body motion:
1.Slider P: rectilinear translation.
2.Link BP: plane motion (does not have a grounded pin and does notremain parallel to a
fixed line throughout the motion), i.e., translation plus rotation.
3.Link AB: rotation about the fixed axis through A and perpendicular on the plane of the
diagram (the motion plane).
4.Coordinate frame for vector representation: Origin at A, xaxis horizontal, yaxis
vertical, zaxis out of the page.
DC MECH 242Rigid Body Kinematics I17
5.Velocity analysis:
1.P is the point with known motion on the link BP:
2.From P, move to B (path of B is circular, with centre at A). UseChasles’theorem
to relate the motion of the points P and B on the link BP and solve:
3.B is the connection between links AB and BC and its motion has been
determined. Use Chasles’stheorem to relate the motion of the points A and B on
the link AB and solve to compute the angular velocity of the link AB:
Link AB is instantaneouslyat rest.
P
v
r
0)
ˆ
)sin(
ˆ
)(cos(
ˆ
00
/
=⇒+×+=
⇒
×
+
=
ABAB
ABABAB
jiABk
rvv
ωθθω
ω
r
r
r
r
−=
=
⇒
=−
⋅+=
⇒×+=−
⇒×+=
BP
v
v
v
BPvv
jPBkivjiv
rvv
P
BP
B
B
BPPB
BPPB
PBBPPB
ω
θ
ωθ
ωθθ
ω
0
0)cos(
)sin(
ˆˆˆ
)
ˆ
)cos(
ˆ
)(sin(
/
r
r
r
r
DC MECH 242Rigid Body Kinematics I18
6.Acceleration analysis:
1.P is the point with known motion on the link BP:
2.From P, move to B (path of B is circular, with centre at A). UseChasles’theorem to
relate the motion of the points P and B on the link BP:
The one vector equation above is equivalent to 2 scalar equations (plane motion), and
contains 3 unknowns: the magnitude and the direction of the acceleration of B, and the
magnitude of the angular acceleration of the link BP. Additionalinformation about the
path of P needs to be added to allow the equation above to be solved.
The additional information about the path of B is provided by the link AB. Link AB rotates
about a fixed axis through A. Hence, B moves on a circle with the centre at A.
Mathematically, this information is available in Chasles’theorem applied to the link AB.
P
a
r
jBPiBPaa
jBPjBPkiaa
rraa
BPBPPB
BPBPPB
PBBPPBBPPB
ˆˆ
)(
ˆˆˆˆ
2
2
/
2
/
⋅−⋅−−=
⇒
⋅−×+−=
⇒
−×+=
ωα
ωα
ωα
r
r
r
r
r
r
r
DC MECH 242Rigid Body Kinematics I19
3.Use Chasles’theorem to relate the motion of the points A and B on the link AB:
Combining the two 2 vector equations (i.e., the 4 scalar equations) above, the
magnitude and the direction of the acceleration of B, and the angular accelerations of
the links BP and AB can be determined. Solving for the angular accelerations of the
links BP and AB:
jABiABa
jiABka
rraa
ABABB
ABB
ABABABABAB
ˆ
)cos(
ˆ
)sin(
)
ˆ
)sin(
ˆ
)(cos(
ˆ
/
2
/
θαθα
θθα
ωα
⋅⋅+⋅⋅−=
⇒+×=
⇒−×+=
r
r
r
r
r
r
r
⋅
⋅
−=
⋅⋅+
−=
⇒
⋅⋅=⋅−
⋅⋅−=⋅−−
)cos(
)tan(
)cos(
)sin(
2
2
2
θ
ω
α
θω
α
θαω
θαα
AB
BP
BP
BPa
ABBP
ABBPa
BP
AB
BPP
BP
ABBP
ABBPP
DC MECH 242Rigid Body Kinematics I20
2.8. Road up to here:
Dynamics
Particles
Rigid bodies
Kinematics
Kinetics
I. Relative motion –analysis using a
translating frame of reference
II. Relative motion –analysis using
a rotating frame of reference
Forcemassacceleration method
Workenergy method
Impulsemomentum method
Notes:
We have used two reference systems:
one fixed coordinate frame
Oxyz
one translating coordinate frame
Ax’y’z’
with axes parallel to the axes of
the fixed frame (x’IIx, y’IIy, z’IIz) and with origin attached to point A on the
body whose motion we have been interested in
and the following equations:
(i.e., Chasles’theorem) to establish relationships between:
the motion of points on the same rigid body;
the motion of connected rigid bodies that do not slip with respect to each
other at the point where they connect and/or that are connected with sliders
that slide on the ground or on translating rigid bodies.
A and P in the equations above are points on the same rigid body.
AnPAtPAAPAPAPAPAPAP
APAAPAPAP
aaarraa
vvrvv
////
//
)(
rrrr
rr
r
r
rr
r
r
r
r
r
r
++=××+×+=
+
=
×
+
=
ωωα
ω
DC MECH 242Rigid Body Kinematics I21
Consider a third coordinate frame, glued to the rigid body,
Ax’’y’’z’’.
An observer sitting in the
fixed frameOxyz
and looking at P sees the
absolute motion of P
(translation of P with A and rotation of P about A):
An observer translating with A (sitting in the
translating frameAx’y’z’
) and
looking at P sees only the
rotation of P about A
:
An observer translating with A and rotating with the body (sitting in the
frame
Ax’’y’’z’’glued to the rigid body
) and looking at P sees
P at a fixed location
(A and P are on the same rigid body):
)(
//
/
APAPAPAPAPAP
APAPAP
rraa
rvv
r
rr
r
r
rr
r
r
r
r
××+×+=
×
+
=
ωωα
ω
)(
/////
//
APAPAPAPAPAnPAtPAP
APAPAP
rraaa
rv
r
rr
r
r
rrr
r
r
r
××+×=+=
×
=
ωωα
ω
AP
r
/
r
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