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Eng3934 - Dynamics
1
1
Introduction
Mechanics is concerned with the the states of rest and motion of bodies subjected to
forces. Mechanics is divided into two areas:
– Statics - study of the equilibrium of a body at rest or moving with a constant
velocity.
– Dynamics - study of the accelerated motion of a body.
Dynamics is divided into two areas:
– Kinematics - study of the motion of a body (i.e. position, velocity, and acceleration).
– Kinetics - study of the forces causing the motion.
The concepts of dynamics will be developed in the context of the motion of a particle,
and then they will be extended to the motion of a rigid body.
2
Kinematics of a Particle
A particle has …nite mass, but negligible size and shape. A body of …nite size and shape
may be modelled as a particle if the body does not rotate and its mass is considered
to act at a mass center.
2.1
General Curvilinear Motion
A particle experiences curvilinear motion when it moves in a curved path. We would
be interested in the position, displacement, velocity and acceleration of the particle.
The position of the particle is de…ned by the position vector, ~r. Consider a particle
located at point P on a space curve de…ned by the path function s. The position
vector, ~r, is measured from a …xed point O to the point P . Both the magnitude and
direction of ~r may change as the particle moves along the curve, therefore, ~r = ~r(t).
The displacement, ~r, of the particle is the vector di¤erence between two position
vectors drawn to the particle at di¤erent instants in time. For example, in a small
time interval t the particle can move a distance s along the curve to a new position
P 0 located by ~r 0 . The displacement during the time interval t is ~r = ~r 0 ~r.
Eng3934 - Dynamics
During the time interval
2
t the average velocity of the particle is:
~vavg =
~r
t
The instantaneous velocity, ~v , is determined by letting the time interval approach 0
( t ! 0), then the direction of ~r (and ~v ) approaches the tangent to the curve at
point P .
~r
d~r
~v = lim
=
= ~r_
(1)
t!0
t
dt
The magnitude of ~v , i.e. the speed of the particle may be obtained as follows:
v = lim
t!0
since
r approaches the arc length
r
= lim
t!0
t
s as
s
ds
=
t
dt
(2)
t ! 0.
The velocity of a particle may change, i.e. ~v 0 = ~v + ~v as it moves from point P to
P 0 in the time interval t. The average acceleration experienced by the particle is:
~aavg =
~v
t
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The instantaneous acceleration is obtained as follows:
~a = lim
t!0
~v
d~v
=
t
dt
(3)
Since ~v = d~r=dt, the instantaneous acceleration can be written as:
~a =
d2~r
= ~r•
dt2
(4)
Note: ~a is not tangent to the path of the particle in general curvilinear motion. If it
was tangent to the path, only a change in magnitude of the velocity would be possible.
To change the direction of the particle a component of acceleration normal to the path
of the particle is required.
2.1.1
Rectilinear Motion
Rectilinear motion can be considered a special case of general curvilinear motion,
where the radius of curvature of the path function s is in…nite, i.e. the particle travels
in a straight line.
All equations in the previous section are still valid for the position, displacement,
velocity and acceleration of the particle, but remember that the particle can only
travel in a straight line, therefore, the velocity and acceleration will be tangent to the
path of the particle.
Special equations for rectilinear motion can be derived if the particle experiences
constant acceleration, ac (e.g. a gravity …eld).
Since ac = dv=dt, and assuming v = v0 at t = 0:
Z v
Z t
dv =
ac dt
v0
0
and
v = v0 + ac t
Since v = ds=dt = v0 + ac t, and assuming s = s0 at t = 0:
Z s
Z t
ds =
(v0 + ac t) dt
s0
0
(5)
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and
1
s = s0 + v0 t + ac t2
2
(6)
Finally, since v = ds=dt and ac = dv=dt, then v dv = ac ds, and assuming v = v0 at
s = s0 :
Z v
Z s
v dv =
ac ds
v0
s0
and
v 2 = v02 + 2ac (s
s0 )
(7)
Note: Be careful of the signs on s, v, and ac .
2.2
Curvilinear Motion: Rectangular Components
Often it is convenient to describe the motion of a particle in terms of the co-ordinates
of a …xed Cartesian (i.e. x, y, z) reference frame.
The position vector, ~r, of the particle P on path s would be de…ned as:
~r = x^{ + y^
| + z k^
where the magnitude of ~r is:
r=
p
x2 + y 2 + z 2
(8)
(9)
and the direction of ~r is given by the unit vector u
^r = ~r=r.
The instantaneous velocity of the particle can be obtained by evaluating the …rst
derivative of ~r.
~v
d~r
dt
dx
dy
dz
=
^{ +
|^ + k^
dt
dt
dt
= vx^{ + vy |^ + vz k^
= x^
_ { + y^
_ | + z_ k^
=
(10)
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The magnitude of ~v is given by:
v=
q
vx2 + vy2 + vz2
(11)
and the direction of ~v is de…ned by the unit vector u
^v = ~v =v which is always tangent
to the path.
The instantaneous acceleration of the particle can be obtained from the …rst derivative
of ~v or the second derivative of ~r:
~a =
=
=
=
=
d~v
dt
d2~r
dt2
dvy
dvz ^
dvx
^{ +
|^ +
k
dt
dt
dt
ax^{ + ay |^ + az k^
v_ x^{ + v_ y |^ + v_ z k^
(12)
= x
•^{ + y•|^ + z•k^
The magnitude of the acceleration is given by:
q
a = a2x + a2y + a2z
(13)
and the direction of the acceleration is given by the unit vector ~ua = ~a=a. In general
the direction of the acceleration is not tangent to the path.
2.3
Curvilinear Motion: Normal and Tangential Components
Normal and tangential components are used when the path of a particle is known. The
normal and tangential components of velocity and acceleration are evaluated using an
n; t co-ordinate system that is normal and tangential to the path of the particle, and
the origin is located at the particle at a given instant in time.
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Consider the particle at point P on the path s. The n; t co-ordinate system has
its origin located at point P . The t axis is tangent to the path and positive in the
direction of increasing s, and designated by the unit vector u
^t . The normal, n, axis
is perpendicular to the path of the particle, and is positive when directed toward
the center of curvature, i.e. towards the concave side of the path. This direction is
indicated by the unit vector u
^n .
Since the velocity is tangent to the path, it may be de…ned as:
~v = v^
ut
(14)
where the magnitude v = s.
