Download Dynamics ME-240 - Department of Aerospace Engineering

Dynamics AE-001
Sharif University-Aerospace Dep. Fall 2004
Rectangular Components
Chapter 2
STATICS OF PARTICLES
Forces are vector quantities; they add according to the
parallelogram law. The magnitude and direction of the
resultant R of two forces P and Q can be determined either
graphically or by trigonometry.
R
P
A
Q
Any given force acting on a particle can be resolved into
two or more components, i.e.., it can be replaced by two
or more forces which have the same effect on the particle.
Q
F
A
P
A force F can be resolved
into two components P
and Q by drawing a
parallelogram which has
F for its diagonal; the
components P and Q
are then represented by
the two adjacent sides
of the parallelogram
and can be determined
either graphically or by
trigonometry.
A force F is said to have been resolved into two rectangular
components if its components are directed along the coordinate
axes. Introducing the unit vectors i and j along the x and y axes,
F = Fx i + Fy j
y
Fx = F cos q
Fy = F sin q
Fy
tan q =
Fx
Fy = Fy j
F
j
q
i
F=
Fx = Fx i
x
2
Fx
+
2
Fy
When three or more coplanar forces act on a particle, the
rectangular components of their resultant R can be obtained
by adding algebraically the corresponding components of the
given forces.
Rx = S Rx
Ry = S Ry
The magnitude and direction of R can be determined from
Ry
tan q =
Rx
R=
2
Rx
+
2
Ry
y
y
B
B
Fy
Fy qy
A
D
O
Fz
E
F
qx
Fx
F
O
x
Fz
E
C
A force F in three-dimensional space
can be resolved into components
Fx
x
y
B
Fy F
Fx = F cos qx Fy = F cos qy
O
Fz = F cos qz
qz
A
Fx
E
z
D
C
z
z
A
Fz
C
D
x
y
l (Magnitude = 1)
Fy j
F=Fl
cos qy j
Fx i
cos qz k
x
The cosines of
qx , qy , and qz
are known as the
direction cosines of
the force F. Using
the unit vectors i , j,
and k, we write
Fz k
z
F = Fx i + Fy j + Fz k
cos qx i
or
F = F (cosqx i + cosqy j + cosqz k )
y
l (Magnitude = 1)
cos qy j
Fy j
cos qz k
F=Fl
Since the magnitude
of l is unity, we have
Fx i
x
Fz k
cos qx i
z
F=
2
Fx +
Fx
cosqx =
F
2
Fy +
cosqy
l = cosqx i + cosqy j
+ cosqz k
cos2qx + cos2qy
+ cos2qz = 1
In addition,
Fz
=
2
Fy
F
cosqz =
Fz
F
y
N (x2, y2, z2)
F
l
dy = y2 - y1
dz = z2 - z1
<0
dx = x2 - x1
M (x1, y1, z1)
x
z
MN
A force vector F
in three-dimensions
is defined by its
magnitude F and
two points M and
N along its line of
action. The vector
MN joining points
M and N is
= dx i + dy j + dz k
The unit vector l along the line of action of the force is
l
=
MN
MN
=
1
( dx i + dy j
d
+ dz k
)
y
N (x2, y2, z2)
d=
2
2
dx + dy + dz
2
dy = y2 - y1
M (x1, y1, z1)
dz = z2 - z1
<0
dx = x2 - x1
x
z
F=Fl
=
A force F is
defined as the
product of F and
l. Therefore,
F
( d x i + dy j
d
From this it follows that
Fdx
Fx =
d
Fdy
Fy = d
Fdz
Fz =
d
+ dz k
)
When two or more forces act on a particle in threedimensions, the rectangular components of their resultant
R is obtained by adding the corresponding components of
the given forces.
Rx = S Fx
Ry = S Fy
Rz = S Fz
The particle is in equilibrium when the resultant of all
forces acting on it is zero.
Vector Algebra
V =PxQ
The vector product of two
vectors is defined as
Q
q
V = Px Q
P
The vector product of P and
Q forms a vector which is
perpendicular to both P and Q, of magnitude
V = PQ sin q
This vector is directed in such a way that a person located at the
tip of V observes as counterclockwise the rotation through q
which brings vector P in line with vector Q. The three vectors P,
Q, and V - taken in that order - form a right-hand triad. It
follows that
Q x P = - (P x Q)
j
k
i
It follows from the definition of the vector
product of two vectors that the vector
products of unit vectors i, j, and k are
ixi=jxj=kxk=0
ixj=k , jxk=i, kxi=j , ixk=-j , jxi=-k, kxj=-i
The rectangular components of the vector product V of two
vectors P and Q are determined as follows: Given
P = Px i + Py j + Pz k
Q = Qx i + Qy j + Qz k
The determinant containing each component of P and Q is
expanded to define the vector V, as well as its scalar
components
P = Px i + Py j + Pz k
i
V = P x Q = Px
Qx
Q = Q x i + Qy j + Q z k
j
k
Py Pz = Vx i + Vy j + Vz k
Q y Qz
where
Vx = Py Qz - Pz Qy
Vy = Pz Qx - Px Qz
Vz = Px Qy - Py Qx
Mo
The moment of force F
about point O is defined
as the vector product
F
MO = r x F
where r is the position
r
O
q
vector drawn from point O
d
to the point of application
A
of the force F. The angle
between the lines of action
of r and F is q.
