•
SECTION 1: Introduction, Motivation and Review
INTRODUCTION
How do we predict/analyse/understand the behaviour of a body/object/system of
bodies subjected to some disturbance?
REALITY
•
(MATHEMATICAL)
MODEL OF REALITY
Applied Mathematics
RESULTS FROM MODEL
Good �nough?
NO
•
YES
Interpret 'mathematical' results
in physical co ext
r
PREDICT/UNDERSTAND/ANALYSE
BEHAVIOUR OF BODY/SYSTEM
•
So far, in your experience, everything has been modelled as a particle or collection
of particles - not terribly realistic e.g. consider the motion of a ball
- it rotates (motion of body about its own center of mass)
- it deforms i.e. changes shape (elastic)
- etc etc
•
How important is the deformation ? - not as important as the fact that the ball
rotates. So, in this case, a more accurate prediction of the ball's behaviour might
come from the modelling of the ball as a 'rigid body' as opposed to a particle - it's
the next stage in complexity of the models listed on the previous page - (remember,
if you go to a further stage e.g. elastic body, what you gain in 'closeness to reality'
you might lose in ability to analyse the model).
So, for now, we will assume that all 'other effects' e.g. deformation, temp. changes
etc are negligible and concentrate on what we believe to be a significant addition
to the model of the ball as particle i.e. rotation:
Example 1.1
Suppose we throw a ball with initial velocity v0 , at height h and angle 0.
Regarding the ball as a particle, we use our model (equations) of projectile motion
to predict H (maximum height) and L (range .. where the ball hits the ground).
Further, if we know "e" (coefficient of restitution - ratio of relative velocity of the
particles's separation just after impact to the relative velocity of the particle's
approach just before impact) between the ball and the ground, we can.. Continue the
analysis
, to predict the motion during the next bounce:
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SECTION 2
PLANAR KINEMATICS
OF A RIGID BODY
PLANAR RIGID-BODY MOTION
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PLANAR RIGID BODY MOTION
(continued)
An example of bodies
undergoing the three types
of motion is shown in this
mechanism.
The wheel and crank undergo rotation about a fixed axis. In this
case, both axes of rotation are at the location of the pins and
perpendicular to the plane of the figure.
The piston undergoes rectilinear translation since it is constrained
to slide in a straight line.
The connecting rod undergoes curvilinear translation, since it will
remain horizontal as it moves along a circular path.
The connecting rod undergoes general plane motion, as it will both
translate and rotate.
We would like to build a mathematical model (some
equations) which describes (and therefore predicts) the
motion of the rigid body executing plane motion.
We expect two things to come into play:
1. Since the body rotates about only ONE axis, we
expect one angular velocity and one angular
acceleration to enter into the model...these will be
VECTORS.
2. Different points on the body will exhibit different
types of motion ….e.g. a pin won’t move….points at
different distances from the pin will move with
different speeds etc etc
We have noted the different types of PLANE MOTIONS
We treat the two simplest cases first:
(i) Translation,
(ii) Rotation about a fixed axis
then ‘combine’ the theories to describe:
(ii) General Plane Motion
RIGID-BODY MOTION: TRANSLATION
The positions of two points A and B
on a translating body can be related by
rB = rA + rB/A
where rA & rB are the absolute
position vectors defined from the
fixed x-y coordinate system, and rB/A
is the relative-position vector between
B and A.
The velocity at B is vB = vA+ drB/A/dt .
Now KEY POINT: drB/A/dt = 0 since rB/A is constant (M + D!)
So, vB = vA, and by following similar logic, aB = aA.
CONCLUSION: ALL POINTS ON A RIGID BODY IN PURE
TRANSLATION MOVE WITH THE SAME VELOCITY AND
ACCELERATION (e.g. as the mass center - so the motion of the body
is essentially represented by the motion of its mass center)N: All
RIGID-BODY MOTION: ROTATION ABOUT A FIXED AXIS
NOTE THAT ANGULAR VELOCITY AND
ANGULAR ACCELERATION are
VECTORS…more importantly, THEY ARE
WHOLE BODY PROPERTIES…i.e. they do
not vary from point to point in the body but are
the same for the entire body.
What would happen if they varied from point to
point?
RIGID-BODY MOTION: ROTATION ABOUT A FIXED AXIS
(continued)
If the angular acceleration of the body is
constant with respect to time i.e. magnitude
α = αC, the equations for angular velocity
and acceleration can be integrated to yield
the following set of algebraic equations for
the magnitudes ω and αC :
ω = ω0 + αC t
θ = θ0 + ω0 t + 0.5 αC t2
ω2 = (ω0)2 + 2αC (θ – θ0)
θ0 and ω0 are the initial values of the body’s
angular position and angular speed. Note
these equations are very similar to the
constant acceleration relations developed for
the rectilinear motion of a particle.
RIGID-BODY MOTION: ROTATION ABOUT A FIXED AXIS
(continued)
So far we have equations which describe the
body’s angular velocity and angular
acceleration in terms of the angular position
θ.
We know however that different points on
the body will move with different velocities
and accelerations depending on their
position in the body e.g. for a given ω , the
speed of P will increase with the
radius of the circle on which it lies….the
speed of point O is zero etc. To get an
equation for the motion of a typical point P,
we need to involve distance from the axis….
RIGID-BODY ROTATION: VELOCITY OF POINT P
The key point is that all points are
moving with circular motion. The only
difference is the radius of the circle on
which the points lie. Thus, we should be
able to use what we already know about
circular motion to describe the motion of
a typical point P on a body executing
plane RB rotation: In fact, recall:
ds = rdθ so
ds/dt = rdθ/dt = ωr
RIGID-BODY ROTATION: VELOCITY OF POINT P
So the magnitude of the velocity of P is
equal to
v = rω.
The velocity’s direction is tangent to the
circular path of P(SeeSection12.4 Hibbeler).
In the vector formulation, the
magnitude and direction of the velocity
v can be determined from the cross
product of ω and rp (here rp is a vector
from any point on the axis of rotation to
P ) i.e.
v = ω × rp
The direction in which v points is determined by the righthand rule (so the order of the cross product is important:
i.e. ω × rp is not the same as rp × ω)
ROTATION ABOUT A FIXED AXIS: SUMMARY
• Establish a sign convention along the axis of rotation.
• If a relationship is known between any two of the variables (α,
ω, θ, or t), the other variables can be determined from the
equations: ω = dθ/dt α = dω/dt α dθ = ω dω
• If α is constant, use the equations for constant angular
acceleration.
• To determine the motion of a point, the scalar equations v = ω r,
at = α r, an = ω2r , and a = (at)2 + (an)2 can be used.
• Alternatively, the vector form of the equations can be used
v =ω × r
a = at + an = α × r+ ω × (ω × r)
EXAMPLE
Given: The motor turns gear A with a
constant angular acceleration, αA = 4.5
rad/s2 CW, starting from rest. The cord
is wrapped around pulley D which is
rigidly attached to gear B.
Find: The velocity of cylinder C and
the distance it travels in 3 seconds.
Solution
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WORKED EXAMPLES
FOLLOW IN LECTURE 3
ABSOLUTE MOTION ANALYSIS
We mentioned earlier that general plane motion is really a
‘combination’ of planar translation and rotation. The description of
‘planar translation’ was really simple: all points on the rigid body
(including its mass center) move with the same velocity and
acceleration – much like particle dynamics.
‘Planar rotation’ of a rigid body was a bit more complicated
essentially invoking the idea that points move in circles around the
axis of rotation giving rise to BODY properties such as ‘angular
velocity’ and ‘angular acceleration’.
Before we attempt to ‘combine’ these theories in a (vector)
model describing GENERAL PLANE MOTION, we pause to look
at another way of describing GENERAL PLANE MOTION. This
method is known as ABSOLUTE MOTION ANALYSIS and uses a
blend of physics, geometry and calculus. It’s not as elegant as
‘vector methods’ (see later) but very useful in some cases.
ABSOLUTE MOTION ANALYSIS
GOAL: To determine the velocity and acceleration of a rigid body
undergoing general plane motion using an absolute motion analysis.
As the name suggests, NO VECTORS INVOLVED HERE:
This method uses the geometry of the problem to relate the position
of a point P on a rigid body to the angle θ(t) of a line contained in
the body. (Often this line is a link in a machine.) Once a ‘formula’
in the form of sP = f(θ)= position function is established, the speed
and magnitude of acceleration of point P are obtained in terms of the
magnitude ω of the angular velocity and α of the angular
acceleration of the rigid body by taking the first and second time
derivatives of the position function. Usually the chain rule must be
used when taking the time derivatives of the position coordinate
equation.
Note that only magnitudes are found from absolute (!) motion
analysis….the ‘directions’ of velocities, accelerations etc are
deduced from the physics of the motion…if possible.
Absolute Motion Analysis
The position of the piston, x, can be defined as a function of
the angular position of the crank, θ. By differentiating x with
respect to time, the velocity of the piston can be related to the
magnitude ω of the angular velocity of the crank (which is
clearly clockwise or –k). This is necessary when designing
an engine.
The stroke of the piston is defined as the total distance moved
by the piston as the crank angle varies from 0 to 180°.
EXAMPLE
Given: Crank AB rotates at a constant velocity of ω = 150
rad/s CCW.
Find: The velocity of point P when θ = 30°.
Plan: Define x as a function of θ and differentiate with
respect to time. Then add direction.
EXAMPLE II (continued)
Solution: xP = 0.2 cos θ +
(0.75)2 – (0.2 sin θ)2
vP = -0.2ω sin θ + (½)[(0.75)2
– (0.2sin θ)2]-½(-2)(0.2sin θ)(0.2cos θ ) ω
vP = -0.2ω sin θ – [0.5(0.2)2 sin2θ ω] / (0.75)2 – (0.2 sin θ)2
At θ = 30°, ω = 150 rad/s and vP = -18.5 ft/s = 18.5 ft/s ←
PROCEDURE FOR ABSOLUTE MOTION ANALYSIS
The velocity and acceleration of a point undergoing rectilinear
motion can be related to the angular velocity and angular
acceleration of a line contained within a body using the
following procedure.
1. Locate point on the body using position coordinate s, which
is measured from a fixed origin.
2. From a fixed reference line, measure the angular position θ
of a line lying in the body. Using the dimensions of the
body, relate s to θ, e.g., s = f(θ).
3. Take the first time derivative of s = f(θ) to get a
relationship between v and ω.
4. Take the second time derivative to get a relationship
between a and α.
•
Example 2.4
Determine the velocity nnd acceleration of the peg A which is confined
between the vertical guide nnd the rotating slotted rod.
SOLUTION
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Ans:
1
V,. wb SCC 6
a a: b sec2B(2cJ tan 9 + a)
•
Example 2.5
y
The crnnk AB has a constant angular velocity "'.
Determine the velocity and acceleration of the slider a\ C
ns a function of 9. Suggestion: Use the x coordinate to
express the motion of C and the cfi coordinate for CB. x -:.
0 when cfi "" 0°.
SOLUTION
•
Ans:
·vc
= bw[
(%)sin9cos9
j)
�1 - (
2
silt' 9
] + bwsin9
2
9)
(7)
)
+
28
cos
(7
(
[
ac=bw2
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(7)
sin4
21
\n2 8
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+cos9
]
0 Example 2.6: Rolling Wheel
Use absolute motion analysis to derive an expression relating the position
of a point on the rim of a wheel and the rotation of the wheel as it rolls
WITHOUT SLIPPING on a stationary horizontal surface.
SOLUTION
Let A,B,C,D be points on the rim of the wheel
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When the wheel rotates through an angle 8, the center O will translate to a
new position o l and point C will rotate down to contact the surface at cl .