_
The acceleration, ~a, of the particle is the …rst time derivative of the velocity vector, ~v :
d^
u
~a = ~v_ = v^
_ ut + v
dt
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To determine d^
ut =dt consider the motion of the particle along the arc ds in the time
dt. At time t + dt the velocity has direction u
^0t and d^
ut = u
^0t u
^t . Since the length of
u
^t is one (unit vector), and d^
ut is very small then dut = (1)d ( i.e. arc length), and
d^
ut has the direction of u
^n . Then d^
ut = d u
^n , and the time derivative d^
ut =dt = _ u
^n .
_
Also, ds = d , then = s=
_ .
s_
v
d^
ut
= _u
^n = u
^n = u
^n
dt
The acceleration vector can be written as:
~a = v^
_ ut +
v2
u
^n = at u
^t + an u
^n
(15)
where the tangential component is:
at = v_
(16)
and the normal component is:
an =
v2
(17)
The magnitude of the acceleration is:
a=
q
a2t + a2n
The direction of ~a is not generally directed along the path of the particle.
(18)
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Note: the tangential component of acceleration changes the magnitude of the velocity
vector. The normal component of acceleration changes the direction of the velocity
vector.
2.4
Curvilinear Motion: Cylindrical Components
For certain problems it may be convenient to use the cylindrical (r, ,z) co-ordinate
system, or, if motion is restricted to a plane, the polar (r, ) co-ordinate system.
2.4.1
Polar Co-ordinates
The position of the particle P can be located by the position vector ~r de…ned in the
polar co-ordinate system by the radial co-ordinate r, which is measured positive outward from a …xed origin to P , and the transverse co-ordinate which is measured
counterclockwise from a …xed reference line to r. The transverse co-ordinate is measured in radians. The unit vectors u
^r and u
^ are mutually perpendicular, and directed
from P in the direction of increasing r and , respectively.
Using the de…nition of the r, co-ordinate system, the position vector, ~r, can be de…ned
as:
~r = r^
ur
(19)
The instantaneous velocity can be obtained from the …rst time derivative of the position vector:
d^
ur
~v = ~r_ = r^
_ ur + r
dt
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Since u
^r is a unit vector, its length is constant, but its direction can change due to
in the time interval t, resulting in u
^0r , and u
^r = u
^0r u
^r . For small angles
,
ur = (1)
and is in the direction of u
^ . Therefore, u
^r =
u
^ , and:
d^
ur
= lim
t!0
dt
u
^r
=
t
lim
t!0
t
u
^
d^
ur
= _u
^
dt
The velocity vector can then be written in the following form:
~v = vr u
^r + v u
^
(20)
vr = r_
(21)
v = r_
(22)
with the radial:
and transverse components of velocity.
Where _ is the angular velocity, and has units rad/s.
The magnitude of the velocity is:
q
_ 2 + (r _ )2
v = (r)
(23)
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The instantaneous acceleration of the particle is obtained from the …rst time derivative
of the velocity vector:
d^
u
d^
ur
~a = ~v_ = r•u
^r + r_
+ r_ _ u
^ + r•u
^ + r_
dt
dt
To evaluate d^
u =dt consider the change in direction of u
^0 as the particle moves
in
0
t. Since u
^ =u
^
u
^ , and for small angles u = (1) , and acts in the direction
of u
^r , then u
^ =
u
^r and:
d^
u
= lim
t!0
dt
u
^
=
t
lim
t!0
u
^r
t
d^
u
= _u
^r
dt
The acceleration vector can then be written as:
~a = ar u
^r + a u
^
(24)
with components in the radial:
ar = r•
r_
2
(25)
and transverse directions.
a = r• + 2r_ _
(26)
Where • is the angular acceleration, and has units of rad/s2 .
The magnitude of the acceleration is:
r
a=
r•
r_
2 2
+ r• + 2r_ _
2
(27)
Eng3934 - Dynamics
2.4.2
11
Alternate Method for Polar Co-ordinates
The position vector to a particle P can be de…ned in polar notation as:
~r = r ej
(28)
Since the velocity is the …rst time derivative of the position vector:
~v
= ~r_
= r_ ej + j _ r ej
(29)
with radial and transverse components.
The acceleration is the …rst time derivative of the velocity vector:
~a = ~v_
= ~r•
= r• ej + j _ r_ ej + j •r ej + j _ r_ ej
_ 2 r ej + j2 _ r_ ej + j •r ej
= r• ej
(30)
_ 2 r ej
The …rst term on the LHS is called a slip (or relative) acceleration, the second term
is the normal acceleration (directed towards the origin (i.e. instantaneous center of
rotation), the third term is the coriolis acceleration (in the transverse direction), and
the fourth term is a transverse acceleration resulting from the angular acceleration.
Eng3934 - Dynamics
2.4.3
12
Cylindrical Co-ordinates
If a particle follows a space curve, its position can be de…ned by the cylindrical (r, ,z)
co-ordinate system. The position, velocity, and acceleration vectors are obtained as
follows:
~r = r^
ur + z u
^z
(31)
~v = r^
_ ur + r _ u
^ + z_ u
^z
~a = (•
r
2
r _ )^
ur + (r• + 2r_ _ )^
u + z•u
^z
(32)
(33)
Eng3934 - Dynamics
2.5
13
Constrained Motion of Connected Particles
Sometimes the motions of particles are interrelated due to constraints imposed by
interconnecting members. A typical example is particles connected by inextensible
cords which are wrapped around pulleys.
De…ne a …xed datum (usually a …xed point) from which the position co-ordinates sA
and sB are measured. Then the total length of the cord, lT , is:
lT = sA +
r2
+ 2sB + r1 + b
2
Taking the …rst time derivative of this equation gives:
0=
dsB
dsA
+2
= s_ A + 2s_ B
dt
dt
or:
vA =
2vB
i.e. if block A moves left (increasing sA , positive vA ) then block B will move up with
speed vA =2 (decreasing sB , negative vB ) and vice versa.
The second time derivative gives:
aA =
2aB
i.e. if block A accelerates left, block B accelerates up at a rate half that of A.
Consider an example with three pulleys:
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Note: It is not necessary for the datum to have the same origin, but the co-ordinate
axes must be directed in the direction of motion of each block.
Since the length of cord wrapped around a pulley is constant (and we will be taking
time derivatives), we can write an equation for the total length of the cord less the
length of cord wrapped around each pulley:
l = 2sB + h + sA
Then:
vA
aA
=
=
2vB
2aB
i.e. if A moves right (increasing sA ), B moves up (decreasing sB ).