The magnitude of the moment of F about O can be expressed as
MO = rF sin q = Fd
where d is the perpendicular distance from O to the line of action
of F.
y
A (x , y, z )
Fy j
yj
Fx i
r
O
Fz k
xi
x
zk
z
The rectangular
components of the
moment Mo of a
force F are
determined by
expanding the
determinant of r x F.
i j k
Mo = r x F = x y z = Mx i + My j + Mzk
Fx Fy Fz
where
Mx = y Fz - z Fy
Mz = x Fy - y Fx
My = zFx - x Fz
y
A (x A, yA, z A)
Fy j
B (x B, yB, z B)
Fx i
r
Fz k
O
x
z
In the more general
case of the moment
about an arbitrary
point B of a force F
applied at A, we have
i
j
where
MB = rA/B x F = xA/B yA/B
Fx Fy
rA/B = xA/B i + yA/B j + zA/B k
and
xA/B = xA- xB
yA/B = yA- yB
k
zA/B
Fz
zA/B = zA- zB
y
F
Fy j
rA/B
(yA - yB ) j
A
Fx i
B
(xA - xB ) i
O
z
MB = M B k
x
In the case of problems
involving only two
dimensions, the force F
can be assumed to lie
in the xy plane. Its
moment about point B
is perpendicular to that
plane. It can be
completely defined by
the scalar
MB = (xA- xB )Fy + (yA- yB ) Fx
The right-hand rule is useful for defining the direction of the
moment as either into or out of the plane (positive or negative
k direction).
Q
q
The scalar product of two
vectors P and Q is denoted
as P Q ,and is defined as
P Q = PQ cos q
P
where q is the angle between
the two vectors
The scalar product of P and Q is expressed in terms of the
rectangular components of the two vectors as
P Q = Px Qx + Py Qy + Pz Qz
y
L
qy
l
qx P
O
z
A
x
qz
The projection of a vector
P on an axis OL can be
obtained by forming the
scalar product of P and the
unit vector l along OL.
POL = P l
Using rectangular components,
POL = Px cos qx + Py cos qy + Pz cos qz
The mixed triple product of three vectors S, P, and Q is
S (P x Q ) =
Sx Sy Sz
Px Py Pz
Qx Qy Qz
The elements of the determinant are the rectangular
components of the three vectors.
y
L
MO
l
F
C
r
O
A (x, y, z)
x
z
MOL = l
MO = l (r x F) =
The moment of a
force F about an axis
OL is the projection
OC on OL of the
moment MO of the
force F. This can be
written as a mixed
triple product.
lx ly lz
x y z
Fx Fy Fz
lx, ly , lz = direction cosines of axis OL
x, y , z = components of r
Fx, Fy , Fz = components of F
Kinematics of Particles
Chapter 11
P
O
x
x
The velocity v of the particle is equal to the time derivative of
the position coordinate x,
dx
v=
dt
and the acceleration a is obtained by differentiating v with
respect to t,
dv
a=
dt
or
we can also express a as
dv
a=v
dx
d 2x
a= 2
dt
Two types of motion are frequently encountered: uniform
rectilinear motion, in which the velocity v of the particle is
constant and
x = xo + vt
and uniformly accelerated rectilinear motion, in which the
acceleration a of the particle is constant and
v = vo + at
x = xo + vot +
1
2
at2
v2 = vo2 + 2a(x - xo )
O
A
B
x
xB/A
xA
xB
When particles A and B move along the same straight line, the
relative motion of B with respect to A can be considered.
Denoting by xB/A the relative position coordinate of B with respect
to A , we have
xB = xA + xB/A
Differentiating twice with respect to t, we obtain
vB = vA + vB/A
aB = aA + aB/A
where vB/A and aB/A represent, respectively, the relative velocity
and the relative acceleration of B with respect to A.
xC
xA
xB
C
A
B
When several blocks are are connected by inextensible cords,
it is possible to write a linear relation between their position
coordinates. Similar relations can then be written between
their velocities and their accelerations and can be used to
analyze their motion.
y
It is sometimes convenient to resolve
C
the velocity and acceleration of a
v2
particle P into components other
an =
e
r n
than the rectangular x, y, and z
components. For a particle P moving
dv
along a path confined to a plane, we
at = dt et
attach to P the unit vectors et
P
tangent to the path and en normal to
x
O
the path and directed toward the
center of curvature of the path.