0
RELATIVE MOTION ANALYSIS: VELOCITY
Recall in the last section we used ABSOLUTE motion analysis.
In this section we use RELATIVE motion analysis to write down the
velocity of an arbitrary point on the rigid body in terms of the velocity
of another point (hence the word ‘RELATIVE’) whose motion is
KNOWN (or easily found) and the angular velocity of the body.
The big advantage here is that we use VECTORS….which allow us to
CALCULATE BOTH magnitudes AND directions of various
quantities.
RELATIVE MOTION ANALYSIS
(Velocity)
When a body is subjected to general plane motion, it undergoes a
combination of translation and rotation.
Point A is called the base point in this analysis. It generally has a
known motion. The x’- y’ frame translates with the body, but does not
rotate. The displacement of point B can be written:
Disp. due to translation
drB = drA + drB/A
Disp. due to translation and rotation
Disp. due to rotation
RELATIVE MOTION ANALYSIS: VELOCITY
The velocity at B is given as : (drB/dt) = (drA/dt) + (drB/A/dt)
or
vB = vA + vB/A
Since vB/A is essentially the body rotating about A,
vB/A = drB/A/dt = ω × rB/A
Here ω will have only a k component since the axis of rotation
is perpendicular to the plane of translation.
RELATIVE MOTION ANALYSIS: VELOCITY (continued)
vB = vA + ωΑΒ × rB/A
or
(vB = vC + ωΒC × rB/C)
When using the relative velocity equation the ‘base or pivot
point’ A (or C) should generally be a point on the body with a
known motion.
This is important because the vector equation is really just two
scalar equations so you are allowed at most two pieces of
UNKNOWN information (why?)
This will often require that you ALSO know something about
point B, the point of interest…and/or the angular velocity of the
body on which the two points lie….
RELATIVE MOTION ANALYSIS: VELOCITY (continued)
vB = vA + ωΑΒ × rB/A
vB = vC + ωΒC × rB/C
For example, point A on link AB must
move along a horizontal path, whereas
point C is stationary.
The directions of vA and vB are known
since they are always tangent to their
paths of motion.
RELATIVE MOTION ANALYSIS: VELOCITY
(continued)
vB = vA + ω × rB/A
When a wheel rolls without slipping, point A is often selected
to be at the point of contact with the ground.
Since there is no slipping, point A has zero velocity.
Furthermore, point B at the center of the wheel moves along a
horizontal path. Thus, vB has a known direction, e.g., parallel
to the surface.
PROCEDURE FOR ANALYSIS
The relative velocity equation can be applied using scalar x and y
component equations or via a Cartesian vector analysis.
Scalar Analysis:
1. Establish the fixed x-y coordinate directions and draw a
kinematic diagram for the body. Then establish the
magnitude and direction of the relative velocity vector vB/A.
2. Write the equation vB = vA + vB/A. In the kinematic diagram,
represent the vectors graphically by showing their
magnitudes and directions underneath each term.
3. Write the scalar equations from the x and y components of
these graphical representations of the vectors. Solve for
the unknowns.
PROCEDURE FOR ANALYSIS
(continued)
Vector Analysis:
1. Establish the fixed x - y coordinate directions and draw the
kinematic diagram of the body, showing the vectors vA, vB,
rB/A and ω. If the magnitudes are unknown, the sense of
direction may be assumed.
2. Express the vectors in Cartesian vector form (CVN) and
substitute them into vB = vA + ω × rB/A. Evaluate the cross
product and equate respective i and j components to obtain
two scalar equations.
3. If the solution yields a negative answer, the sense of
direction of the vector is opposite to that assumed.
EXAMPLE I
Given: Roller A is moving to the
right at 3 m/s.
Find: The velocity of B at the
instant θ = 30°.
Plan:
1. Establish the fixed x - y directions and draw a kinematic
diagram of the bar and rollers.
2. Express each of the velocity vectors for A and B in terms
of their i, j, k components and solve vB = vA + ω × rB/A.
EXAMPLE I (continued)
Solution:
Kinematic diagram:
y
y
Express the velocity vectors in CVN
vB = vA + ω × rB/A
-vB j = 3 i + [ ωk ×
(-1.5cos30i +1.5sin 30j )]
-vB j = 3 i – 1.299 ω j – 0.75 ω i
Equating the i and j components gives:
0 = 3 – 0.75 ω
-vB = – 1.299 ω
Solving: ω = 4 rad/s or ω = 4 rad/s k
vB = 5.2 m/s or vB = -5.2 m/s j
EXAMPLE II
Given: Crank rotates OA with an
angular velocity of 12 rad/s.
Find: The velocity of piston B
and the angular velocity of
rod AB.
Plan:
Notice that point A moves on a circular path.
The direction of vA is tangent to its path of
motion. Draw a kinematic diagram of rod AB
and use vB = vA + ω AB × rB/A.
EXAMPLE II (continued)
Solution:
Kinematic diagram of AB:
Since crack OA rotates with an angular
velocity of 12 rad/s, the velocity at A
will be: vA = -0.3(12) i = -3.6 i m/s
Rod AB. Write the relative-velocity
equation:
vB = vA + ωAB × rB/A
vB j = -3.6 i + ωAB k × (0.6cos30 i − 0.6sin30 j )
vB j = -3.6 i + 0.5196 ωAB j + 0.3 ωAB i
By comparing the i, j components:
⇒ ωAB = 12 rad/s
i: 0 = -3.6 + 0.3 ωAB
j: vB = 0.5196 ωAB
⇒ vB = 6.24 m/s
Example 2.7
If the flywheel is rotating with an angular velocity of
vA = 6 rad>s, determine the angular velocity of rod BC at
the instant shown.
1.5 m
A
0.3 m
vA ⫽ 6 rad/s
O
C
B
SOLUTION
60⬚ 0.6 m
D
INSTANTANEOUS CENTER OF ZERO VELOCITY
Recall that the relative velocity equation relied on a base/pivot point
around which to base the calculation. It would certainly be nice if we
could always (easily!?) find a point with ZERO velocity (at the instant
of interest) on the rigid body – we could then base our relative velocity
calculation around this point. Such a point does exist – it’s called
the instantaneous center of zero velocity (IC)
The only difficulty arises when the IC is difficult to identify…. making
it easier (when you have a choice!) to use another point instead.
In this section we’ll look at the whole business of IC and show how we
can use the IC as a base point to determine the velocity of any point
on a rigid body in general plane motion.
APPLICATIONS
The instantaneous center (IC)
of zero velocity for this
bicycle wheel is at the point
in contact with ground. The
velocity direction at any point
on the rim is perpendicular to
the line connecting the point
to the IC.
Which point on the wheel has the maximum velocity?
APPLICATIONS (continued)
As the board slides down the wall (to
the left), it is subjected to general
plane motion (both translation and
rotation).
Since the directions of the velocities
of ends A and B are known, the IC is
located as shown.
How can this result help you analyze
other situations?
What is the direction of the velocity of the center of gravity of
the board?
INSTANTANEOUS CENTER OF ZERO VELOCITY
For any body undergoing planar motion, there always exists a
point in the plane of motion at which the velocity is
instantaneously zero .
This point is called the instantaneous center (IC) of zero
velocity. It may or may not lie on the body!
(in the case of the latter we say that the IC lies on the
‘extended body’ which inherits the properties of the real body)
If the location of this point can be determined, the velocity
analysis can be simplified because the body appears to rotate
about this point at that instant.
LOCATION OF THE INSTANTANEOUS CENTER
To locate the IC, we use the fact that the velocity of a point on a
body is always perpendicular to the relative position vector from
the IC to the point. Several possibilities exist.
First, consider the case when velocity vA
of a point A on the body and the angular
velocity ω of the body are known.
In this case, the IC is located along the
line drawn perpendicular to vA at A, a
distance rA/IC = vA/ω from A.
Note that the IC lies up and to the right
of A since vA must cause a clockwise
angular velocity ω about the IC.
LOCATION OF THE INSTANTANEOUS CENTER
(continued)
A second case occurs when the
lines of action of two nonparallel velocities, vA and vB,
are known.
First, construct line segments
from A and B perpendicular to
vA and vB. The point of
intersection of these two line
segments locates the IC of the
body.
LOCATION OF THE INSTANTANEOUS CENTER
(continued)
A third case is when the magnitude and direction of two parallel
velocities at A and B are known. Here the location of the IC is
determined by proportional triangles.
As a special case, note that if the body is translating only
(vA = vB), then the IC would be located at infinity. Then ω
equals zero, as expected.
NOTE:
If two pairs of corresponding angles in a pair of triangles are congruent,
then the triangles are SIMILAR (PROPORTIONAL). We know this because
if two angle pairs are the same, then the third pair must also be equal.
When the three angle pairs are all equal, the three pairs of sides must
also be in proportion.
VELOCITY ANALYSIS
The velocity of any point on a body undergoing general plane
motion can be determined easily, often with a scalar approach,
once the instantaneous center of zero velocity of the body is
located.
Since the body seems to rotate about
the IC at any instant, as shown in this
kinematic diagram, the magnitude of
velocity of any arbitrary point is
v = ω r, where r is the radial distance
from the IC to the point.
The velocity’s line of action is
perpendicular to its associated radial
line.
EXAMPLE I
Given: A linkage undergoing
motion as shown. The
velocity of the block, vD,
is 3 m/s to the right.
Find: The angular velocities
of links AB and BD.
Plan: Locate the instantaneous center of zero velocity of link
BD and then solve for the angular velocities.
EXAMPLE I
Solution: We are asked for angular velocities of links AB and
BD. If we can locate the IC of each bar and deduce the
direction of rotation of each bar from the motion, then all we
need are the equations: ωBD = v(.)/r(.)/IC and ωAB = v(.)/r(.)/IC
where (.) stands for a point on the respective bar whose speed
is known or easily found e.g. point D for bar BD and point B
for bar AB (why not use point A?)
EXAMPLE I
Motion: Since D moves to the right, it causes link AB to
rotate clockwise about point A. The instantaneous center of
velocity for BD is located at the intersection of the line
segments drawn perpendicular to vB and vD. Note that vB is
perpendicular to link AB. Therefore we can see that the IC is
located along the extension of link AB.
EXAMPLE I (continued)
Using these facts,
rB/IC = 0.4 tan 45° = 0.4 m
rD/IC = 0.4/cos 45° = 0.566 m
Since the magnitude of vD is known, the
angular velocity of link BD can be found
from vD = ωBD rD/IC .
ωBD = vD/rD/IC = 3/0.566 = 5.3 rad/s ccw
Link AB is subjected to rotation about A.
So it’s IC is at point A.
ωAB = vB/rB/A = (rB/IC)ωBD/rB/A = 0.4(5.3)/0.4 = 5.3 rad/s cw
EXAMPLE 2
Given: The disk rolls without
slipping between two
moving plates.
vB = 2v
vA = v
Find: The angular velocity of the disk.
Plan: This is an example of the third case discussed in the
theory. Locate the IC of the disk using geometry and
trigonometry. Then calculate the angular velocity.
EXAMPLE 2 (continued)
Solution:
v
A
Using similar triangles:
x
IC
ω
x/v = (2r-x)/(2v)
or x = (2/3)r
O
Therefore ω = v/x = 1.5(v/r)
r
B
2v
Example 2.10.
0.3 m 0.3 m
If link CD has an angular velocity of vCD = 6 rad>s,
determine the velocity of point B on link BC and the
angular velocity of link AB at the instant shown.
C
B
E
0.6 m
A
SOLUTION
30
vCD
6 rad/s
D
RELATIVE MOTION
ANALYSIS:
ACCELERATION
RELATIVE MOTION ANALYSIS: ACCELERATION
We extend the previous theories on relative velocity to
determine the acceleration of a point on a body using a
relative acceleration analysis.