Eng3934 - Dynamics
2.6
15
Relative Motion Analysis: Translating Axes
Thus far, the absolute motion of a particle has been de…ned with respect to a single
…xed reference frame. For many problems, however, it is more feasible to de…ne more
than one frame of reference. For example, the absolute position of a point on the edge
of a wheel rolling with constant velocity along a ‡at surface is more easily described
with two frames of reference than one. One frame (x,y,z) is de…ned …xed with the
surface, O, and the other frame (x0 ,y 0 ,z 0 ) is de…ned …xed at the center of the wheel,
A, (this frame does not rotate).
The absolute position vector, ~rB , to B can then be de…ned as:
~rB = ~rA + ~rB=A
(34)
where ~rA is the position of A from O and ~rB=A is the position of B measured from
the frame …xed at A, or vector B from A.
Di¤erentiation of the position vector leads to the velocity:
~vB = ~vA + ~vB=A
(35)
~aB = ~aA + ~aB=A
(36)
and acceleration vectors:
The terms ~rB=A , ~vB=A and ~aB=A are the relative position, velocity, and acceleration
vectors, i.e. they are how an observer moving with A would see B moving in space.
In the above example, an observer at O would see B perform a cycloid motion, while
the observer at A would see B rotate around A.
Eng3934 - Dynamics
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3.1
16
Kinetics of a Particle: Force and Acceleration
Newton’s Laws of Motion
Newton’s three laws of motion are:
1. A particle remains at rest or continues to move with uniform velocity (in a straight
P~
line with constant speed) if there is no unbalanced force acting on it, i.e.
F =0
(statics).
2. The acceleration of a particle is proportional
P ~ to the resultant force acting on it,
and is in the direction of the force, i.e.
F = m~a (dynamics).
3. The forces of action and reaction between two interacting bodies are equal in
magnitude, opposite in direction, and collinear.
3.2
Equation of Motion
If more than one force acts on a particle, then a vector summation is performed to
determine the resultant force acting on the particle, and Newton’s Second Law, or the
equation of motion, is written in the following form:
X
F~ = m~a
(37)
For example:
Use of Newton’s Second Law requires the de…nition of a Newtonian or inertial frame
of reference. An inertial frame of reference does not rotate, and is either …xed or
translates with a constant velocity (i.e. there can be no acceleration).
An inertial reference frame is chosen so that two observers using di¤erent inertial
frames will observe the same acceleration of a particle.
We usually choose an inertial frame …xed to the surface of the earth.
The equation of motion for a collection of particles can be written as follows:
X
F~ = m~aG
(38)
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i.e. the vector sum of the external forces acting on a collection of particles is equivalent
to the product of the sum of the masses of all particles and the acceleration of the
center of mass of the collection of particles. This is the form of the equation of motion
that is applied to a body of …nite size when it is treated as a particle.
3.3
Equations of Motion: Rectangular Co-ordinates
If a particle moves relative to an inertial (x,y,z) frame of reference, the forces acting
on the particle and the acceleration of the particle may be resolved into the x,y,z
components, i.e. the Cartesian co-ordinate system may be used. The equation of
motion is:
X
F~ = m~a
(39)
X
X
X
^
Fx^{ +
Fy |^ +
Fz k^ = m(ax^{ + ay |^ + az k)
The equations of motion can be rewritten in the following scalar form:
X
Fx = max
+ .
X
+
!
Fy = may
X
+
"
Fz = maz
(40)
Eng3934 - Dynamics
3.4
18
Equations of Motion: Normal and Tangential Co-ordinates
If a particle moves along a known curvilinear path, it can be convenient to use normal
and tangential co-ordinates. The equation of motion becomes:
X
F~ = m~a
(41)
X
X
X
Ft u
^t +
Fn u
^n +
Fb u
^b = m(at u
^t + an u
^n )
Since the particle moves along the path, there is no acceleration in the binormal
direction, therefore, the following scalar equations may be used.
X
+ %
Ft = mat
(42)
X
+ Fn = man
X
+ "
Fb = 0
Remember that the tangential acceleration, at , acts to change the magnitude of the
velocity, and the normal component of acceleration, an , acts to change the direction
of the velocity vector. The normal acceleration, and therefore, the normal force, F~n ,
always act towards the center of curvature of the path, and the normal force is often
called the centripetal force.
Note: the forces in the binormal, b, direction sum to zero due to the de…nition of the
n,t co-ordinate system (i.e. there is no acceleration in the b-direction).
Eng3934 - Dynamics
3.5
19
Equations of Motion: Cylindrical Co-ordinates
In certain problems it may be convenient to resolve the accelerations and forces into
cylindrical (r, ,z) co-ordinates (e.g. rotation about a …xed or moving axis). The
equation of motion can then be rewritten in the cylindrical co-ordinate system as:
X
F~ = m~a
(43)
X
X
X
Fr u
^r +
F u
^ +
Fz u
^z = m(ar u
^r + a u
^ + az u
^z )
the following scalar form of the equations of motion may also be used:
X
+ &
Fr = mar
X
+ %
F = ma
X
+ "
Fz = maz
(44)
If the path of a particle has been described in the polar or cylindrical co-ordinate
system it is easy to obtain the force components in the r, , and z directions. If
the normal and tangential components are required, de…ne the angle between the
extended radial line, and the tangent to the path.
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To evaluate consider the displacement of a particle along the path from P to P 0 .
The angle can be determined from:
tan
=
rd
r
=
dr
dr=d
(45)
Note: is measured positive from the extension of r to the tangent in the positive
(ccw) direction.
Eng3934 - Dynamics
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4.1
21
Planar Kinematics of a Rigid Body
Rigid Body Motion
The purpose of this section is to determine the planar kinematics of a rigid body. This
material will be useful when analysing the motion of gears, cams and planar mechanisms (e.g. the crank-slider in an engine). Following determination of the motion, the
forces and moments causing (or resulting) from the motion can be evaluated.
Planar motion - all particles of a rigid body move along paths that are equidistant
from a plane.
There are three types of planar rigid body motion:
1. Translation - The body does not change orientation during motion (i.e. every
line segment in the body remains parallel to its original orientation). Rectilinear translation occurs when the paths traced by any two particles remain along
equidistant straight lines. Curvilinear translation occurs when the paths of motion of two particles remain on equidistant curved lines.
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2. Rotation about a …xed axis - All particles of the rigid body (except those at the
axis of rotation) move along circular paths.
3. General plane motion - is a combination of translation and rotation.
Consider a slider-crank mechanism (e.g. from the lawnmower engine).
Eng3934 - Dynamics
4.2
23
Translation
Consider a rigid body which undergoes translation in the x,y plane:
De…ne the positions of points A and B with respect to the …xed x,y reference frame
by the position vectors ~rA and ~rB .