The velocity and acceleration are expressed in terms of tangential
and normal components. The velocity of the particle is
v = vet
The acceleration is
v2
dv
a=
et +
en
r
dt
y
C
v = vet
v2
an =
e
r n
P
O
dv
at = dt et
v2
dv
a=
et +
en
r
dt
x
In these equations, v is the speed of the particle and r is the
radius of curvature of its path. The velocity vector v is directed
along the tangent to the path. The acceleration vector a
consists of a component at directed along the tangent to the
path and a component an directed toward the center of
curvature of the path,
eq
When the position of a particle moving
er
in a plane is defined by its polar
coordinates r and q, it is convenient to
r = r er
P
use radial and transverse components
directed, respectively, along the
position vector r of the particle and in
q
the direction obtained by rotating r
x
O
through 90o counterclockwise. Unit
vectors er and eq are attached to P and are directed in the radial
and transverse directions. The velocity and acceleration of the
particle in terms of radial and transverse components is
.
.
v = rer + rqeq
..
.. . 2
..
a = (r - rq )er + (rq + 2rq)eq
eq
r = r er
er
P
.
.
v = rer + rqeq
..
.. . 2
..
a = (r - rq )er + (rq + 2rq)eq
q
O
x
In these equations the dots represent differentiation with
respect to time. The scalar components of of the velocity
and acceleration in the radial and transverse directions are
therefore
.
.
vr = r
.. . 2
ar = r - rq
vq = rq
..
..
aq = rq + 2rq
It is important to note that ar is not equal to the time derivative
of vr, and that aq is not equal to the time derivative of vq.
Newton’s Second Law
Chapter 12
ay
y
To solve a problem involving the motion of a
particle, S F = ma should be replaced by
equations containing scalar quantities. Using
rectangular components of F and a, we have
P
az
ax
x
z
S Fx = max
S Fy = may
S Fz = maz
Using tangential and normal components,
y
an
at
P
x
O
O
r
Using radial and transverse components,
aq
r
dv
S Ft = mat = m
dt
v2
S Fn = man = m
q
P
ar
x
.. . 2
S Fr = mar= m(r - rq )
..
..
S Fq = maq = m(rq + 2rq)
mv
f
P
mv0
r
f0
O
r0
P0
When the only force acting on a
particle P is a force F directed
toward or away from a fixed
point O, the particle is said to be
moving under a central force.
Since S MO = 0 at any .given
instant, it follows that HO = 0 for
all values of t, and
HO = constant
We conclude that the angular momentum of a particle moving
under a central force is constant, both in magnitude and
direction, and that the particle moves in a plane perpendicular
to HO .
mv
f
P
mv0
r
f0
O
r0
P0
Recalling that HO = rmv sin f, we
have, for points PO and P
rmv sin f = romvo sin fo
for the motion of any particle under
a central force.
.
Using polar coordinates and recalling that vq = rq and HO =
we have
.
mr2q,
.
r2 q = h
where h is a constant representing the angular momentum per
unit mass Ho/m, of the particle.
Work Energy
Chapter 13
Chapter 13 KINETICS OF PARTICLES:
ENERGY AND MOMENTUM METHODS
s2
A
dr
a
F
A1
s1
A2
ds
Consider a force F acting on
a particle A. The work of F
corresponding to the small
displacement dr is defined as
dU = F dr
s
Recalling the definition of scalar product of
two vectors,
dU = F ds cos a
where a is the angle between F and dr.
s2
A2
ds
A
a
The work of F during a finite
F
displacement from A1 to A2 ,
denoted by U1 2 , is obtained
by integrating along the path described by
the particle.
A2
dr
A1
s1
dU = F dr = F ds cos a
s
U1
2
=

F dr
A1
For a force defined by its rectangular components, we write
U1
2
=

A2
A1
(Fxdx + Fydy + Fzdz)
A2
The work of the weight
W of a body as its
center of gravity moves
from an elevation y1 to
y2 is obtained by setting
Fx = Fz = 0 and
Fy = - W .
W
dy
y2
A
A1
y1
y
U1
2=
-

y2
Wdy = Wy1 - Wy2
y1
The work is negative when the elevation increases, and
positive when the elevation decreases.
The work of the force F exerted by a
spring on a body A during a finite
displacement of the body from A1 (x = x1)
to A2 (x = x2) is obtained by writing
spring undeformed
B
AO
dU = -Fdx = -kx dx
B
U1
A1
x1
2=

x2
1
2
2
kx1
-
x1
F
B
=
A
x
x2
k x dx
A2
-
2
1
kx
2 2
The work is positive when
the spring is returning to.
its undeformed position.