THEORY
The equation relating the accelerations of two points on the
body is determined by differentiating the velocity equation
with respect to time.
dvB
dvA
dv
=
+ B/ A
dt
dt
dt
These are absolute accelerations
of points A and B. They are
measured from a set of fixed
x,y axes.
This term is the acceleration
of B with respect to A and
includes both tangential and
normal components.
The result is aB = aA + (aB/A)t + (aB/A)n
RELATIVE MOTION ANALYSIS: ACCELERATION
(continued)
aB = aA + (aB/A)t + (aB/A)n
Graphically:
The relative tangential acceleration component (aB/A)t is (α ×rB/A)
and perpendicular to rB/A.
The relative normal acceleration component ( aB/A)n is ω × (ω × rB/A)
= (−ω2 rB/A when ω = ωk )&the direction is always from B towards A.
RELATIVE MOTION ANALYSIS: ACCELERATION
(continued)
Since the relative acceleration components can be expressed
as (aB/A)t = α × rB/A and (aB/A)n = ω × (ω × rB/A), the relative
acceleration equation becomes
aB = aA + α × rB/A + ω × (ω × rB/A)
Or…if ω = ωk
aB = aA + α × rB/A − ω2 rB/A
Note that the last term in the relative acceleration equation is
not a cross product. It is the product of a scalar (square of
the magnitude of angular velocity, ω2) and the relative
position vector, rB/A.
APPLICATION OF THE RELATIVE ACCELERATION
EQUATION
In applying the relative acceleration equation, the two points used in the
analysis (A and B) should generally be selected as points which have a
known motion, such as pin connections with other bodies.
In this mechanism, point B is known to travel along a circular path, so
aB can be expressed in terms of its normal and tangential components.
Note that point B on link BC will have the same acceleration as point B
on link AB.
Point C, connecting link BC and the piston, moves along a straight-line
path. Hence, aC is directed horizontally.
PROCEDURE FOR ANALYSIS
1. Establish a fixed coordinate system.
2. Draw the kinematic diagram of the body.
3. Indicate on it aA, aB, ω, α, and rB/A. If the points A and B
move along curved paths, then their accelerations should
be indicated in terms of their tangential and normal
components, i.e., aA = (aA)t + (aA)n and aB = (aB)t + (aB)n.
4. Apply the relative acceleration equation:
aB = aA + α × rB/A + ω × (ω × rB/A) = aA + α × rB/A − ω2 rB/A (if ω = ωk)
5. If the solution yields a negative answer for an unknown
magnitude, this indicates that the sense of direction of the
vector is opposite to that shown on the diagram.
EXAMPLE I
Given: Point A on rod AB has an
acceleration of 5 m/s2 and a
velocity of 6 m/s at the instant
shown.
Find: The angular acceleration of
the rod and the acceleration at
B at this instant.
Plan: Follow the problem solving
procedure.
Solution:
Apply the relative acceleration equation
aB = aA + α × rB/A – ω2rB/A
What do we know?
EXAMPLE I
Since points A and B both move along straight-line paths,
aA = -5 j m/s2
aB = aB i m/s2
Also:
α and ω are ccw.
EXAMPLE I
IC
Next, we need to find the angular velocity of the rod at this
instant. Locating the instant center (IC) for rod AB, we can
determine ω: ω = vA/rA/IC = vA / (3) = 2 rad/s
EXAMPLE I (continued)
Applying the relative acceleration equation
aB = aA + α × rB/A – ω2rB/A
aB i = - 5 j + α k × (3 i – 4 j) – 22 (3 i – 4 j)
aB i = - 5 j + 4 α i + 3 α j – (12 i – 16 j)
EXAMPLE I (continued)
Applying the relative acceleration equation
aB = aA + α × rB/A – ω2rB/A
aB i = - 5 j + α k × (3 i – 4 j) – 22 (3 i – 4 j)
aB i = - 5 j + 4 α i + 3 α j – (12 i – 16 j)
EXAMPLE I (continued)
So with aB i = - 5 j + 4 α i + 3 α j – (12 i – 16 j) , we can solve
for aB and α.
By comparing the i, j components;
aB = 4 α – 12
0 = 11 + 3α; aB= -26.7; α = −3.67
Solving:
aB = 26.7 m/s2
α = 3.67 rad/s2
BODIES IN CONTACT
Consider two rotating bodies in contact with one another without
slipping, where the points in contact move along different circular
paths.
The velocity and the tangential components of acceleration of the
points will be the same, i. e.,
vA = vA’ ; (aA)t = (aA’)t (which implies αB rB = αC rC ).
The normal components of acceleration will not be the same.
(aA)n ≠ (aA’)n so aA ≠ aA’
ROLLING MOTION
Another common type of problem encountered in dynamics
involves rolling motion without slip; e.g., a ball, cylinder, or disk
rolling without slipping. This situation can be analyzed using
relative velocity and acceleration equations.
As the cylinder rolls, point G (center) moves along a straight line.
If ω and α are known, the relative velocity and acceleration
equations can be applied to A, at the instant A is in contact with
the ground. The point A is the instantaneous center of zero
velocity, however it is not a point of zero acceleration.
• Velocity:
ROLLING MOTION
(continued)
Since no slip occurs, vA = 0 when A is in contact
with ground. From the kinematic diagram:
vG = vA + ω × rG/A
vG i = 0 + (-ωk) × (r j)
vG = ωr or vG = ωr i
• Acceleration: Since G moves along a straight-line path, a is
G
horizontal and can be determined from the time
derivative of its speed. So aG = αr i
aA = aG + α × rA/G – ω2rA/G ⇒ aA= αr i + (-α k) × (-r j) – ω2(-r j)
Evaluating :
aA = ω2r j
EXAMPLE II
Given: The gear with a center
at O rolls on the fixed
rack.
Find: The acceleration of point
A at this instant.
Plan:
Follow the solution procedure.
Solution: Since the gear rolls on the fixed rack without
slipping, aO is directed to the right with a magnitude
of
aO = αr = (6 rad/s2)(0.3 m)=1.8 m/s2.
EXAMPLE II (continued)
So now with aO = 1.8 m/s2, we can apply the relative
acceleration equation between points O and A.
aA = aO + α × rA/O – ω2 rA/O
aA = 1.8i + (-6k)×(0.3j) –122 (0.3j)
= (3.6 i – 43.2j) m/s2
1.8 m/s2
y
x
Example 2.12
Crank AB is rotating with an angular velocity of
vAB = 5 rad>s
and
an
angular
acceleration
of
aAB = 6 rad>s2. Determine the angular acceleration of BC
and the acceleration of the slider block C at the instant shown.
0.5 m
C
0.3 m
SOLUTION
.
45°
B
vAB ⫽ 5 rad/s
aAB ⫽ 6 rad/s2
A
Example 2.13
As the cord unravels from the cylinder, the cylinder has an
angular acceleration of a = 4 rad>s2 and an angular velocity
of v = 2 rad>s at the instant shown. Determine the
accelerations of points A and B at this instant.
A
α = 4 rad/s2
ω = 2 rad/s
C
B
SOLUTION
0.75 ft
SECTION 3
Planar Kinetics of
a Rigid Body
Force and
Acceleration
MASS MOMENT OF INERTIA
To determine the mass moment of inertia of a rigid body or a
system of rigid bodies.
Mass moment of inertia is a measure of the resistance of a
body to angular acceleration.
Mass moment of inertia is always a positive quantity.
MASS MOMENT OF INERTIA
Consider a rigid body with a center of
mass at G. It is free to rotate about the z
axis, which passes through G. Now, if we
apply a torque T about the z axis to the
body, the body begins to rotate with an
angular acceleration of α.
We have seen in our discussion that the body resists rotational
motion and hence α via the mass moment of inertia (MMI) I
about the z axis. Later we shall see that T and α are related by
the equation T = I α .
MASS MOMENT OF INERTIA
(continued)
Consider a rigid body and the arbitrary axis
P shown in the figure. The MMI about the
P axis is defined as IP = ∫m r2 dm, where r,
the “moment arm,” is the perpendicular
distance from the axis to the arbitrary
element dm and m is the mass of the body
i.e. integration takes place over the
ENTIRE RIGID BODY.
MASS MOMENT OF INERTIA
(continued)
If we write dm = ρdV where ρ is the
density at dm and dV is the volume of the
element dm then The MMI about the P axis
can then also be written as
IP = ∫V r2 ρdV ,
Where we now integrate over the entire
VOLUME of the rigid body.
The mass moment of inertia is always a
positive quantity and has a unit of
(SI) kg ·m2 or (FPS) slug · ft2.
MASS MOMENT OF INERTIA
(continued)
Note that it makes no sense to talk about
‘moment of inertia’ unless you specify an
axis: the same rigid body can have
different moments of inertia about
different axes.
Most of the time you can look up moments of
inertia for different bodies about different axes from
tables (see back cover of Hibbeler) but in some
cases it may be necessary to calculate from the
fundamental integral definition above. However, the
fundamental definition may lead to double or even
triple integration. Are there any shortcuts?
MASS MOMENT OF INERTIA
(continued)
The figures below show the mass moment of inertia formulations for two
shapes commonly used when working with three dimensional bodies. These
shapes are often used as the differential element being integrated over an
entire body. Note the dependency on the axis (we calculated Izz (about zaxis) for the uniform disk previously using the fundamental integral
definition when the disk thickness is taken to be h = constant)
PROCEDURE FOR ANALYSIS
The above result for the thin uniform disk can be used as the basis for
finding the moment of inertia for other more complicated geometries
from the fundamental integral definition:.
Disk element
• If a disk element having a radius y and a thickness dz is
chosen for integration, then the volume element can be
taken as dV = (πy2)dz.
• Using the moment of inertia of the disk element, we
can integrate over z to determine the moment of inertia
of the entire body.
EXAMPLE I
Given: The volume shown with
ρ = 5 slug/ft3.
Find: The mass moment of inertia
of this body about the yaxis.
Plan: Find the mass moment of
inertia of a disk element about
the y-axis, dIy, and integrate.
EXAMPLE I (continued)
Solution:
The moment of inertia of a disk about
an axis perpendicular to its plane is
I = ½ m r2.
Thus, for the disk element, we have
dIy = ½ (dm) x2 ;
Iy = ∫dIy
where the differential mass
dm = ρ dV = ρπx2 dy.
ρπx4
π(5)
ρπ 8
=
=
=
Iy= ∫dIy
∫ 2 dy 2 ∫ y dy 18 = 0.873 slug•ft2
0
0
1
1
PROCEDURE FOR ANALYSIS
We can do the same with a shell element
Shell element
• If a shell element having a height z, radius r = y, and
thickness dy is chosen for integration, then the volume
element is dV = (2πy)(z)dy.
• This element may be used to find the moment of inertia
Iz since the entire element, due to its thinness, lies at the
same perpendicular distance y from the z-axis.
See the following table for the moments
of inertia of common geometries….but remember to pick
the correct M of I for the correct axis!
PARALLEL-AXIS THEOREM
If the mass moment of inertia of a body about an axis passing
through the body’s mass center is known, then the moment of
inertia about any other parallel axis may be determined by using
the parallel axis theorem,
I = IG + md2
where IG = mass moment of inertia about the body’s mass center
m = mass of the body
d = perpendicular distance between the parallel axes
PARALLEL-AXIS THEOREM
Caution:
I ≠ IB + md2
If B is another arbitrary point NOT EQUAL TO G!!!
The formula works only when B ≡ G !!!
RADIUS OF GYRATION AND COMPOSITE BODIES
Radius of Gyration
The mass moment of inertia of a body about a specific axis can be
defined using the radius of gyration (k). The radius of gyration has
units of length and is a measure of the distribution of the body’s
mass about the axis at which the moment of inertia is defined.