De…ne the x0 ,y 0 translating co-ordinate system that is …xed to the rigid body with
origin at A. The relative position vector ~rB=A de…nes the position of B in the x0 ,y 0
system, i.e. B with respect to A.
By vector addition:
~rB = ~rA + ~rB=A
(46)
and di¤erentiation of the position vector equation gives:
~vA = ~vB
(47)
where ~vA and ~vB are the absolute velocities measured in the …xed x,y reference frame.
Note:
1. ~vB=A = 0 since the body is translating and it is rigid, therefore, there can be no
change in ~rB=A .
2. To facilitate vector addition the x,y and x0 ,y 0 co-ordinate systems are aligned and
have the same units.
Di¤erentiation of the velocities gives:
~aA = ~aB
(48)
These equations apply for both rectilinear and curvilinear translation. The equations
also indicate that the equations for the kinematics of a particle can be applied to the
kinematics of a rigid body in translation.
Eng3934 - Dynamics
4.3
24
Rotation about a Fixed Axis
Consider a rigid body of arbitrary shape rotating about a …xed axis.
A line r to the point P undergoes rotation or angular motion (the point P does not
follow angular motion, because it has no dimension).
The angular position of r is de…ned by the angle measured from a …xed reference
line. has units of degrees, radians or revolutions.
The angular displacement d~ is the change in angular position of r as the body rotates.
It is a vector quantity since it has magnitude and direction. The motion is about a
…xed axis, therefore, d~ is always along the axis with the direction determined from
the right hand rule. The standard convention is to de…ne counterclockwise motion as
positive. Angular displacement has units of degrees or radians.
The angular velocity, !
~ , is the time rate of change of angular position.
!
~ =
d~
dt
(49)
Angular velocity has units of radians/s. The direction of !
~ is along the axis of rotation
with the direction de…ned positive in the direction of positive , i.e. the right hand
rule is used.
The angular acceleration, ~ , is the time rate of change of angular velocity.
~ =
d~
!
d2~
= 2
dt
dt
(50)
and has units of radians/s2 . Again, ~ is directed along the axis of rotation, but the
direction depends on the change in !
~ . ~ is positive when in the direction of !
~.
By eliminating dt, we get an equation of a form familiar from kinematics of a particle:
d = ! d!
i.e. a ds = v dv.
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If the angular acceleration is constant, ~ = ~ c , then the equations of motion can be
integrated to give equations of a form familiar from kinematics of a particle.
!
!2
4.3.1
= !0 +
ct
1
= 0 + !0 t +
2
= ! 20 + 2 c (
(51)
ct
2
0)
Vector Formulation
Consider the motion of point P about O (the axis).
The position of P is identi…ed by the position vector ~r.
The velocity of P could be written using the Polar co-ordinates: vr = r_ and v =
r _ = r!. Since r is constant (rigid body):
v = !r
(52)
where ~v is tangent to the path of P .
This equation for velocity can be written in vector form. Since !
~ is perpendicular to
the plane of P (and ~r), then !
~ ~r will give a vector of magnitude ! r perpendicular
to !
~ and ~r, i.e. in the plane of, and perpendicular to, ~r, and the right hand rule would
de…ne its direction as that of ~v , therefore:
~v = !
~
~r
(53)
Note: The order of the vectors is important in cross products, i.e. !
~ ~r 6= ~r !
~ (the
magnitude is the same but ~r !
~ would give the opposite direction by the right hand
rule).
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The acceleration of P can be expressed in its n and t components:
at
=
an
=
dv
d(! r)
=
= r
dt
dt
v2
(! r)2
=
= !2 r
r
(54)
Remember: the tangential acceleration acts to change the magnitude of ~v , and the
normal acceleration acts to change the direction of ~v . Also, ~an is always directed
towards the center of the circular path.
The acceleration can also be written in vector form:
~a =
d~v
d~
!
=
dt
dt
~r + !
~
d~r
=~
dt
~r + !
~
(~
!
~r)
(55)
The cross product ~ ~r will give a vector of magnitude r perpendicular to the plane
of ~ and ~r (in the plane of motion of P ), i.e. the tangential acceleration. The cross
product !
~ (~
! ~r) will give a vector of magnitude ! 2 r directed in the ~r direction,
i.e. the normal acceleration.
Breaking the acceleration into its tangential and normal components:
~a = ~at + ~an
= ~ ~r ! 2~r
(56)
where the simpler notation for the normal acceleration has been used.
The magnitude of ~a is:
a=
q
a2t + a2n
(57)
Eng3934 - Dynamics
4.4
27
Relative Motion Analysis: Velocity and Acceleration
General plane motion of a rigid body can be considered to occur as a translation
followed by rotation about a …xed point in the body.
De…ne the x0 ,y 0 translating co-ordinate system that is …xed to the rigid body with
origin at A, i.e. the base point. The relative position vector ~rB=A de…nes the position
of B in the x0 ,y 0 system, i.e. B with respect to A.
By vector addition:
~rB = ~rA + ~rB=A
(58)
and di¤erentiation of the position vector equation gives:
~vB = ~vA + ~vB=A
(59)
where ~vA and ~vB are the absolute velocities measured in the …xed x,y reference frame,
and ~vB=A is the relative velocity of B with respect to A, measured in the translating
x0 ,y 0 reference frame. Here, ~vB=A 6= 0 since the relative position vector may rotate
(it does not change length), i.e. the body is considerd to rotate about point A. The
relative velocity ~vB=A will then have magnitude ! rB=A and direction perpendicular to
~rB=A . In vector form:
~vB = ~vA + !
~ ~rB=A
(60)
Eng3934 - Dynamics
28
Consider the simple crank-slider mechanism:
The crank rotates about point A at constant angular velocity !
~ 1 . The point C is
constrained to travel in rectlinear translation. Then:
~vB
~vC
= ~vA + ~vB=A = !
~ 1 ~rB=A
= ~vB + ~vC=B = ~vB + !
~ 2 ~rC=B
Note: an observer …xed at A sees B rotate about A. An observer …xed at B sees C
rotate about B (C actually oscillates about B). When solving problems try to de…ne
points for which the motion is known (this will give directions for unknown velocities
and accelerations).
Di¤erentiation of the velocities gives:
~aB = ~aA + ~aB=A
(61)
where ~aA and ~aB are the absolute accelerations measured in the …xed x,y reference
frame, and ~aB=A is the relative acceleration of B with respect to A, measured in the
translating x0 ,y 0 reference frame (…xed at A).
Since ~aB=A results from the rotation of the body about point A, then it can have two
components (normal and tangential).