The kinetic energy of a particle of mass m moving with a
velocity v is defined as the scalar quantity
T=
1
2
mv2
From Newton’s second law the principle of work and energy
is derived. This principle states that the kinetic energy of a
particle at A2 can be obtained by adding to its kinetic energy at
A1 the work done during the displacement from A1 to A2 by the
force F exerted on the particle:
T1 + U1
2=
T2
The power developed by a machine is defined as the time rate
at which work is done:
dU
Power =
=F v
dt
where F is the force exerted on the particle and v is the velocity
of the particle. The mechanical efficiency, denoted by h, is
expressed as
power output
h = power input
When the work of a force F is independent of the path followed,
the force F is said to be a conservative force, and its work is
equal to minus the change in the potential energy V associated
with F :
U1
2=
V1 - V2
The potential energy associated with each force considered
earlier is
Force of gravity (weight):
Vg = Wy
Gravitational force:
GMm
Vg = r
Elastic force exerted by a spring:
Ve =
1
2
kx2
U1
2=
V1 - V2
This relationship between work and potential energy, when
combined with the relationship between work and kinetic
energy (T1 + U1 2 = T2) results in
T1 + V 1 = T2 + V 1
This is the principle of conservation of energy, which states that
when a particle moves under the action of conservative forces,
the sum of its kinetic and potential energies remains constant.
The application of this principle facilitates the solution of
problems involving only conservative forces.
Impulse & Momentum
Chapter 13
The linear momentum of a particle is defined as the product mv
of the mass m of the particle and its velocity v. From Newton’s
second law, F = ma, we derive the relation
mv1 +

t2
t1
F dt = mv2
where mv1 and mv2 represent the momentum of the particle at a
time t1 and a time t2 , respectively, and where the integral defines
the linear impulse of the force F during the corresponding time
interval. Therefore,
mv1 + Imp1
2=
mv2
which expresses the principle of impulse and momentum for a
particle.
When the particle considered is subjected to several forces, the
sum of the impulses of these forces should be used;
mv1 + SImp1
2=
mv2
Since vector quantities are involved, it is necessary to consider
their x and y components separately.
The method of impulse and momentum is effective in the study
of impulsive motion of a particle, when very large forces, called
impulsive forces, are applied for a very short interval of time Dt,
since this method involves impulses FDt of the forces, rather
than the forces themselves. Neglecting the impulse of any
nonimpulsive force, we write
mv1 + SFDt = mv2
In the case of the impulsive motion of several particles, we write
Smv1 + SFDt = Smv2
where the second term involves only impulsive, external forces.
In the particular case when the sum of the impulses of the
external forces is zero, the equation above reduces to
Smv1 = Smv2
that is, the total momentum of the particles is conserved.
Line of
Impact
vB
B
A
Before Impact
vA
v’B
B
A
v’A
After Impact
In the case of direct central impact,
two colliding bodies A and B move
along the line of impact with
velocities vA and vB , respectively.
Two equations can be used to
determine their velocities v’A and v’B
after the impact. The first represents
the conservation of the total
momentum of the two bodies,
mAvA + mBvB = mAv’A + mBv’B
Line of
Impact
mAvA + mBvB = mAv’A + mBv’B
vB
B
A
Before Impact
vA
v’B
B
A
v’A
After Impact
The second equation relates the
relative velocities of the two bodies
before and after impact,
v’B - v’A = e (vA - vB )
The constant e is known as the
coefficient of restitution; its value lies
between 0 and 1 and depends on the
material involved. When e = 0, the
impact is termed perfectly plastic; when
e = 1 , the impact is termed perfectly
elastic.
Line of
Impact
n
t
vB
B
A
In the case of oblique central impact,
the velocities of the two colliding
bodies before and after impact are
resolved into n components along the
line of impact and t components
along the common tangent to the
surfaces in contact. In the t direction,
Before Impact
vA
v’B
n
t
v’A
B
vB
A
vA After Impact
(vA)t = (v’A)t
(vB)t = (v’B)t
while in the n direction
mA (vA)n + mB (vB)n =
mA (v’A)n + mB (v’B)n
(v’B)n - (v’A)n = e [(vA)n - (vB)n]
Line of
Impact
n
t
(vA)t = (v’A)t
mA (vA)n + mB (vB)n =
mA (v’A)n + mB (v’B)n
vB
B
A
Before Impact
vA
v’B
B
(v’B)n - (v’A)n = e [(vA)n - (vB)n]
n
t
v’A
(vB)t = (v’B)t
vB
A
vA After Impact
Although this method was
developed for bodies moving freely
before and after impact, it could be
extended to the case when one or
both of the colliding bodies is
constrained in its motion.
System of Particle
Chapter 14
The theory in this chapter serves as a support to Chapter 16.
Chapter 14 SYSTEMS OF PARTICLES
The effective force of a particle Pi of a given system is the
product miai of its mass mi and its acceleration ai with respect
to a newtonian frame of reference centered at O. The system
of the external forces acting on the particles and the system
of the effective forces of the particles are equipollent; i.e.,
both systems have the same resultant and the same moment
resultant about O :
n
n
S Fi =iS=1miai
i =1
n
n
S (ri x Fi ) =iS=1(ri x miai)
i =1
The linear momentum L and the angular momentum Ho about
point O are defined as
n
L = S mivi
i =1
It can be shown that
.
S F=L
n
Ho = S (ri x mivi)
i =1
.