I = m k2 or k = (I/m)
The radius of gyration is a defined geometrical quantity which, if
measured experimentally, can be useful for the determination of
M of I for non-uniform bodies or those of irregular shape.
Composite Bodies
If a body is constructed of a number of simple shapes,
such as disks, spheres, or rods, the mass moment of
inertia of the body about any axis can be determined by
algebraically adding together all the mass moments of
inertia, found about the same axis, of the different
shapes.
EXAMPLE II
q
r
Given: The pendulum consists of a
slender rod with a mass 10 kg
and sphere with a mass of 15 kg.
Find: The pendulum’s MMI about an
p
axis perpendicular to the screen
and passing through point O.
Plan: Follow steps similar to finding
the MoI for a composite area (as
done in statics). The pendulum
can be divided into a slender rod
(r) and sphere (s).
EXAMPLE II (continued)
Solution:
1. The center of mass for rod is at point Gr, 0.225m
from Point O. The center of mass for sphere is at
Gs, 0.55 m from point O.
2. The MMI data for a slender rod and sphere are
given on the inside back cover of the textbook.
Using those data and the parallel-axis theorem,
calculate the following.
IO = IG + (m) (d) 2
IOr = (1/12) (10)(0.45)2 +10 (0.225)2 = 0.675 kg·m2
IOs = (2/5) (15) (0.1)2 + 15 (0.55)2 = 4.598 kg·m2
3. Now add the two MMIs about point O.
IO = IOr + IOs = 5.27 kg·m2
Example 3.1
The slender rods have a weight of 3 lb>ft. Determine the
moment of inertia of the assembly about an axis
perpendicular to the page and passing through point A.
A
2 ft
1 ft
SOLUTION
1.5 ft
1.5 ft
•
(JL
The slender rods have II weight or 3 lb/rt. Determine the
moment of inertia or the assembly about on 11xis
perpendicular to the page and passing through point A.
.rP\
--= J- [( 3 )3 J Cs{+
l '2,,
'3 '2 ,
i.,
(
-z.{
3 2.?.
3 ( 1)
vV\
-=- �") C9
•
cA 2-
Example 3.2
Determine the moment of inertia about an axis perpendicular
to the page and passing through the pin at O. The thin plate
has a hole in its center. Its thickness is 50 mm, and the
material has a density r = 50 kg>m3.
O
150 mm
SOLUTION
1.40 m
1.40 m
&,yN\�L� '.S · '2._
•
Determine the moment or inertia about an axis perpendicular
to the page and passing through the pin at 0. The thin plate
has a hole in its center. Its thickness is 50 mm and the
material has a density p = 50 kg/ml .
,('
150 mm0
�
f').
'/
'�\)�.<Om
So�
rU\'li't • (IG)1 = Izz = (i Co.. 1-� 1o1.)
W\
•
:::: 0 [
t;o (l·4)(H•Xo·O� (Cl·ll\\(l·U�
M- ":_ \) �\ �
(1- G-) �� �
W1 -
Cle-\_::
,Awll):J
I· sq;t- kj •t'Vl t.
JL• -ro � �
d\
_L G-- \- V\A.-
M- ::: �0(1·4 '( \ -�rn.a;) - �O (a)(O ,1,f (O·D5)
:::: lt- t-2- 3 1 ':_:J
I
Example 3.2A
0.4 m
The pendulum consists of two slender rods AB and OC
which have a mass of 3 kg/m. The thin plate has a mass of
12 kg>m2. Determine the moment of inertia of the
pendulum about an axis perpendicular to the page and
passing through the pin at O.
0.4 m
A
B
O
–y
G
SOLUTION
C
0.1 m
0.3 m
1.5 m
•
Example 3.2A
The pendulum consists of two slender rods AB nnd OC
which hove o moss or 3 kg/m. The thin plnte hns a moss of
12 kg/m2• Determine the moment of inertia of the
pendulum about an oxis perpendiculnr to the page nnd
passing through the pin at O.
s�L�
•
RD"v f'"tln ·.
l\
\
l 2:,
M. A z_
I�
lJooJ- i- �v�k o)
c JioJ- t- �loJ
PLANAR KINETIC EQUATIONS OF MOTION
• We will limit our study of planar kinetics to rigid bodies which,
along with their loadings, are considered to be symmetric with
respect to a fixed reference plane.
Since the motion of the body can be viewed within the reference
plane, all the forces (and couple moments) acting on the body
can then be projected onto the plane. This means that the
THREE-DIMENSIONAL problem can then be viewed as a
TWO-DIMENSIONAL problem in the reference plane!
PLANAR KINETIC EQUATIONS OF MOTION
The more general case of body shape and loading involves the
study of 3-D motion (not planar motion) of rigid bodies
(NOT IN THIS COURSE)
As discussed in previous lectures on kinematics, when a body
is subjected to general plane motion, it undergoes a combination
of translation and rotation. We expect therefore that the planar
kinetic equations of motion will involve BOTH translational and
rotational equations
PLANAR KINETIC EQUATIONS OF MOTION
To try and see what these equations look like, we draw on what
we know from particle dynamics.
• First, an inertial frame of
reference xyz has its origin
coincident with the arbitrary
point P in the (reference plane
of the) rigid body. By
definition, these axes do not
rotate and are either fixed or
translate with constant velocity.
EQUATIONS OF TRANSLATIONAL MOTION
Recall that for a system of particles
ΣFi = Σ mi ai,
where the sum is over all particles and the LHS represents the vector sum of all external forces
acting on all particles. This can also be written as (see, for example, Section 13.3 Hibbeler):
ΣFi = maG,
where G is the center of mass of the system of particles.
(mrG=Σ mi ri differentiate twice w.r.t. time)
Let’s denote ΣFi by FR .
EQUATIONS OF TRANSLATIONAL MOTION
The equation
FR = maG,
(*)
is a mathematical expression of the “principle of motion of the mass
center” of a system of particles.
It is identical in form to that for a single particle. This
correspondence indicates that a system of particles can be treated
as a single particle with the mass of the system concentrated at the
mass center G
PROVIDED:
The resultant FR of the external forces applied to the
particles of the system has a line of action that passes
through the mass center G of the system.
EQUATIONS OF TRANSLATIONAL MOTION
In general, however, the line of action of the resultant force FR
will not pass through the mass center of the system.
What then? We recall from our study of ‘statics’
(e.g. Hibbeler Section 4.7) that a general system of forces
acting on a body may be replaced by a resultant force applied at
a chosen point and a corresponding couple.
By replacing FR by its equivalent force-couple system with the
resultant force acting through the mass center, we may
then visualize the action of FR as a superposition of a translation
and a rotation
Keep this in mind….
EQUATIONS OF TRANSLATIONAL MOTION
Now…back to RIGID BODIES.
So…if we consider a rigid body as being equivalent to a system of
particles, given that the line of action of the resultant FR will not, in
general, pass through the mass center of the system we will replace
FR by an equivalent single resultant force (ΣF ) acting at the mass
center and a resultant couple moment.
Hence, there will be two parts to the description of its motion:
1. A translation caused by ΣF (resultant force applied at mass
center).
2. A rotation caused by the resultant couple moment.
EQUATIONS OF TRANSLATIONAL MOTION
Now…back to RIGID BODIES.
So…if we consider a rigid body as being equivalent to a system
of particles.
Then, for the rigid body we can write:
ΣF = maG
(Vector Form)
or
Σ Fx = m(aG)x
and
Σ Fy = m(aG)y
(Scalar form):
where ∑F represents the resultant force applied at the mass center i.e. the
vector sum of all external forces acting on the rigid body APPLIED AT ITS
MASS CENTER.
THIS IS THE TRANSLATIONAL EQUATION(S) OF MOTION FOR
THE MASS CENTER OF A RIGID BODY
EQUATIONS OF ROTATIONAL MOTION
Next, we need to determine the effects caused by the accompanying
couple moment.
To do this, we examine the effects of the moments of a general external
force system.
EQUATIONS OF ROTATIONAL MOTION
Σ Mp = rG/p × maG + IGα
OR Σ Mp = Σ(Mk)p
EQUATIONS OF ROTATIONAL MOTION
Here Σ Mp is the resultant moment about P due to all the
external forces.
The term Σ(Mk)p is called the kinetic moment about point P
and is the moment generated about P by the acceleration of
the mass center and the contribution to moment of the body’s
angular acceleration.
EQUATIONS OF ROTATIONAL MOTION (continued)
Recall that the rotational equation is a vector equation but since in
planar motion rotation takes place only about one axis, this vector
equation is easily written as a scalar equation with the direction
of rotation specified separately.
Σ Mp = Σ (Mk)p
(CCW+)
EQUATIONS OF ROTATIONAL MOTION (continued)
If point P coincides with the mass center G, this equation reduces
to the scalar equation of ΣMG = IG α . Why?
In words: the resultant (summation) moment about the mass
center due to all the external forces is equal to the moment of
inertia about G times the angular acceleration of the body.
Thus, three independent scalar equations of motion may be used
to describe the general planar motion of a rigid body. These
equations are:
Σ Fx = m(aG)x
Σ Fy = m(aG)y
and Σ MG = IGα or Σ Mp = Σ (Mk)p
EQUATIONS OF ROTATIONAL MOTION (continued)
When applying these equations DRAW A FREE-BODY
DIAGRAM to identify all forces and moments
Σ Fx = m(aG)x
Σ Fy = m(aG)y
and Σ MG = IGα or Σ Mp = Σ (Mk)p
EQUATIONS OF MOTION: TRANSLATION
When a rigid body undergoes only translation, all the particles of
the body have the same acceleration so aG = a and α = 0.
The equations of motion become:
Σ Fx = m(aG)x
Σ Fy = m(aG)y
Σ MG = 0
Note that, if it makes the problem easier, the moment equation
can be applied about another point instead of the mass center.
For example, if point A is chosen,
ΣMA = (m aG) d .
EQUATIONS OF MOTION: TRANSLATION
(continued)
When a rigid body is subjected to
curvilinear translation, it is best to use
an n-t coordinate system.
Then apply the equations of motion, as
written below, for n-t coordinates.
Σ Fn = m(aG)n
Σ Ft = m(aG)t
Σ MG = 0 or
Σ MB = e[m(aG)t] – h[m(aG)n]
PROCEDURE FOR ANALYSIS
Problems involving kinetics of a rigid body in only translation
should be solved using the following procedure:
1. Establish an (x-y) or (n-t) inertial coordinate system and specify
the sense and direction of acceleration of the mass center, aG.
2. Draw a FBD and kinetic diagram showing all external forces,
couples and the inertia forces and couples.
3. Identify the unknowns.
4. Apply the three equations of motion (one set or the other):
Σ Fx = m(aG)x
Σ Fy = m(aG)y
Σ Fn = m(aG)n Σ Ft = m(aG)t
Σ MG = 0 or
Σ MP = Σ (Mk)P
Σ MG = 0 or Σ MP = Σ (Mk)P
5. Remember, friction forces always act on the body opposing the
motion of the body.
EXAMPLE
Given: The cart and its load have a
total mass of 100 kg and
center of mass at G. A force
of P = 100 N is applied to
the handle. Neglect the
mass of the wheels.
Find: The normal reactions at each of the two wheels at A and B.
Plan: Follow the procedure for analysis.
EXAMPLE (continued)
Solution: The cart will move along a rectilinear path.
Draw the FBD and kinetic diagram.
=
Apply the equation of motion in the x-direction first:
+
→ Σ Fx = m(aG)x
100 (4/5) = 100 aG
aG = 0.8 m/s2
EXAMPLE (continued)
Then apply the equation of motion in the y-direction and sum
moments about G.