~aB
= ~aA + ~aB=A t + ~aB=A n
= ~aA + ~ ~rB=A + !
~ (~
!
= ~aA + ~
~rB=A
(62)
~rB=A )
2
! ~rB=A
Considering the crank-slider again. Assume the crank has angular velocity !
~ 1 and
angular acceleration ~ 1 .
~aB
~aC
= ~aA + ~ 1 ~rB=A + !
~ 1 (~
! 1 ~rB=A )
= ~ 1 ~rB=A + !
~ 1 (~
! 1 ~rB=A )
= ~aB + ~ 2 ~rC=B + !
~ 2 (~
! 2 ~rC=B )
Eng3934 - Dynamics
5
5.1
29
Planar Kinetics of a Rigid Body: Force and Acceleration
Equations of Motion
A rigid body can rotate, therefore, when dealing with forces applied to a rigid body
the direction and points of application of the forces become important. When approximating a real body as a particle, all forces are assumed to act through the center
of mass of the body. For a rigid body, which has …nite shape and size, the forces do
not necessarily act through the center of mass, therefore, an applied force can supply
a moment about the center of mass and lead to rotation of the body.
This section will be divided into subsections concerned with rigid bodies in translation,
rotating about a …xed axis, and in general plane motion. Techniques used to locate
the center of mass of a rigid body will be covered, and the moment of inertia will be
introduced.
5.1.1
Translational Motion
A rigid body may be subjected to a system of forces and couple moments:
All particles of a rigid body in translation experience the same acceleration, therefore,
~a = ~aG . A moment about the center of mass G would cause rotation, but since ~ = 0
for a body in translation, then the sum of all moments about G must be zero. The
equations that govern a rigid body in translation are:
X
F~ = m~aG
(63)
X
~G = 0
M
(64)
In scalar form:
+
+
+
!
"
X
Fx = m (aG )x
X
MG = 0
X
Fy = m (aG )y
(65)
Eng3934 - Dynamics
30
It may be convenient to sum moments about a point other than G (to eliminate
unknowns):
Then the moment equation can be replaced by:
X
X
+
MA =
(Mk )A
(66)
i.e. the sum of moments about A on the free body diagram is equivalent to the sum
of the kinetic moments about A on the kinetic diagram.
Eng3934 - Dynamics
5.2
5.2.1
31
Center of Gravity
Center of Gravity for a System of Particles
The center of gravity, G, is the point which locates the resultant weight of a system
of particles.
~ i . The weight of each particle is a parallel force
Consider the n particles of weight W
~ R located at G.
that can be replaced by a single resultant W
Of course the resultant will be:
WR =
n
X
Wi
i=1
But how to determine the location point of the resultant, i.e. x, y, z?
The sum of the moment of each weight about an axis is equal to the moment of the
resultant about the axis. For example, about the x axis:
xWR = x
~ 1 W1 + x
~ 2 W2 + : : : + x
~ n Wn
A similar procedure is followed for the y and z axes. Remember replacing a system
of forces with a single force in Statics?
In general, the location of the center of gravity for a collection of particles is evaluated
as follows:
P
P
P
y~i Wi
z~i Wi
x
~ i Wi
y= P
z= P
(67)
x= P
Wi
Wi
Wi
where the summation is over all n particles.
The center of mass for a collection of particles is at the same location as the center of
gravity, since, each particle will be exposed to the same gravity …eld.
P
P
P
x
~i mi
y~i mi
z~i mi
x= P
y= P
z= P
(68)
mi
mi
mi
Eng3934 - Dynamics
5.2.2
32
Center of Gravity for a Rigid Body
A rigid body is composed of an in…nite number of particles, so the summations in the
previous equations would become integrals.
R
x
~dW
x= R
dW
R
y~dW
y= R
dW
R
z~dW
z= R
dW
To perform the integral, de…ne dW = dV , where is the speci…c weight of a substance
N= m3 .
R
R
R
x
~ dV
y~ dV
z~ dV
V
V
R
R
x=
y=
z = RV
(69)
dV
dV
dV
V
V
V
where the integral is over the total volume of the body.
Again, since gravity would be constant for each small volume dV , the center of mass
would be located at the same point as the center of gravity. The location of the
center of mass would be obtained by replacing speci…c weight, , with the density,
kg= m3 .
The centroid is the geometric center of a rigid body.
Eng3934 - Dynamics
33
If the density of a body is uniform, then the speci…c weight is also uniform. Then the
centroid of a body can be evaluated as:
R
R
R
x
~dV
y~dV
z~dV
V
V
R
R
x=
y=
z = RV
(70)
dV
dV
dV
V
V
V
And the centroid of an area is:
R
x
~dA
x = RA
dA
A
R
y~dA
y = RA
dA
A
R
z~dA
z = RA
dA
A
(71)
The centroids of some common shapes are tabulated inside the back cover of the text.
5.2.3
Composite Bodies
A composite body can be constructed through the addition of several simpler shapes:
The center of gravity of the composite body can be located by the same equation used
to determine the center of gravity for a collection of particles:
P
P
P
y~i Wi
z~i Wi
x
~ i Wi
y= P
z= P
x= P
Wi
Wi
Wi
Eng3934 - Dynamics
34
where x
~i ; y~i ; z~i are the co-ordinates for the center of gravity for the i-th body.
5.3
Mass Moment of Inertia
The mass moment of inertia, I, is a property of a body that measures the resistance of
the body to angular motion. It is analogous to mass, which measures the resistance
of a body to acceleration. For general plane motion we will use two equations:
X
F~ = m~a
X
~ = I~
M
where the second equation is a form of Newton’s second law for rotational motion.
The mass moment of inertia is de…ned as the integral of the second moment about an
axis of all elements of mass dm that compose a body. For example the mass moment
of inertia of the body below about the z axis is:
I=
Z
r2 dm
(72)
m
The moment arm r is the perpendicular distance from the axis to the element dm.
Usually the axis chosen passes through the center of gravity and we obtain IG . The
moment of inertia is always positive and has units of mass times length squared ( kg m2
or slug ft2 ).
The integral may be expressed as a volume integral by replacing dm with dV .
Z
I=
r2 dV
(73)
V
and if density is uniform:
I=
Z
r2 dV
(74)
V
The mass moments of inertia for some common shapes are tabulated inside the back
cover of the text.
Eng3934 - Dynamics
5.3.1
35
Parallel Axis Theorem
If the mass moment of inertia is known for an axis passing through the center of gravity
of a body, the parallel axis theorem is used to determine the mass moment of inertia
about another parallel axis.
Consider an arbitrary body with center of gravity G. The z 0 axis passes through G.