S Mo = Ho
This expresses that the resultant and the moment resultant
about O of the external forces are, respectively, equal to the
rates of change of the linear momentum and of the angular
momentum about O of the system of particles.
The mass center G of a system of particles is defined by a
position vector r which satisfies the equation
n
mr = S miri
i =1
n
where m represents the total mass S mi. Differentiating both
i =1
members twice with respect to t, we obtain
L = mv
.
L = ma
where v and a are the .velocity and acceleration of the mass
center G. Since S F = L, we obtain
S F = ma
Therefore, the mass center of a system of particles moves as if
the entire mass of the system and all the external forces were
concentrated at that point.
y’
miv’i
y
r’i
G
Pi
x’
O
z’
x
The kinetic energy T of a system of
particles is defined as the sum of
the kinetic energies of the particles.
n
1
2
T = 2 S mivi
i=1
Using the centroidal reference
frame Gx’y’z’ we note that the
kinetic energy of the system can also be obtained by adding the
1
kinetic energy 2 mv2 associated with the motion of the mass
center G and the kinetic energy of the system in its motion
relative to the frame Gx’y’z’ :
z
1
2
T = mv 2 +
n
2
1
miv’i
2 iS
=1
y’
miv’i
y
r’i
G
Pi
x’
O
z
z’
x
n
1
2
2
T = mv + 2 S miv’i
i=1
The principle of work and energy
can be applied to a system of
particles as well as to individual
particles
1
2
T1 + U1
2=
T2
where U1 2 represents the work of all the forces acting on the
particles of the system, internal and external.
If all the forces acting on the particles of the system are
conservative, the principle of conservation of energy can be
applied to the system of particles
T1 + V1 = T2 + V2
y
(mAvA)1
y
t2
S  Fdt
y
(mAvA)2
t1
(mCvC)2
(mBvB)1
O
x
(mCvC)1
(mBvB)2
O
S
t2
t1
x
MOdt
O
x
The principle of impulse and momentum for a system of particles
can be expressed graphically as shown above. The momenta of
the particles at time t1 and the impulses of the external forces
from t1 to t2 form a system of vectors equipollent to the system of
the momenta of the particles at time t2 .
y
(mAvA)1
y
(mBvB)2
(mAvA)2
(mCvC)2
(mBvB)1
O
x
(mCvC)1
O
x
If no external forces act on the system of particles, the systems
of momenta shown above are equipollent and we have
L1 = L2
(HO)1 = (HO)2
Many problems involving the motion of systems of particles can
be solved by applying simultaneously the principle of impulse
and momentum and the principle of conservation of energy or
by expressing that the linear momentum, angular momentum,
and energy of the system are conserved.
Smivi
B
S
A
(D m)vB
S F Dt
S
S M Dt
B
Smivi
S
A
(D m)vA
For variable systems of particles, first consider a steady stream of
particles, such as a stream of water diverted by a fixed vane or
the flow of air through a jet engine. The principle of impulse and
momentum is applied to a system S of particles during a time
interval Dt, including particles which enter the system at A during
that time interval and those (of the same mass Dm) which leave
the system at B. The system formed by the momentum (Dm)vA of
the particles entering S in the time Dt and the impulses of the
forces exerted on S during that time is equipollent to the
momentum (Dm)vB of the particles leaving S in the same time Dt.
Smivi
B
S
A
(D m)vB
S F Dt
S
S M Dt
B
Smivi
S
A
(D m)vA
Equating the x components, y components, and moments about
a fixed point of the vectors involved, we could obtain as many
as three equations, which could be solved for the desired
unknowns. From this result, we can derive the expression
dm
SF =
(vB - vA)
dt
where vB - vA represents the difference between the vectors vB
and vA and where dm/dt is the mass rate of flow of the stream.
Kinematics of Rigid Bodies
Chapter 15
Chapter 15 KINEMATICS OF RIGID BODIES
z
A’
B
q
In rigid body translation, all points of
the body have the same velocity and
the same acceleration at any given
instant.
P
f
Considering the rotation of a rigid
body about a fixed axis, the position
O
x
of the body is defined by the angle q
that the line BP, drawn from the axis
y
A
of rotation to a point P of the body,
forms with a fixed plane. The magnitude of the velocity of P is
r
.
.
ds
v=
= rq sin f
dt
where q is the time derivative of q.
.
ds
v=
= rq sin f
dt
z
A’
The velocity of P is expressed as
B
q
dr
v=
=wxr
dt
P
f
r
O
x
A
y
where the vector
.
w = wk = qk
is directed along the fixed axis of rotation and represents the
angular velocity of the body.
z
A’
dr
v=
=wxr
dt
B
q
Denoting by a the derivative dw/dt of
the angular velocity, we express the
acceleration of P as
P
f
r
O
x
A
.
w = wk = qk
y
a = a x r + w x (w x r)
differentiating w and recalling that k is constant in magnitude
and direction, we find that
..