=
+↑ Σ Fy = 0 ⇒ NA + NB –981 –100 (3/5) = 0
NA + NB = 1041 N
(1)
+ ΣMG = 0
⇒ NA(0.6) – NB(0.4) + 100(3/5) 0.7– 100(4/5)(1.2-0.5) = 0
0.6 NA − 0.4 NB = 14 N m (2)
Using Equations (1) and (2), solve for the reactions, NA and NB
NA = 430 N and NB = 611 N
Another Example
Given: The lift truck has a mass of
70 kg and mass center at G. It
lifts the 120-kg spool with an
acceleration of 3 m/s2. The
spool’s mass center is at C.
You can neglect the mass of
the movable arm CD.
Find: The normal reactions at each of the four wheels.
Plan: Follow the procedure for analysis.
Another Example (continued)
Solution: Draw FBD and kinetic diagram.
=
Applying the equations of motion:
+ Σ MB = Σ (Mk)B
⇒ 70(9.81)(0.5) + 120(9.81)(0.7) − 2 NA(1.25) = -120(3)(0.7)
NA = 568 N
Another Example (continued)
=
+↑ Σ Fy = m(aG)y
⇒ 2 NA + 2 NB –120(9.81) – 70 (9.81) = 120 (3)
2 NA + 2 NB = 2224 N
Since NA = 568 N, NB = 544 N
Example 3.3
6 ft
6 ft
A
The door has a weight of 200 lb and a center of gravity at G.
Determine how far the door moves in 2 s, starting from rest,
if a man pushes on it at C with a horizontal force F = 30 lb.
Also, find the vertical reactions at the rollers A and B.
B
G
C
12 ft
F
SOLUTION
3 ft
5 ft
Example3.3
j--6ft:--a..- 6fl --1
The door has a weight of 200 lb and u center of gravity at G.
Detcnninc how for the door moves in 2 s, starting from rest,
if a man pushes on it at C with a horizontal force F • 30 lb.
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EQUATIONS OF MOTION:
ROTATION ABOUT A FIXED AXIS
Analyze the planar kinetics of a rigid body
undergoing rotational motion.
EQUATIONS OF MOTION:
ROTATION ABOUT A FIXED AXIS
When a rigid body rotates about a fixed axis
perpendicular to the plane of the body at
point O, the body’s center of gravity G moves
in a circular path of radius rG. Thus, the
acceleration of point G can be represented by
a tangential component (aG)t = rG α and a
normal component (aG)n = rG ω2.
Since the body experiences an angular acceleration, its inertia
creates a moment of magnitude, IGα, balanced by the moment of
the external forces about point G. Thus, the scalar equations of
motion can be stated as:
∑ Fn = m (aG)n = m rG ω2
∑ Ft = m (aG)t = m rG α
∑ MG = I G α
EQUATIONS OF MOTION (continued)
Note that the ∑MG moment equation may be replaced by a
moment summation about any arbitrary point. Summing the
moment about the center of rotation O yields
∑MO = IGα + rG m (aG) t = [IG + m(rG)2] α
From the parallel axis theorem,
IO = IG + m(rG)2, therefore the term
in parentheses represents IO.
Consequently, we can write the three
equations of motion for the body as:
∑Fn = m (aG)n = m rG ω2
∑Ft = m (aG)t = m rG α
∑MO = IO α
PROCEDURE FOR ANALYSIS
Problems involving the kinetics of a rigid body rotating about
a fixed axis can be solved using the following process.
1. Establish an inertial coordinate system and specify the sign and
direction of (aG)n and (aG)t.
2. Draw a free body diagram accounting for all external forces
and couples. Show the resulting inertia forces and couple
(typically on a separate kinetic diagram).
3. Compute the mass moment of inertia IG or IO.
4. Write the three equations of motion and identify the
unknowns. Solve for the unknowns.
5. Use kinematics if there are more than three unknowns (since
the equations of motion allow for only three unknowns).
EXAMPLE
Given:The uniform slender rod has a
mass of 15 kg and its mass
center is at point G.
•G
Find: The reactions at the pin O and
the angular acceleration of the
rod just after the cord is cut.
Plan: Since the mass center, G, moves in a circle of radius
0.15 m, it’s acceleration has a normal component toward
O and a tangential component acting downward and
perpendicular to rG.
Apply the problem solving procedure.
EXAMPLE (continued)
Solution:
FBD
&
Kinetic Diagram
rG
=
Equations of motion:
+ ∑Fn = man = mrGω2
⇒ Ox = 0 N
+ ∑Ft = mat = mrGα
⇒ -Oy + 15(9.81) = 15(0.15)α (1)
+ ∑MO = IG α + m rG α (rG) ⇒ (0.15) 15(9.81)= IG α + m(rG)2 α
Using IG = (ml2)/12 and rG = (0.15), we can write:
IG α + m(rG)2 α = [(15×0.92)/12 + 15(0.15)2] α = 1.35 α
EXAMPLE (continued)
FBD
&
Kinetic Diagram
rG
=
After substituting:
22.07 = 1.35 α ⇒ α = 16.4 rad/s2
From Eq (1) :
-Oy + 15(9.81) = 15(0.15)α
⇒ Oy = 15(9.81) − 15(0.15)16.4 = 110 Ν
Example 3.8
Cable is unwound from a spool supported on small rollers
at A and B by exerting a force of T = 300 N on the cable in
the direction shown. Compute the time needed to unravel
5 m of cable from the spool if the spool and cable have a
total mass of 600 kg and a centroidal radius of gyration of
kO = 1.2 m. For the calculation, neglect the mass of the
cable being unwound and the mass of the rollers at A and B.
The rollers turn with no friction.
T
300 N
1.5 m
30
0.8 m
O
A
B
SOLUTION
1m
N
Example 3.9
0.6 m
s
29
.4
Determine the vertical and horizontal components of
reaction at the pin support A and the angular acceleration
of the 12-kg rod at the instant shown, when the rod has an
angular velocity of v = 5 rad>s.
SOLUTION
A
v ⫽ 5 rad/s
B
EQUATIONS OF MOTION:
GENERAL PLANE MOTION
Analyze the planar kinetics of a rigid
body undergoing general plane
motion.
GENERAL PLANE MOTION
When a rigid body is subjected to
external forces and couple
moments, it can undergo both
translational motion as well as
rotational motion. This
combination is called general
plane motion.
Using an x-y inertial coordinate
system, the equations of motions about
the center of mass, G, may be written
as
∑ Fx = m (aG)x
P
∑ Fy = m (aG)y
∑ MG = IG α
GENERAL PLANE MOTION
(continued)
Sometimes, it may be convenient to write
the moment equation about some point P
other than G. Then the equations of
motion are written as follows.
∑ Fx = m (aG)x
∑ Fy = m (aG)y
∑ MP = ∑ (Mk )P
P
In this case, ∑ (Mk )P represents the sum of the
moments of IGα
and
maG about
point P.
FRICTIONAL ROLLING PROBLEMS
When analyzing the rolling motion of wheels, cylinders, or disks,
it may not be known if the body rolls without slipping or if it
slides as it rolls.
For example, consider a disk with mass m
and radius r, subjected to a known force P.
The equations of motion will be
∑ Fx = m(aG)x => P - F = maG
∑ Fy = m(aG)y => N - mg = 0
∑ MG = IGα
=> F r = IGα
There are 4 unknowns (F, N, α and aG) in
these three equations.
FRICTIONAL ROLLING PROBLEMS
(continued)
Hence, we have to make an assumption
to provide another equation. Then we
can solve for the unknowns.
The 4th equation can be obtained from
the slip or non-slip condition of the disk.
Case 1(see Lecture 9):
Assume no slipping and use aG = rα as the 4th equation and
DO NOT use Ff = µkN. After solving, you will need to verify
that the assumption was correct by checking if Ff ≤ µ sN.
Case 2:
Assume slipping and use Ff = µkN as the 4th equation. In
this case, aG ≠ rα
PROCEDURE FOR ANALYSIS
Problems involving the kinetics of a rigid body undergoing
general plane motion can be solved using the following procedure.
1. Establish the x-y inertial coordinate system. Draw both the
free body diagram and kinetic diagram for the body.
2. Specify the direction and sense of the acceleration of the
mass center, aG, and the angular acceleration α of the
body. If necessary, compute the body’s mass moment of
inertia IG.
3. If the moment equation ΣMp= Σ(Mk)p is used, use the
kinetic diagram to help visualize the moments developed by
the components m(aG)x, m(aG)y, and IGα .
4. Apply the three equations of motion.
PROCEDURE FOR ANALYSIS
(continued)
5. Identify the unknowns. If necessary (i.e., there are four
unknowns), make your slip-no slip assumption (typically no
slipping, or the use of aG = rα, is assumed first).
6. Use kinematic equations as necessary to complete the
solution.
7. If a slip-no slip assumption was made, check its validity!!!
Key points to consider:
1. Be consistent in assumed directions. The direction of aG
must be consistent with α .
2. If Ff = µkN is used, Ff must oppose the motion. As a test,
assume no friction and observe the resulting motion. This
may help visualize the correct direction of Ff.
EXAMPLE 1
Given: A spool has a mass of 8 kg and a radius of gyration (kG)
of 0.35 m. Cords of negligible mass are wrapped around
its inner hub and outer rim. There is no slipping.
Find: The angular acceleration (α ) of the spool.
Plan: Focus on the spool. Follow the solution procedure (draw
a FBD, etc.) and identify the unknowns.
Solution:
EXAMPLE 1
(continued)
The moment of inertia of the spool is
IG = m (kG)2 = 8 (0.35)2 = 0.980 kg·m 2
Method I (Take CW and “up” +ve)
Equations of motion:
∑ Fy = m (aG)y
T + 100 -78.48 = 8 aG
∑MG = IG α
100 (0.2) – T(0.5) = 0.98 α
There are 3 unknowns, T, aG, α.
We need one more equation
to solve for 3 unknowns. Since the spool rolls on the cord at point
A without slipping, aG = rα So the third equation is aG = 0.5α
Solving these three equations, we find:
α =10.3 rad/s2, aG = 5.16 m/s2, T = 19.8 N
EXAMPLE 1
(continued)
Method II
Now, instead of using a moment
equation about G, a moment equation
about A will be used. This approach will
eliminate the unknown cord tension (T).
Σ MA= Σ (Mk)A: 100 (0.7) - 78.48(0.5) = 0.98 α
+ (8 aG)(0.5)
Using the non-slipping condition again yields aG = 0.5α .
Solving these two equations, we get
α = 10.3 rad/s2, aG = 5.16 m/s2
Example 2
Given: A 50 lb wheel has a radius
of gyration kG = 0.7 ft.
Find: The acceleration of the
mass center if M = 35 lb-ft
is applied. µs = 0.3, µk =
0.25.
Plan: Follow the problem solving procedure.
Solution:
The moment of inertia of the wheel about G is
IG = m(kG)2 = (50/32.2)(0.7)2 = 0.761 slug·ft2
Example 2
(continued)
FBD:(Take positive as up, to the right and CW)
Equations of motion:
∑ Fx = m(aG)x
FA = (50/32.2) aG
∑ Fy = m(aG)y
NA – 50 = 0
∑ MG = IGα
35 - 1.25 FA= 0.761 α
We have 4 unknowns: NA, FA,
aG and α .
Another equation is needed before solving for the unknowns.
Example 2
(continued)
The three equations we have now are:
FA = (50/32.2) aG
NA - 50 = 0
35 – (1.25)FA = 0.761 α
First, assume the wheel is not slipping.
Thus, we can write
aG = r α = 1.25 α
Now solving the four equations yields:
NA = 50.0 lb, FA = 21.3 lb, α
= 11.0 rad/s2 , aG = 13.7 ft/s2
The no-slip assumption must be checked.