The z axis is parallel to z 0 and located at a distance d from z 0 .
The mass moment of inertia of the body about the z axis is:
Z
I =
r2 dm
m
Z h
i
2
=
(d + x0 ) + y 02 dm
Zm
Z
Z
=
x02 + y 02 dm + 2d
x0 dm + d2
dm
m
m
m
But
x02 + y 02 = r02
So the …rst term represents IG .
The second term is zero since the z 0 axis passes through G.
Z
Z
x0 dm = x
dm
m
m
but x = 0, since we are on an axis through G.
So the moment of inertia about the z axis is de…ned by the parallel axis theorem:
I = IG + md2
(75)
where IG is the mass moment of inertia about an axis through G, m is the mass of
the body, and d is the distance between the parallel axes.
Eng3934 - Dynamics
5.3.2
36
Radius of Gyration
The mass moment of inertia is sometimes tabulated using the radius of gyration, k:
r
I
k=
(76)
m
where k has units of length.
The mass moment of inertia is found from k as follows:
I = mk 2
5.3.3
(77)
Composite Bodies
The mass moment of inertia of a composite body about an axis may be obtained by
algebraically adding the mass moments of inertia of simpler bodies about the same
axis. Note: this would probably require use of the parallel axis theorem.
5.3.4
Confusion
Another term that is de…ned in mechanics is the moment of inertia:
Z
I=
y 2 dA
A
which is actually the second moment of area about the y axis. Inertia is a misnomer,
as inertia requires mass, but there is no mass in this term. Also, it is given the same
symbol, I. This is too stoopid! And it leads to much confusion in students. And it
is stoopid! So, the real moment of inertia is sometimes called the mass moment of
inertia (as I have done) to try to reduce the misunderstanding.
This would never happen in thermodynamics!
The second moment of area arises in Solid Mechanics when determining normal stresses
resulting from moments, and de‡ection of beams. See Term 4.
Eng3934 - Dynamics
5.4
37
General Plane Motion
Consider the plane motion of a rigid body. A translating reference frame, x,y, is …xed
at the center of mass G of the body. The acceleration of the center of mass is ~aG , and
the body has angular velocity and acceleration !
~ and ~ , respectively.
Consider the forces that act on a particle of mass mi located by position vector ~ri
measured from the center of mass, G.
The acceleration of mi is:
~ai
= ~aG + (~ai=G )t + (~ai=G )n
= ~aG + ~
~ri
! 2~ri
The resultant of all forces acting on mi has the components mi aG , mi ri ! 2 , and mi ri .
the sum of the moments of these forces about G is:
X
+
(MG )i = mi ri2 + (mi aG sin )xi (mi aG cos )yi
Summing over all the particles in the body:
X
X
X
+
MG =
mi ri2 + aG sin
mi xi
aG cos
Since the origin of the co-ordinate system is taken at G, then:
X
mi xi = mx = 0
X
mi yi = my = 0
X
mi yi
Eng3934 - Dynamics
and the sum of moments about G becomes:
X
X
+
MG =
mi ri2 = IG
38
(78)
This is why moment of inertia is de…ned, it arises whenever rotational acceleration
exists.
It may be convenient to determine the moments about an axis through a point P that
is parallel to the axis through G. This would be done to eliminate unknowns.
Using the kinetic diagram, the sum of moments about P is:
X
MP =
ym(aG )x + xm(aG )y + IG
X
=
(Mk )P
(79)
i.e. the sum of the moments of the external forces shown on an FBD about point P
is equivalent to the sum of the “kinetic moment" of the components of m~aG and the
“kinetic moment" of IG ~ . This equation is useful when the acceleration of a body is
known, or when a point P may be chosen at the intersection of two unknown forces
to eliminate them as unknowns.
Eng3934 - Dynamics
5.4.1
39
Summary
The equations describing the general plane motion of a rigid body are:
X
F~G = m~aG
X
X
X
~ G = IG ~ or
~P =
~ k )P
M
M
(M
5.5
These equations can be written in the following scalar form:
X
+
!
Fx = m(aG )x
X
+
"
Fy = m(aG )y
X
X
X
+
MG = IG
or +
MP =
(Mk )P
(80)
(81)
Translation
When a rigid body undergoes translation, all particles in the body undergo the same
acceleration, ~aG = ~a, and the body cannot rotate, ~ = 0.
5.5.1
Rectilinear Translation
In rectilinear translation, all particles travel along parallel straight lines.
The equations of motion are:
+
+
+
!
"
X
Fx = m(aG )x
X
MG = 0
X
(82)
Fy = m(aG )y
The sum of moments about a point A in the body is:
X
X
+
MA =
(Mk )A = (maG )d
(83)
Eng3934 - Dynamics
5.5.2
40
Curvilinear Translation
In curvilinear translation, all particles travel along parallel curved lines.
In curvilinear translation it is often convenient to use the n,t sytem. The equations of
motion are:
X
+ .
Fn = m(aG )n
(84)
X
+ Ft = m(aG )t
X
+
MG = 0
The sum of moments about a point B is:
X
X
+
MB =
(Mk )B = e(m(aG )t )
h(m(aG )n )
(85)
Eng3934 - Dynamics
5.6
41
Rotation about a Fixed Axis
Consider the rigid body that is constrained to rotate about a …xed axis at the point O.
The center of mass G will travel in a circular path about O, therefore, it is convenient
to use the n,t system at G to evaluate the acceleration. The tangential acceleration
will have magnitude rG and the normal acceleration will have magnitude ! 2 rG .
The equations governing this motion are:
X
+ .
Fn = m(aG )n = m! 2 rG
X
+ Ft = m(aG )t = m rG
X
+
MG = IG
(86)
As before, it may be convenient to sum moments about another point other than G.
Here, the axis of rotation O would be an obvious choice, as taking moments about O
would eliminate the (generally unknown) reaction force F~O .
Using the kinetic diagram:
X
X
+
MO =
(Mk )O
= rG m(aG )t + IG
= rG m(rG ) + IG
= (md2 + IG )
= IO
where IO = md2 + IG from the parallel axis theorem.
(87)
Eng3934 - Dynamics
5.7
So an alternate set of equations governing motion about a …xed axis is:
X
+ .
Fn = m(aG )n = m! 2 rG
X
+ Ft = m(aG )t = m rG
X
+
MO = IO
42
(88)
General Plane Motion
The equations governing general plane motion are:
X
+
!