.
a = ak = wk = qk
The vector a represents the angular acceleration of the body
and is directed along the fixed axis of rotation.
v = wk x r Consider the motion of a representative
slab located in a plane perpendicualr to
P
the axis of rotation of the body. The
angular velocity is perpendicular to the
O
r
slab, so the velocity of point P of the
x
slab is
w = wk
y
v = wk x r
y
O
a = ak
at = ak x r
P
an= -w2 r
w = wk x
where v is contained in the plane of the
slab. The acceleration of point P can
be resolved into tangential and normal
components respectively equal to
at = ak x r
at = ra
an= -w2 r
an = rw2
The angular velocity and angular acceleration of the slab can
be expressed as
dq
w=
dt
d2q
dw
a=
= 2
dt
dt
or
dw
w=a
dq
Two particular cases of rotation are frequently encountered:
uniform rotation and uniformly accelerated rotation. Problems
involving either of these motions can be solved by using
equations similar to those for uniform rectilinear motion and
uniformly accelerated rectilinear motion of a particle, where x,
v, and a are replaced by q, w, and a.
vA
vA
vB
A
A
y’
wk
(fixed)
vA
A
x’
rB/A
vB/A
B
B
Plane motion
=
Translation with A
B
+
Rotation about A
The most general plane motion of a rigid slab can be considered
as the sum of a translation and a rotation. The slab shown can
be assumed to translate with point A, while simultaneously
rotating about A. It follows that the velocity of any point B of the
slab can be expressed as
vB = vA + vB/A
where vA is the velocity of A and vB/A is the relative velocity of B
with respect to A.
vA
vA
vB
A
A
y’
vA
vB/A
wk
(fixed)
A
x’
rB/A
vA
vB/A
B
Plane motion =
B
Translation with A
B
+
Rotation about A
vB = vA + vB/A
Denoting by rB/A the position of B relative to A, we note that
vB/A = wk x rB/A
vB/A = (rB/A )w = rw
The fundamental equation relating the absolute velocities of
points A and B and the relative velocity of B with respect to A
can be expressed in the form of a vector diagram and used to
solve problems involving the motion of various types of
mechanisms.
vB
Another approach to the solution of
problems involving the velocities of
the points of a rigid slab in plane
motion is based on determination of
the instantaneous center of rotation
C of the slab.
C
vB
B
A
C
vB
vA
vA
y’
aA
aB
A
B
=
ak
aB/A
A
A
B
Plane motion
wk
aA
(aB/A)n
aA
Translation with A
x’
(aB/A)t
B
+
Rotation about A
The fact that any plane motion of a rigid slab can be considered
the sum of a translation of the slab with reference to point A and
a rotation about A is used to relate the absolute accelerations of
any two points A and B of the slab and the relative acceleration
of B with respect to A.
aB = aA + aB/A
where aB/A consists of a normal component (aB/A )n of magnitude
rw2 directed toward A, and a tangential component (aB/A )t of
magnitude ra perpendicular to the line AB.
y’
aA
aB
A
B
=
ak
aB/A
A
A
B
Plane motion
wk
aA
Translation with A
(aB/A)t
(aB/A)n
aA
x’
B
+
Rotation about A
aB = aA + aB/A
The fundamental equation relating the
absolute accelerations of points A and B
and the relative acceleration of B with
respect to A can be expressed in the form of
a vector diagram and used to determine the
accelerations of given points of various
mechanisms.
(aB/A)n
aB
aB/A
(aB/A)t
aA
y’
aA
aB
A
B
=
ak
aB/A
A
A
B
Plane motion
wk
aA
Translation with A
(aB/A)t
(aB/A)n
aA
x’
B
+
Rotation about A
aB = aA + aB/A
(aB/A)n
The instantaneous center of rotation C
cannot be used for the determination of
accelerations, since point C , in general,
does not have zero acceleration.
aB
aB/A
(aB/A)t
aA
Inertia
Chapter 9
It is unlikely that you will calculate the mass moment of inertia. But the theory behind
it has to be covered.
Chapter 9 DISTRIBUTED FORCES:
MOMENTS OF INERTIA
y
The rectangular moments of
inertia Ix and Iy of an area are
defined as
y
x
dx
x
Ix =  y 2dA
Iy =  x 2dA
These computations are reduced to single integrations by
choosing dA to be a thin strip parallel to one of the coordinate
axes. The result is
dIx =
1
3
y 3dx
dIy = x 2ydx
y
The polar moment of
inertia of an area A with
respect to the pole O is
defined as
dA
r
O
y
x
x
JO =  r 2dA
A
The distance from O to the element of area dA is r. Observing
that r 2 =x 2 + y 2 , we established the relation
J O = I x + Iy
y
The radius of gyration of
an area A with respect to
the x axis is defined as
the distance kx, where
2
Ix = kx A. With similar
definitions for the radii of
gyration of A with respect
to the y axis and with
respect to O, we have
A
kx
O
x
kx =
Ix
A
ky =
Iy
A
kO =
JO
A
The parallel-axis theorem
states that the moment of
c
inertia I of an area with
B
B’
respect to any given axis
d
AA’ is equal to the moment
A’
A
of inertia I of the area with
respect to the centroidal
axis BB’ that is parallel to AA’ plus the product of the area A
and the square of the distance d between the two axes:
I = I + Ad 2
This expression can also be used to determine I when the
moment of inertia with respect to AA’ is known:
I = I - Ad 2
A similar theorem can be
used with the polar moment
c
of inertia. The polar
moment of inertia
d
JO of an area about O and
the polar moment of inertia
o
JC of the area about its
centroid are related to the distance d between points C and O
by the relationship
JO = JC + Ad 2
The parallel-axis theorem is used very effectively to compute
the moment of inertia of a composite area with respect to a
given axis.