Is FA = 21.3 lb ≤ µs NA= 15 lb?
No! Therefore, the wheel slips as it rolls.
Example 2
(continued)
Therefore, aG = 1.25α
is not a
correct assumption. The slip
condition must be used.
Again, the three equations of
motion are:
FA = (50/32.2) aG
NA – 50 = 0
35 – (1.25)FA = 0.761 α
Since the wheel is slipping, the fourth equation is
FA = µk NA = 0.25 NA
Solving for the unknowns yields:
NA = 50.0 lb
FA = 0.25 NA = 12.5 lb
aG = 8.05 ft/s2
α = 25.5
rad/s2
Example 3.13
The uniform 150-lb beam is initially at rest when the forces
are applied to the cables. Determine the magnitude of the
acceleration of the mass center and the angular acceleration
of the beam at this instant.
FA ⫽ 100 lb
FB ⫽ 200 lb
A
B
12 ft
SOLUTION
60⬚
Example 3.14
The spool has a mass of 100 kg and a radius of gyration of
kG = 0.3 m. If the coefficients of static and kinetic friction
at A are ms = 0.2 and mk = 0.15, respectively, determine the
angular acceleration of the spool if P = 50 N.
P
250 mm
G
A
SOLUTION
400 mm
Example 3.15
A uniform rod having a weight of 10 lb is pin supported at A
from a roller which rides on a horizontal track. If the rod is
originally at rest, and a horizontal force of F = 15 lb is
applied to the roller, determine the acceleration of the
roller. Neglect the mass of the roller and its size d in the
computations.
SOLUTION
d
A
F
2 ft
SECTION 4
PLANAR KINETICS
OF A RIGID BODY
WORK AND ENERGY
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Example 4.1
The double pulley consists of two parts that are attached to
one another. It has a weight of 50 lb and a radius of gyration
about its center of kO = 0.6 ft. If it rotates with an angular
velocity of 20 rad>s clockwise, determine the kinetic energy
of the system. Assume that neither cable slips on the pulley.
v
20 rad/s
0.5ft
1 ft
O
SOLUTION
B 30 lb
A 20 lb
Example 4.2
At the instant shown, link AB has an angular velocity
vAB = 2 rad>s. If each link is considered as a uniform
slender bar with a weight of 0.5 lb>in., determine the total
kinetic energy of the system.
vAB
A
2 rad/s
3 in.
4 in.
SOLUTION
C
B
5 in.
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Example: Work of forces and couple moment.
The bar has a mass of 10 kg and is subjected to a couple moment of magnitude M
::: SON.m and a force of magnitude P::: 80N, which is always applied
perpendicular to the end of the bar. Also, the spring has an unstretched length of
0.5 m and remains in the vertical position due to the roller guide at B. Determine
the total work done by all forces and the couple moment acting on the bar when it
has rotated downward from e::: o 0 to e::: 90°
Planar Kinetics of a Rigid Body:
Work and Energy
1. Define the various ways a
force and couple do work.
2. Apply the principle of work
and energy to a rigid body.
3. To show how conservation of
energy can be used to solve
rigid-body planar kinetic
problems
PRINCIPLE OF WORK AND ENERGY
Recall the statement of the principle of work and energy
used earlier:
T1 + ΣU1-2 = T2
In the case of general plane motion, this equation states
that the sum of the initial kinetic energy (both
translational and rotational) and the work done by all
external forces and couple moments equals the body’s
final kinetic energy (translational and rotational).
This equation is a scalar equation. It can be applied
to a system of rigid bodies by summing contributions
from all bodies.
EXAMPLE: Rolling Wheel
Given: The 50 kg wheel is subjected to
a force of 50 N. The radius of
gyration of the wheel about its
mass center O is kO = 0.3 m.
Find:
The angular velocity of the wheel after it has rotated
10 revolutions. The wheel starts from rest and rolls without
slipping.
Plan: Use the principle of work and energy to solve the problem
since distance is the primary parameter. Draw a free body
diagram of the wheel and calculate the work of the external
forces.
EXAMPLE: Rolling Wheel (continued)
Solution:
Free body diagram of the wheel:
Since the wheel rolls without slipping on
a horizontal surface, only the force P's
horizontal component does work.
Why don't forces Pv, Ff and N do any
work?
N
EXAMPLE: Rolling Wheel (continued)
Work:
U1-2 = ∫ F•dr = ∫ (P cos 30) ds.
U1-2 = (50 cos 30) 10(π 0.8) = 1088 N·m
Kinematic relation: vO = r ω = 0.4 ω
Kinetic energy:
T1 = 0
T2 = ½ m (vO)2 + ½ IO ω2
= ½(50) (0.4 ω)2 + ½ (50) (0.3)2 ω2
T2 = 6.25 ω2
Work and energy:
T1 + U1-2 = T2
0 + 1088 = 6.25 ω2
ω = 13.2 rad/s (-k)
N
Example:Crate on Platform
Given: The combined weight of
the load and the platform is
200 lb, with the center of
gravity located at G. A
couple moment is applied
to link AB. The system is at
rest when θ = 0. Neglect
the weight of the links.
Find: The angular velocity of
links AB & CD at θ = 60.
Plan: Since the problem involves distance, the principle of
work and energy is an efficient solution method.
Example: Crate on Platform (continued)
Solution: Work done by the external loads on the system
Calculate the vertical distance the mass center moves.
4 sinθ
From θ = 0° to θ = 60° the mass
center G rises :
∆y = 4 sin θ where θ = 60°
Then, determine the work due to the
weight.
Uw = -W ∆y = - W (4 sin θ)
Uw = - 200 (4 sin 60)
= - 692.8 ft·lb
Work due to the moment (careful of θ’s units!):
UM = M⋅θ = 900 (60°) (π/180) = 942.5 ft·lb
(What about pin-reactions at B and D?)
Therefore, U1-2 = - 692.8 + 942.5 = 249.7 ft·lb
Example: Crate on Platform (continued)
Kinetic energy:
T1 = 0
T2 = ½ m (vG)2
= ½ (200/32.2) (vG)2
= 3.1056(vG)2 ft·lb
Now apply the principle of work and energy
equation:
T1 + U1-2 = T2
0 + 249.7 = 3.1056(vG)2
vG = 8.966ft/s
Example: Crate on Platform (continued)
As links AB and CD rotate, the platform will be subjected to
only curvilinear translational motion with a speed of 4ω.
Now apply the principle of work and energy
equation:
(vG) = 8.966 = 4 ω
ω = 2.24 rad/s (-k)
EXAMPLE 4.1
Given: The disk weighs 40 lb and has a
radius of gyration (kG) of 0.6 ft. A
15 ft·lb moment is applied and the
spring has a spring constant of 10
lb/ft.
Find: The angular velocity of the wheel when point G moves
0.5 ft. The wheel starts from rest and rolls without
slipping. The spring is initially unstretched.
Plan:
Use the principle of work and energy since distance
is the primary parameter. Draw a free body diagram
of the disk and calculate the work of the external
forces.
EXAMPLE 4.1 (continued)
Solution:
Free body diagram of the disk:
Since the body rolls without
slipping on a horizontal surface,
only the spring force and couple
moment M do work. Why don’t
forces FB and NB do work?
Since the spring is attached to
the top of the wheel, it will
stretch twice the amount of
displacement of G, or 1 ft.
EXAMPLE 4.1 (continued)
Work: U1-2 = -½k[(s2)2 – (s1)2] + M(θ2 – θ1)
U1-2 = -½(10)(12 – 0) + 15(0.5/0.8) = 4.375 ft·lb
Kinematic relation: vG = r ω = 0.8ω
Kinetic energy: T1 = 0
T2 = ½ m (vG)2 + ½ IG ω2
T2 = ½(40/32.2)(0.8ω)2 + ½(40/32.2)(0.6)2ω2
T2 = 0.621 ω2
Work and energy: T1 + U1-2 = T2
0 + 4.375 = 0.621 ω2
ω = 2.65 rad/s (-k)
Example 4.2
Given: A sphere weighing 10 lb rolls
along a semicircular hoop. Its
ω equals 0 when θ = 0.
Find The angular velocity of the sphere when θ = 45° if
the sphere rolls without slipping.
:
Plan Since the problem involves distance, the principle of
:
work and energy is a good solution method.The
only force involved that does work is the weight, so
only its work need be determined.
••
•
Example 4.2 ( continued)
Solution: Draw a FBD
and calculate the vertical
distance the mass center
moves.
(R + r)(l -cos9)
Now calculate the work due to the weight:
U 1 _2 = W (R + r) (1 - cos 0)
= 10 (10 + 0.5) (1 - cos 45° )
= 30.75 ft·lb
-
Example 4.2 (continued)
A kinematic equation for finding the velocity of the
mass center is needed. It is vG = r ω = 0.5 ω
Kinetic energy:
T1 = 0
T2 = 0.5m(vG)2 + 0.5IGω2
= 0.5(10/32.2)(0.5ω)2 + 0.5(0.4)(10/32.2)(0.5)2ω2
= 0.054 ω2
Now apply the principle of work and energy equation:
T1 + U1-2 = T2
0 + 30.75 ft·lb = 0.054 ω2
ω = 23.8 rad/s (-k)
Example 4.3
The double pulley consists of two parts that are attached to
one another. It has a weight of 50 lb and a centroidal radius
of gyration of kO = 0.6 ft and is turning with an angular
velocity of 20 rad> s clockwise. Determine the angular
velocity of the pulley at the instant the 20-lb weight moves
2 ft downward.
v ⫽ 20 rad/s
0.5 ft
1 ft
O
SOLUTION
B 30 lb
A 20 lb
Example 4.4
The wheel is made from a 5-kg thin ring and two 2-kg
slender rods. If the torsional spring attached to the wheel’s
center has a stiffness k = 2 N # m>rad, and the wheel is
rotated until the torque M = 25 N # m is developed,
determine the maximum angular velocity of the wheel if it
is released from rest.
SOLUTION
0.5 m
O
M
Example 4.5
A ball of mass m and radius r is cast onto the horizontal
surface such that it rolls without slipping. Determine its
angular velocity at the instant u = 90°, if it has an initial
speed of vG as shown.
SOLUTION
u R
r
G
vG
PLANAR KINETICS OF A RIGID BODY:
CONSERVATION OF ENERGY
(a) Determine the potential energy of conservative
forces.
(a) Apply the principle of conservation of energy.
APPLICATIONS
The torsional spring located at the top
of the garage door winds up as the
door is lowered. When the door is
raised, the potential energy stored in
the spring is transferred into the
gravitational potential energy of the
door’s weight, thereby making it easy
to open.
CONSERVATION OF ENERGY
The conservation of energy theorem is a “simpler” energy
method (recall that the principle of work and energy is also an
energy method) for solving problems. Once again, the problem
parameter of distance is a key indicator of when conservation of
energy is a good method for solving the problem.
If it is appropriate, conservation of energy is easier to use than the
principle of work and energy. This is because the calculation of
the work of a conservative force is simpler. But, what makes a
force conservative?
CONSERVATIVE FORCES
A force F is conservative if the work done by the force is
independent of the path. In this case, the work depends only
on the initial and final positions of the object with the path
between positions of no consequence.
Typical conservative forces encountered in dynamics are
gravitational forces (i.e., weight) and elastic forces (i.e.,
those due to springs).
What is a common force that is not conservative?
CONSERVATION OF ENERGY
When a rigid body is acted upon by a system of conservative
forces, the work done by these forces is conserved. Thus,
the sum of kinetic energy and potential energy remains
constant. This principle is called conservation of energy and
is expressed as
T1 + V1 = T2 + V2 = Constant
In other words, as a rigid body moves from one position to
another when acted upon by only conservative forces, kinetic
energy is converted to potential energy and vice versa.