Fx = m(aG )x
X
+
"
Fy = m(aG )y
X
X
X
+
MG = IG
or +
MP =
(Mk )P
(89)
Eng3934 - Dynamics
6
43
Kinetics of a Particle: Work and Energy
The principle of work and energy will provide a means of solving problems involving
force, velocity and displacement.
The principle of conservation of energy will be used to solve problems involving velocity, displacement, and conservative force systems.
6.1
Work of a Force
Consider the force F~ acting on the particle P moving along the path s. The initial
position of the particle is de…ned by the position vector ~r and the position of the
particle at time t + dt is given by ~r 0 . The displacement of the particle is d~r = ~r 0 ~r,
the magnitude of which can be expressed as the di¤erential length of the path, dr = ds.
De…ning the angle between d~r and F~ as
the work dU done by F~ is:
dU = F cos ds
or using vector notation:
dU = F~ d~r
i.e. the product of the component of the force in the direction of the displacement, or
the product of the component of the displacement in the direction of the force.
Work is a scalar and has units of Joule (J) i.e. N m in the SI system and units of ft lb
in the FPS system.
Work is positive when the force acts in the direction of motion of the particle, and
negative when the force acts in the opposite direction of the displacement. No work is
done by a force perpendicular to the path (since = 90o there is no force component
in the direction of motion).
Eng3934 - Dynamics
44
To obtain the total amount of work done during a …nite displacement the above equations must be integrated
U1
2
=
Z
~
r2
~
r1
F~ d~r =
Z
s2
F cos ds
(90)
s1
For a constant force acting on a body moving in a straight line:
U1
2
= Fc cos (s2
s1 )
The work of a weight is:
U1
2
=
Z
~
r2
~
r1
( W ~j) (dx~i + dy ~j + dz ~k) =
Z
y2
W dy =
W (y2
y1 )
y1
therefore, the work of a weight is negative if a body is raised and positive if the weight
lowers.
Eng3934 - Dynamics
45
The work done on a linear spring is
Z s2
Z
U1 2 =
Fs ds =
s1
s2
ks ds =
s1
1
k s22
2
s21
Note: when evaluating the work of a spring force on a particle or body, the direction
of the force is reversed. Therefore, if the spring is being stretched, the force would act
in the opposite direction of the motion of the particle, so:
U1
2
=
1
k s22
2
s21
(91)
Eng3934 - Dynamics
6.2
46
Principle of Work and Energy
Consider a particle travelling along a path s. The initial position of the particle is s1
and the …nal position is s2 . As the particle travels from s1 to s2 a system of external
forces, giving a resultant force F~R , act on it, and the speed of the particle changes
from v1 to v2 .
From Newton’s second law
X Z s2
s1
P
Ft = mat , and from kinematics at = v dv
ds , therefore:
Z v2
X Z s2
Ft ds =
F cos ds =
mv dv
s1
v1
and:
1
1
mv22
mv 2
(92)
2
2 1
i.e. the sum of the work done by the components of all forces acting along the tangent
to the path has increased the kinetic energy of the particle.
U1
2
=
This equation can be written in the following form:
X
T1 +
U1 2 = T2
(93)
Note: force, displacement, and speed are the only variables in this equation, therefore,
it will be convenient in problems where these are the parameters of interest (e.g. the
maximum height of a rollercoaster or stopping distance for a car (i.e. distance at
speed)).
6.3
Power and E¢ ciency
Power can be considered to be the rate at which work is done.
P =
dU
dt
Eng3934 - Dynamics
47
Since dU = F~ d~r, then
P =
dU
F~ d~r
=
dt
dt
d~r
= F~
dt
~
= F ~v
(94)
The units of power in the SI and FPS systems are the watt (W=J/s) and horsepower
(1 hp=550 ft lb/s), respectively.
Note: if you ride a bicycle up Signal Hill you will have to do a …xed amount of work
to raise the weight of your body and bicycle through the change in elevation. To ride
up the hill in a shorter time requires more power.
The mechanical e¢ ciency ( ) of a machine is de…ned as the ratio of output power to
input power:
power output
=
(95)
power input
Since all real machines have friction, < 1, from the 2nd Law of Thermodynamics.
6.4
6.4.1
Conservative Forces and Potential Energy
Conservative Forces
The work done by a conservative force is independent of the path.
Weight is a conservative force since the work depends upon the vertical displacement
only.
A spring force is conservative since the work done is dependent on the compression
(or extension) of the spring only.
Friction forces are nonconservative since the amount of work done will be dependent
on the path followed.
6.4.2
Potential Energy (V )
Potential energy is the capacity to do work due to position. For example, a 10 kg
mass at an elevation of 100 m has the capacity to do mgh = 9810 N m of work.
Consider the position of a weight W measured with respect to a datum:
Eng3934 - Dynamics
48
Consider a linear elastic spring stretched (or compressed) by s. The potential energy
of the spring is:
1
Ve = k s2
2
i.e. the work done to compress (or stretch) the spring to s.
Note: Ve > 0 since the spring can do postive work on a particle attached to the spring.
If a particle is subjected to gravitational and spring forces, the potential energy of the
particle is:
V = V g + Ve
The work required to change the position of the particle with respect to a datum is:
U1
2
= V1
V2
e.g. Raising a mass from y1 to y2 (y2 > y1 ):
U1
2
= mg(y1
y2 ) < 0
since the force W opposes the direction of motion.
6.5
Conservation of Energy
If a particle is acted upon by a system of conservative and nonconservative forces, the
work-energy equation can be written as:
X
X
T1 +
U1 2
+
U1 2
= T2
cons
But:
and:
X
T1 + V1 +
U1
X
2
U1
noncons
cons
2
= V1
noncons
V2
= T2 + V2
(96)
Eng3934 - Dynamics
49
If there are no nonconservative forces:
T 1 + V1 = T 2 + V2
(97)
i.e. Conservation of Mechanical Energy (T + V = const).
To apply conservation of mechanical energy to a system of particles, sum the kinetic
and potential energies of each particle:
X
X
X
X
T1 +
V1 =
T2 +
V2
(98)
Eng3934 - Dynamics
7
50
Kinetics of a Particle: Linear Impulse and Momentum
The principle of linear impulse and momentum will provide a means of solving problems involving: (1) force, velocity and time; (2) impact; and (3) steady ‡uid streams.
7.1
Principle of Linear Impulse and Momentum
The equation of motion for a particle of mass m is:
X
d~v
F~ = m~a = m
dt
Rearranging the equation of motion and integrating from state 1 (t1 ,v1 ) to state 2
(t2 ,v2 ) gives:
Z v2
X Z t2
F~ dt = m
dv
t1
or
XZ
t2
v1
F~ dt = m~v2
m~v1
(99)
t1
which is the principle of linear impulse and momentum.