A’
r1
Dm1
Dm3
r3
r2
Moments of inertia of mass are
encountered in dynamics. They
involve the rotation of a rigid body
about an axis. The mass moment
of inertia of a body with respect
to an axis AA’ is defined as
Dm2
I =  r 2dm
A
where r is the distance from AA’
to the element of mass.
The radius of gyration of the body is defined as
k=
I
m
The moments of inertia of mass with respect to the coordinate
axes are
Ix =  (y 2 + z 2 ) dm
Iy =  (z 2 + x 2 ) dm
Iz = (x 2 + y 2 ) dm
A’
d
B’
The parallel-axis theorem also
applies to mass moments of inertia.
I = I + d 2m
A
G
B
I is the mass moment of inertia with
respect to the centroidal BB’ axis,
which is parallel to the AA’ axis. The
mass of the body is m.
A’
B’
t
C
C’
B
A
b
The moments of inertia of thin plates
can be readily obtained from the
moments of inertia of their areas. For
a rectangular plate, the moments of
inertia are
1
IAA’ = 12 ma
a
2
IBB’ =
1
mb
12
2
ICC’ = IAA’ + IBB’ = 1 m (a 2 + b 2)
12
A’
B’
For a circular plate they are
r
1
C
t
C’
B
A
IAA’ = IBB’ = 4 mr 2
1
ICC’ = IAA’ + IBB’ = 2 mr 2
PLANE MOTION OF RIGID
BODIES:
FORCES AND ACCELERATIONS
Chapter 16
Chapter 16 PLANE MOTION OF RIGID BODIES:
FORCES AND ACCELERATIONS
.
HG
The relations existing
F4
F1
between the forces acting
ma
on a rigid body, the shape
and mass of the body, and
F3
the motion produced are
G
G
studied as the kinetics of
rigid bodies. In general,
our analysis is restricted
F2
to the plane motion of
rigid slabs and rigid bodies symmetrical with respect to the
reference plane.
.
F1
G
HG
F4
ma
F3
G
The two equations for the
motion of a system of
particles apply to the most
general case of the motion
of a rigid body. The first
equation defines the
motion of the mass center
G of the body.
F2
SF = ma
where m is the mass of the body, and a the acceleration of G. The
second is related to the motion of the body relative to a
centroidal frame of reference.
.
SMG = HG
.
F1
HG
F4
ma
SF = ma
..
SMG = HG
.
G
F2
F3
G
where HG is the rate of
change of the angular
momentum HG of the
body about its mass
center G.
These equations express that the system of the external forces
is equipollent to the system consisting
of the vector ma attached
.
at G and the couple of moment HG.
.
F1
G
HG
F4
ma
F3
G
F2
For the plane motion of
rigid slabs and rigid
bodies symmetrical with
respect to the reference
plane, the angular
momentum of the body is
expressed as
HG = Iw
where I is the moment of inertia of the body about a centroidal
axis perpendicular to the reference plane and w is the angular
velocity of the body. Differentiating both members of this
equation
.
.
HG = Iw = Ia
F1
F4
For the restricted case
ma considered here, the rate
of change of the angular
G
F
3
G
momentum of the rigid
body can be represented
Ia
by a vector of the same
F2
direction as a (i.e.
perpendicular to the plane of reference) and of magnitude Ia.
The plane motion of a rigid body symmetrical with respect to
the reference plane is defined by the three scalar equations
SFx = max
SFy = may
SMG = Ia
The external forces acting on a rigid body are actually equivalent
to the effective forces of the various particles forming the body.
This statement is known as d’Alembert’s principle.
F4
d’Alembert’s principle can
be expressed in the form
ma of a vector diagram, where
G
F3
the effective forces are
G
represented by a vector
Ia
ma attached at G and a
F2
couple Ia. In the case of a
slab in translation, the
(a)
(b)
effective forces (part b of
the figure) reduce to a
single vector ma ; while in the particular case of a slab in
centroidal rotation, they reduce to the single couple Ia ; in any
other case of plane motion, both the vector ma and Ia should
be included.