GRAVITATIONAL POTENTIAL ENERGY
The gravitational potential energy of an object is a
function of the height of the body’s center of gravity
above or below a datum.
The gravitational potential
energy of a body is found
by the equation
Vg = W yG
Gravitational potential energy is positive when yG is
positive, since the weight has the ability to do positive
work when the body is moved back to the datum.
ELASTIC POTENTIAL ENERGY
Spring forces are also conservative forces.
The potential energy of a
spring force (F = ks) is
found by the equation
Ve = ½ ks2
Notice that the elastic potential energy is always positive.
PROCEDURE FOR ANALYSIS
Problems involving velocity, displacement and conservative force
systems can be solved using the conservation of energy equation.
• Potential energy: Draw two diagrams: one with the body located
at its initial position and one at the final position. Compute the
potential energy at each position using
V = Vg+Ve, where Vg=W yG and Ve = 1/2 k s2.
• Kinetic energy: Compute the kinetic energy of the rigid body at
each location. Kinetic energy has two components:
translational kinetic energy (½ m(vG)2 ) and rotational kinetic
energy (½ IG ω2 ).
• Apply the conservation of energy equation.
EXAMPLE: Rod and Spring
Given: The rod AB has a mass of 10 kg.
Piston B is attached to a spring of
constant k = 800 N/m. The spring
is un-stretched when θ = 0°.
Neglect the mass of the pistons.
Find: The angular velocity ω2 of rod AB at
θ = 0° if the rod is released from rest
when θ = 30°.
Plan: Use the energy conservation equation since all forces
doing work are conservative and distance is a parameter
(represented here by θ).The potential energy and kinetic
energy of the rod at states 1 and 2 will have to be
determined.
EXAMPLE Rod and Spring (continued)
Solution:
Initial Position
Final Position
Potential Energy:
Let’s put the datum in line with the rod when θ = 0°.
Then, the gravitational potential energy and the elastic potential
energy will be zero at position 2. => V2 = 0
Gravitational potential energy at 1: - (10)( 9.81) ½ (0.4 sin 30)
Elastic potential energy at 1: ½ (800) (0.4 sin 30)2
So V1 = - 9.81 J + 16.0 J = 6.19 J
EXAMPLE: Rod and Spring (continued)
Initial Position
Final Position
Kinetic Energy:
The rod is released from rest from position 1 (so
vG1 = 0, ω1 = 0).
Therefore, T1 = 0.
At position 2, the rod is in GPM: its angular
velocity has magnitude ω2 and the speed of its
center of mass is vG2 .
EXAMPLE: Rod and Spring (continued)
Therefore,
T2 = ½ (10)(vG2)2 + ½ (1/12)(10)(0.42)(ω2)2
At position 2, point A is the instantaneous center
of rotation. Hence, vG2 = r ω = 0.2 ω2 .
Then, T2 = 0.2 ω22 + 0.067 ω22 = 0.267 ω22
Now apply the conservation of energy equation and solve for
the unknown angular velocity, ω2.
T1 + V1 = T2 + V2
0 + 6.19 = 0.267ω22 + 0
=> ω2 = 4.82 rad/s
ω2 = 4.82k rad/s
EXAMPLE: Rolling Wheel and Spring
Given: The weight of the disk is 30 lb and
its kG equals 0.6 ft. The spring has
a stiffness of 2 lb/ft and an
unstretched length of 1 ft.
Find: The velocity at the instant G
moves 3 ft to the left. The disk is
released from rest in the position
shown and rolls without slipping.
Plan: Since distance is a parameter and all forces doing work are
conservative, use conservation of energy. Determine the
potential energy and kinetic energy of the system at both
positions and apply the conservation of energy equation.
EXAMPLE: Rolling Wheel and Spring (continued)
Solution:
Potential Energy:
There are no changes in the gravitational
potential energy since the disk is moving
horizontally.
The elastic potential energy at position
1 is: V1 = ½ k (s1)2 where s1 = 4 ft.
Thus, V1 = ½ 2 (4)2 = 16 J
(1) Initial Position
(2) Final Position
Similarly, the elastic potential energy at position 2 is
V2 = ½ 2 (3)2 = 9 J
EXAMPLE: Rolling Wheel and Spring (continued)
Kinetic Energy:
The disk is released from rest at position 1, so vG1 = 0
and ω1 = 0. Thus, the kinetic energy at position 1 is
T1 = 0.
At position 2, the angular velocity has magnitude ω2
and the speed of the center of mass is vG2 .
T2 = ½ m (vG2)2 + ½ IG (ω2) 2
= ½ (30/32.2) (vG2)2 + ½ (30/32.2) 0.62 (ω2) 2
The disk is rolling without slipping, so vG2 = (0.75 ω2).
T2 = ½(30/32.2)(0.75 ω2)2 + ½(30/32.2) 0.62 (ω2)2 = 0.43 (ω2)2
EXAMPLE: Rolling Wheel and Spring (continued)
Now all terms in the conservation of energy equation have been
formulated. First, writing the general equation and then
substituting into it yields:
T1 + V1 = T2 + V2
0 +16.0 J = 0.43 ω22 + 9 J
Solving , ω2 = 4.04k rad/s
Example: Bar and Spring
Given: A 50 lb bar is rotating
downward at 2 rad/s.
The spring has an
unstretched length of 2
ft and a spring constant
of 12 lb/ft.
Find: The angle (measured down from the horizontal) to which
the bar rotates before it stops its initial downward
movement.
Plan: Conservative forces and distance (θ) leads to the use of
conservation of energy. First, determine the potential
energy and kinetic energy for both positions. Then
apply the conservation of energy equation.
Example: Bar and Spring (continued)
Solution:
Potential Energy:
Let’s put the datum in line with
the rod when θ = 0.
Then, at position 1, the
gravitational potential energy is
zero and the elastic potential
energy will be
1
θ 3 sin θ
V1 = ½ k (s1)2
6 sin θ
2
= ½ (12) (4 - 2)2
Gravitational potential energy at position 2: - (50) (3 sin θ)
Elastic potential energy at position 2: ½ (12) {4 + (6 sin θ) - 2}2
So, V2 = - (50) (3sin θ) + ½ (12) {4 + (6 sin θ) - 2}2
Example: Bar and Spring (continued)
Kinetic Energy:
At position 1 (when θ = 0), the
rod has a rotational motion
about point A.
T1 = ½ IA (ω2 )
= ½{1/3 (50/32.2) 62} (22 )
At position 2, the rod momentarily has no translation or rotation
since the rod comes to rest.
Therefore, T2 = 0.
Example: Bar and Spring (continued)
Now, substitute into the
conservation of energy
equation.
T1 + V1 = T2 + V2
½{1/3 (50/32.2) 62}( 22 ) + ½ (12) (4 - 2)2
= 0.0 - (50)(3 sin θ) + ½(12){4 + (6 sin θ) - 2}2
Solving for sin θ yields
sin θ = 0.4295.
Thus, θ = 25.4 deg.
Example 4.6
G
An automobile tire has a mass of 7 kg and radius of gyration
kG = 0.3 m. If it is released from rest at A on the incline,
determine its angular velocity when it reaches the horizontal
plane. The tire rolls without slipping.
0.4 m
A
5m
30°
0.4 m
B
SOLUTION
is
Example 4.7
The two bars are released from rest at the position u.
Determine their angular velocities at the instant they become
horizontal. Neglect the mass of the roller at C. Each bar has
a mass m and length L.
SOLUTION
B
L
A
L
C
SECTION 5
PLANAR KINETICS OF A
RIGID BODY
IMPULSE AND
MOMENTUM
A solid ball of mass m and radius r is rolling without slipping on
a flat horizontal surface, at an initial angular velocity w1. It hits a
small bump of height h. What is the angular velocity of the ball
immediately after impact? Also, what is the minimum initial
angular velocity w1 so that the ball just makes it over the bump?
What is the minimum initial speed of the ball?
Assume that the ball pivots about the tip of the bump during,
and after impact.
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EXAMPLE 5.3A
The inner hub of the wheel rests on the horizontal track. If
it does not slip at A, determine the speed of the
10-lb block in 2 s after the block is released from rest.
The wheel has a weight of 30 lb and a radius of gyration
kG = 1.30 ft. Neglect the mass of the pulley and cord.
2 ft
G
1 ft
A
SOLUTION
ISSN 1029-9599, Physical
Mesomechanics,
2016, Vol. 19, No.
2, pp. 115–122.
© Pleiades
Publishing,
Ltd., 2016.
WHAT
DOES FRICTION
REALLY
DEPEND
ON?
ROBUST
GOVERNING PARAMETERS
Original Text © V.L. Popov, 2015, published in Fizicheskaya Mezomekhanika, 2015, Vol. 18, No. 4, pp. 5–11.
115
What Does Friction Really Depend on? Robust Governing
Parameters in Contact Mechanics and Friction
V. L. Popov*
Berlin University of Technology, Berlin, 10623 Germany
* e-mail: v.popov@tu-berlin.de
Received June 13, 2015
Abstract—It is known that the coefficient of friction generally depends on a large number of system and loading
parameters. Already Coulomb presented experimental evidence that the static coefficient of friction may depend
on time, on normal force, on the contact size, on the nature of contacting materials, and on the presence of
intermediate lubricant layers. For the sliding coefficient of friction, he observed the dependence on the sliding
velocity as well as the force and size dependencies. Later research has shown that the friction coefficient is very
sensitive to the presence of oscillations (including self-excited vibrations). In spite of the practical importance of
the problem, no generalized laws of friction or empirical procedures for measuring and representing the law of
friction have been developed so far, which included at least the following four parameters: contacting body
velocity, normal force, shape (and thus implicitly size), and time. In the present paper, we discuss the question of
how the dimension of space of governing parameters can be reduced and if a small set of “robust governing
parameters” of friction can be identified. We argue that one of such robust governing parameters is the indentation depth (or relative approach) of contacting bodies and discuss further candidates for the role of robust
governing parameters.
DOI: 10.1134/S1029959916020016
Keywords: coefficient of friction, generalized law of friction, Amontons–Coulomb law, contact mechanics, influence of vibrations on friction
1. INTRODUCTION
Static and sliding friction are phenomena whose understanding is indispensable for the construction of safe
and energy-saving designs. Knowledge of exact law of
friction is of interest for countless applications, for
example clutches, brakes, tires, bush and ball bearings,
combustion engines, hinges, gaskets, castings, machining, cold forming, ultrasonic welding, electrical
contacts, and many others [1, 2]. Friction is a phenomenon that people have been interested in for over
hundreds and even thousands of years and still today
remains in the mainstream of the development of new
products and technologies.
However, in contrast to many other areas of engineering design, in most engineering applications the
simplest Amontons’ law of friction is used stating that
the force of friction is directly proportional to the normal force [3]. Every tribologist knows that this is only
a very rough “zero approximation”. Already Coulomb
knew that the coefficient of static friction may depend
PHYSICAL MESOMECHANICS
Vol. 19
No. 2
2016 115
on time, normal force, and on the contact size of the
system while the sliding coefficient of friction generally
depends on velocity, normal force and size as well as
on many other factors [4, 5]. Even if many of these
dependencies are relatively weak, they may have a
major influence on the behaviour of tribological systems. Thus, even a weak velocity dependence can lead
to development of dynamic instabilities and stick-slip
movement [2].