~ = m~v , has the same direction as ~v , and units of kg m/s and
The linear momentum, L
slug ft/s in the SI and FPS unit systems, respectively.
R
The linear impulse I~ = F~ dt has the same direction as F~ , and units of N s and lb s
in the SI and FPS unit systems, respectively. The linear impulse gives the e¤ect of a
force during a time interval.
The principle of linear impulse and momentum can be rearranged as follows:
X Z t2
m~v1 +
F~ dt = m~v2
(100)
t1
i.e. the initial linear momentum of a particle plus the sum of all linear impulses acting
on the particle equals the …nal linear momentum of the particle.
Eng3934 - Dynamics
51
You may …nd it useful to draw momentum and impulse diagrams:
Also, the vector form of the principle of linear impulse and momentum may be written
in its scalar components, e.g. the x,y,z components:
X Z t2
mvx;1 +
Fx dt = mvx;2
(101)
t1
t2
mvy;1 +
mvz;1 +
XZ
Fy dt = mvy;2
t1
X Z t2
Fz dt = mvz;2
t1
For a system of particles, the principle of linear impulse and momentum can be written
as follows:
X
X X Z t2
X
mi~vi;1 +
F~i dt =
mi~vi;2
(102)
i
t1
i
i
or, using the center of mass G of the system of particles:
X Z t2
m~vG;1 +
F~ dt = m~vG;2
(103)
t1
7.2
Conservation of Linear Momentum for a System of Particles
The principle of linear impulse and momentum for a system of particles is:
X
X X Z t2
X
mi~vi;1 +
F~ dt =
mi~vi;2
i
t1
i
i
If the sum of all external impulses acting on the system of particles is zero, the principle
of linear impulse and momentum reduces to:
X
X
mi~vi;1 =
mi~vi;2
(104)
i
i
i.e. the conservation of linear momentum.
P
Using the center of mass, m~vG = mi~vi :
~vG;1 = ~vG;2
(105)
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Conservation of linear momentum is often applied in impact problems.
Take care when identifying internal and external impulses. Draw a FBD of the whole
system to identify the impulses. Remember that internal impulses will cancel, since
the forces that cause them are equal in magnitude, act in opposite directions and are
collinear (Newton’s 3rd Law).
Are all external forces impulsive? No. For example, if the time interval (t2 t1 ) is
very small, usually the forces of interest are due to impact or explosion. In such cases
the weight of the particles is not an impulsive force. The spring force of an elastic
spring may also be nonimpulsive (weak spring, small displacement of the spring in the
time interval). In general, any force that is small relative to other forces acting in the
time interval of interest may be neglected. Note: small is a relative term and it would
be problem dependent.
Once conservation of momemtum has been applied to the system of particles, the
principle of linear impulse and momentum may be applied to individual particles in
the system to determine the impulse acting on the individual particles during the time
interval. From the impulse and length of the time interval, the mean force applied to
a particle during the time interval may be determined.
7.3
Impact
Impact refers to the collision between two particles and is characterized by the generation of relatively large contact forces during a relatively short time interval.
Impact is a complex phenomenon involving material deformation and recovery, and
the generation of heat and sound.
Small changes in parameters can have a signi…cant e¤ect on the results, therefore,
impact calculations should be treated as estimates.
Central impact occurs when the direction of motion of the mass centers of the two
colliding particles is directed along a line through the mass centers of the particles (or
contact forces are directed along a line through the path of the mass centers of the
particles).
The line of impact is de…ned as a line through the mass centers of two impacting
objects, which passes through the point of contact of the two objects.
Oblique impact occurs when the motion of one or more mass centers is at an angle to
the line of impact.
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53
Consider the central impact of two deformable particles:
R
R
~ dt i.e. the deformation impulse is always greater than the restituNote: P~ dt > R
tion impulse (except for the special case of perfectly elastic impact).
The typical problem in impact analysis is to determine the …nal velocities of two
particles when the initial velocities, vA;1 and vB;1 , are known.
For the case of central impact above, linear momentum is conserved:
mA vA;1 + mB vB;1 = mA vA;2 + mB vB;2
(106)
since the deformation and restitution impulses are internal (when the particles are in
contact) and will cancel.
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54
Here we have one equation and two unknowns: vA;2 and vB;2 . To obtain a second
equation, the principle of linear impulse and momentum is applied to each particle
separately. During the deformation phase of A:
Z
mA vA;1
P dt = mA v
and during the restitution phase of A:
Z
~ dt = mA vA;2
mA v
R
De…ne the coe¢ cient of restitution, e, as the ratio of the restitution impulse to the
deformation impulse.
R
R dt
v vA;2
R
e=
=
vA;1 v
P dt
During the deformation phase of B:
Z
mB vB;1 + P dt = mB v
and during the restitution phase of B:
Z
mB v + R dt = mB vB;2
and the coe¢ cient of restitution is:
Eliminating the unkown v:
R
R dt
vB;2 v
R
e=
=
v vB;1
P dt
vB;2 vA;2
(107)
vA;1 vB;1
If e is known then we can solve for vA;2 and vB;2 . Note: Be very careful of the sign
convention.
e=
The coe¢ cient of restitution is dependent on the material, size and shape of the
particle, and the speed of impact. Data is reliable only if it is taken for similar
conditions.
– e = 1 - perfectly elastic impact (no permanent deformation)
– e = 0 - plastic or perfectly inelastic (particles deform and stick to each other)
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55
Work energy methods cannot be used in impact problems, however, the energy dissipated during an impact can be determined from the change in kinetic energy of the
particles:
X
X
X
U1 2 =
T2
T1
(108)
No energy is dissipated for a perfectly elastic collision, and the maximum energy is
dissipated for plastic collisions.
In oblique impact there will be four unknowns (the velocity components (vA;2 )x ,
(vA;2 )y , (vB;2 )x , and (vB;2 )y or the speeds and angles vA;2 , vB;2 , 2 , and 2 ). With
the de…nition of the x,y axes shown above (always de…ne the x axis along the line of
impact), conservation of x momentum gives:
mA (vA;1 )x + mB (vB;1 )x = mA (vA;2 )x + mB (vB;2 )x
(109)
The principle of linear impulse and momentum applied in the y-direction gives:
(vA;1 )y
(vB;1 )y
= (vA;2 )y
= (vB;2 )y
(110)
And the coe¢ cient of restitution can be de…ned (as for central impact) as follows:
e=
(vB;2 )x
(vA;1 )x
(vA;2 )x
(vB;1 )x
(111)