F1
F4
Any problem involving the
plane motion of a rigid slab
ma may be solved by drawing a
free-body-diagram equation
G
F3
G
similar to that shown. Three
equations of of motion can
Ia
then be obtained by
F2
equating the x components,
y components, and moments about an arbitrary point A, of the
forces and vectors involved.
F1
This method can be used to solve problems involving the
plane motion of several connected rigid bodies.
Some problems, such as noncentroidal rotation of rods and
plates, the rolling motion of spheres and wheels, and the plane
motion of various types of linkages, which move under
constraints, must be supplemented by kinematic analysis.
PLANE MOTION OF RIGID
BODIES:
ENERGY AND MOMENTUM
METHODS
Chapter 17
Chapter 17 PLANE MOTION OF RIGID BODIES:
ENERGY AND MOMENTUM METHODS
The principle of work and energy for a rigid body is expressed in
the form
T1 + U1
2=
T2
where T1 and T2 represent the initial and final values of the
kinetic energy of the rigid body and U1 2 the work of the
external forces acting on the rigid body.
The work of a force F applied at a point A is
s2
U1
2
=

(F cos a) ds
s1
where F is the magnitude of the force, a the angle it forms with
the direction of motion of A, and s the variable of integration
measuring the distance traveled by A along its path.
The work of a couple of moment M applied to a rigid body during
a rotation in q of the rigid body is
U1
2
=

q2
M ds
q1
The kinetic energy of a rigid body in plane motion is
T=
G
w
v
1
2
2
1
2
mv + Iw2
where v is the velocity of the mass center G of
the body, w the angular velocity of the body,
and I its moment of inertia about an axis
through G perpendicular to the plane of
reference.
T=
1
2
mv 2 + 12 Iw2
G
The kinetic energy of a rigid body in plane
motion may be separated into two parts:
w
v
(1) the kinetic energy 12 mv 2 associated
with the motion of the mass center G of the
1
body, and (2) the kinetic energy 2 Iw2 associated with the rotation
of the body about G.
For a rigid body rotating about a fixed axis through O with an
angular velocity w,
O
1
T = 2 IOw2
w
where IO is the moment of inertia of the body
about the fixed axis.
When a rigid body, or a system of rigid bodies, moves under
the action of conservative forces, the principle of work and
energy may be expressed in the form
T1 + V1 = T2 + V2
which is referred to as the principle of conservation of energy.
This principle may be used to solve problems involving
conservative forces such as the force of gravity or the force
exerted by a spring.
The concept of power is extended to a rotating body subjected
to a couple
dU M dq
Power =
=
= Mw
dt
dt
where M is the magnitude of the couple and w is the angular
velocity of the body.
The principle of impulse and momentum derived for a system of
particles can be applied to the motion of a rigid body.
Syst Momenta1 + Syst Ext Imp1
2=
Syst Momenta2
For a rigid slab or a rigid body symmetrical with respect to the
reference plane, the system of the momenta of the particles
forming the body is equivalent to a vector mv attached to the
mass center G of the body and a couple Iw. The vector mv is
associated with translation of the body with G and represents the
linear momentum of the body, while the couple Iw corresponds
to the rotation of
(Dm)v
the body about G
mv and represents the
P
angular momentum
Iw
of the body about
an axis through G.
The principle of impulse and momentum can be expressed
graphically by drawing three diagrams representing
respectively the system of initial momenta of the body, the
impulses of the external forces acting on it, and the system of
the final momenta of the body. Summing and equating
respectively the x components, the y components, and the
moments about any given point of the vectors shown in the
figure, we obtain three equations of motion which may be
solved for the desired unknowns.
 Fdt
y
mv1
y
y
G
G
Iw2
Iw1
O
mv2
x
O
x
O
x
 Fdt
y
mv1
y
y
G
G
Iw2
Iw1
O
mv2
x
O
x
O
x
In problems dealing with several connected rigid bodies each
body may be considered separately or, if no more than three
unknowns are involved, the principles of impulse and
momentum may be applied to the entire system, considering
the impulses of the external forces only.
When the lines of action of all the external forces acting on a
system of rigid bodies pass through a given point O, the angular
momentum of the system about O is conserved.
The eccentric impact of two rigid
bodies is defined as an impact in
which the mass centers of the
colliding bodies are not located
on the line of impact. In such a
situation a relation for the impact
involving the coefficient of
restitution e holds, and the
velocities of points A and B
where contact occurs during the
impact should be used.
n
B
vB
A
n
vA
(a) Before impact
n
B
A
(v’B)n - (v’A)n = e[(vA)n - (vB)n]
n
v’B
v’A
(b) After impact
n
n
B
A
n
B
vB
vA
(a) Before impact
A
n
v’B
v’A
(b) After impact
(v’B)n - (v’A)n = e[(vA)n - (vB)n]
where (vA)n and (vB)n are the components along the line of impact
of the velocities of A and B before impact, and (v’A)n and (v’B)n
their components after impact. This equation is applicable not
only when the colliding bodies move freely after impact but also
when the bodies are partially constrained in their motion.