In the past there have been many attempts to understand why the law of friction by Amontons and Coulomb is approximately valid [6–8], and such attempts
continue to this day [9]. However, in the last time, the
research becomes more focused on the deviations of
the law of friction from the simple Amontons law [10–
12]. The deviations from the simple Coulomb’s law of
friction can be related either to the dependence of the
coefficient of friction on various system parameters or
to more far reaching changes in the friction law. For
example, since Coulomb, tribologists differentiate be-
116
POPOV
tween the static and sliding friction. However, this differentiation is only relative. In reality, some preliminary sliding always occurs already under forces smaller
than the critical force of gross slip. The phenomenon
of preliminary sliding is of the fundamental importance
for any applications where movements with small amplitudes are in play. These include the nanopositioning
systems as well as systems under vibrations with small
amplitudes. Corresponding empirical laws have been
developed first for nanopositioning applications [13–
15]. It was recently argued that the preliminary sliding
may be just a manifestation of partial sliding under tangential loading [16, 17]. Dieterich, Rice and Ruina [18–
20] went further and replaced the notions of “static”
and “kinetic” friction with the unifying concept of rateand state-dependent friction.
While there exist many investigations of friction in
various particular conditions, the attempts to work out
unifying concepts, which allow to construct “generalized laws of friction” which could be used by engineers
in practical applications, are still rare. A successful
example for formulation of such generalized laws provides the friction of elastomers. Based on the hypothesis
of Grosch [21] about the rheological nature of elastomer
friction, a very powerful “master curve procedure” has
been developed [2, 22], which allows a unifying representation of the coefficient of friction as function of
sliding velocity and temperature in the form of one
single “master curve” in appropriate coordinates. The
master curve procedure is based on the hypothesis that
the coefficient of friction does depend only on the product of the velocity and relaxation time of the elastomer,
while the temperature goes into the dependence only
over the relaxation time. Such parameter combinations,
which the coefficient of friction really depend on, will
be called “robust governing parameters”.
However, velocity and temperature are not the only
parameters which friction of elastomers—to continue
with this example—does depend on. As already shown
by Coulomb and illustrated by simple one-dimensional
simulations in [23, 24], the coefficient of friction also
depends on the normal force and on the size of the contact configuration. In paper [24], a generalized master
curve procedure was suggested and verified experimentally, which includes a unifying description of dependence of the coefficient of friction on velocity, temperature, and normal force. In [24], it was argued that
one of the “robust governing parameters” could be the
indentation depth. In the present paper, we discuss in a
more detail this parameter as well as other candidates
for the role of robust governing parameters in friction.
2. INDENTATION DEPTH AS A ROBUST
GOVERNING PARAMETER OF CONTACT
CONFIGURATION
If a rigid body of an arbitrary shape is pressed against
a homogeneous elastic half-space then the resulting contact configuration is only a function of the indentation
depth d. For example, in the case of the Hertz-type contact of a rigid parabolic indenter with the curvature radius R pressed against an elastic half-space, the contact
configuration is uniquely determined by the contact radius
a = Rd .
(1)
The contact radius does depend on the shape of the
indenter and the indentation depth but it does not depend
on elastic properties of the medium. This is a very general property which is valid for any shape of indenter
in contact with an elastomer with arbitrary linear rheology. Let us illustrate this statement with a simple
example of indentation of a parabolic indenter into an
elastic body and into a linear viscous fluid. The equilibrium condition for an elastic continuum reads
1
∆u +
∇divu = 0,
(2)
1 − 2ν
where ν is Poisson’s ratio. The corresponding “equilibrium” equation for a linearly viscous fluid reads as follows:
1
(3)
∆u +
∇divu = 0.
1 − 2ν
For “zero” initial conditions (plane surface of the medium before the indentation), the integration in time
of (3) leads to (2). Thus, the equations determining the
displacement field will be identical both for the elastic
body and the fluid, and it is easy to see that the boundary
conditions (given normal displacements inside the contact area and zero normal stress outside the contact area)
will be also identical. It follows that both the contact
radius and the complete form of the free surface will
be identical in these cases too. This general behavior
was recognized by Lee [25] and Radok [26] who used
this property to formulate a generalized procedure of
functional equations. The procedure is general and is
applicable to arbitrary profiles. In [27] it was applied
and verified numerically for fractal rough surfaces.
While the dependence of the contact configuration
only on the indentation depth is an exact statement for
a normal indentation of a rigid indenter into a linearly
elastic continuum, it remains approximately valid also
in other situations. For example, in the case of a nonlinear elastic material with the power law stress-strain
PHYSICAL MESOMECHANICS
Vol. 19
No. 2
2016
WHAT DOES FRICTION REALLY DEPEND ON? ROBUST GOVERNING PARAMETERS
relationship ε = ε0 (σ σ0 ) n, it was found [28] that the
penetration depth is given by
( n −1)
 2n + 1 
.
a = Rd 
(4)

 2n 
In the case of the linearly elastic (n = 1) medium we
reproduce the Hertz result (1) while in the general case
there will be some correction factor of the order of unity.
In the limiting case n = ∞, the dependence (4) reads
a ≈ 1.65 Rd .
For tangential contact with a viscoelastic medium
it is not exactly correct but remains qualitatively correct.
The standard rough estimation of the contact configuration in this case [29] is as follows. Due to tangential
motion, the effective elastic modulus will correspond
to the frequency ω ≈ L a , where L is the sliding velocity and a is the contact radius. However, as the relation
between the indentation depth and the contact radius
does not depend on the elastic modulus, this relation
will remain (at least in the frame of this rough estimation) unchanged independently of the sliding velocity.
There are further indications that the contact configuration is a function of solely the indentation depth—
with the same “qualitative precision”—even for spatially heterogeneous systems. For example, in [30], it
was shown that in the case of an elastic medium with a
thin coating, the equation (1) remains valid provided
the properties of the coating differ not too strong from
the properties of the underlying material. The same was
shown in [31] for the case of multi-layer materials. In
[32], it was argued that this is equally valid for media
which are heterogeneous in the lateral direction (along
the contact plane).
We thus can summarize that the relation between
the “contact configuration” and the indentation depth
is relatively insensitive to material properties or the
state of motion. Up to a factor of the order of unity, it
remains the same, as in the case of the normal contact
with a homogeneous elastic half-space.
Along with the contact configuration, all contact properties including the real contact area, the contact size,
the contact stiffness, as well as the rms value of the
surface gradient in the contact area will be unambiguous
functions of the indentation depth. The indentation
depth is thus a convenient and robust “governing parameter” for contact and frictional properties of media
of various physical nature.
Let us now discuss another aspect of the law of friction—the preliminary sliding. In a series of recent papers devoted to the problem of presliding in stick-slip
drives, it was shown that the phenomenon of prelimiPHYSICAL MESOMECHANICS
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117
nary sliding can be easily and precisely understood in
terms of partial sliding of curved elastic contacts. These
results have been summarized in [33]. It was shown
that the value of the presliding distance can be predicted
precisely in dependence of materials properties of both
contacting bodies, the size of the driving sphere, and
the normal force just by assuming that the presliding
distance is nothing else than the displacement till the
start of the gross slip. This assumption was further exploited in simulations and experiments on the influence
of in-plane and out-of-plane oscillations on friction. In
[34], a theoretical analysis of the influence of the inplane oscillations has been carried out. In [35, 36], these
theoretical predictions have been verified in a wide
range of sizes of the contacting bodies and for different
normal forces. Thus, the hypothesis of the nature of
the presliding as a pure partial sliding in the contact
area can be considered as confirmed for both microscopic and macroscopic systems. The direct implication
of this fact is that the presliding, too, is governed by
the indentation depth. Indeed, the theoretical prediction
[37] for the maximum tangential displacement till the
start of the gross sliding is given by
E*
ux(0) = µ * d ,
(5)
G
where
−1
−1
 1 − ν12 1 − ν 22 
2 − ν2 
*  2 − ν1
+
E = 
+
 , G =
 ,

4G2 
E2 
 4G1
 E1
E1 and E2 are the Young moduli of contacting bodies,
ν1 and ν 2 are their Poisson’s ratios, and G1 and G2
are the shear elastic moduli. For isotropic media, the
ratio E * G* is on the order of unity. For example, if
one of the bodies is rigid and the other is incompressible, this ratio is equal to E * G* = 1.5. From this it
follows that for a frictional contact with the coefficient
of friction µ, the maximum tangential displacement to
the onset of complete sliding is determined solely by
the indentation depth and has the order of magnitude
of µd. This result is independent of the form of the
bodies in contact and is valid for arbitrary bodies of
revolution and even for randomly rough fractal surfaces.
The latter fact was validated by direct numerical simulation of tangential contact of rough surfaces [38, 39].
Again, the indentation depth turns out to be the determining (governing) parameter of the frictional law.
Finally, consider the influence of normal oscillations
on friction. It is obvious that in this case the indentation
depth is also the main governing parameter of this process. In particular, the static coefficient of friction be*
118
POPOV
comes zero if the amplitude of normal oscillations
becomes equal to the indentation depth (so that the
sample starts to “jump”).
With the above argumentation we tried to substantiate the following thesis: Independently of the type of
contact (elastic, viscoelastic, plastic), the velocity, and
the particular considered property (coefficient of friction, presliding, influence of vibrations), the indentation
depth is a parameter which provides the most robust
information about the “contact configuration” in a broad
sense, and therefore determines the law of friction.
Thus, this quantity is surely one of the “robust governing parameters” of friction which we search for.
However, the contact configuration is unambiguously determined by the indentation depth only under assumption of the given topography of the contacting bodies. The properties which determine the surface topography are discussed in the next section.
3. SURFACE GRADIENT AND THE SIZE
OF MICROCONTACTS AS MAIN
REPRESENTATIVE SURFACE PARAMETERS
In [24], it was argued that at least for linearly elastic elastomers in most practical situations there are the
largest and the smallest scales in the power spectrum
of the surface roughness which determine their contact
and frictional properties. The long wavelength part of
the spectrum determines the stiffness of the contact
while the short wavelength part determines the surface
gradient playing the most important role in friction.
Indeed, both in a plastic contact and in a viscoelastic
contact in the region of the plateau of the coefficient of
friction (as function of the sliding velocity), almost all
“micro-asperities” have the “one-sided contact”. Under
this condition, the coefficient of friction is just equal
to some weighted average of the surface gradient (slope)
∇z, measured in the whole contact area. Of course, the
average slope may depend on the normal force. Even
if this dependence is very weak, we will discuss it later
in this paper.
As has been shown already by Archard in 1950th
[7] the main effect of changing the normal force is the
changing number of asperities coming into contact,
while the local conditions in the real contact area, including the surface gradient, depend only weakly on
the normal force. Thus, the pure “rheological” contribution to the coefficient of friction—independently on
whether we have to do with linearly or nonlinearly elastic or plastic body—is given by the characteristic surface gradient in the area of contact. Another important
contribution to the force of friction can be the “adhesive” contribution which was introduced already by
Coulomb [5] and later was considered by Bowden and
Tabor as the main contribution in the case of metallic
partners [6]. Recently, it was suggested that shearing
of surface layers may give an additional contribution
to the force of friction of elastomers, too [40, 41]. It is
easy to see that this contribution is also governed by
the surface gradient. Indeed, one of the most robust
results of the contact mechanics of rough surfaces states
that the real contact area A is determined by the following equation:
F
A≈2 * .
(6)
E ∇z
This result was first found by Hyun et al. [42] and was
confirmed in numerous subsequent studies. If, for the
sake of simplicity, we consider a square rigid block of
the size L in a contact with an elastic half-space and
assume that the normal force is large enough so that
the contact stiffness has already achieved its plateau
[43], then the normal force is given by the equation
F ≈ E * Ld and Eq. (6) takes the form
Ld
(7)
A≈2 .
∇z
This is again a “pure geometrical” relation which does
not contain elastic properties of the system and thus is
independent of the details of the process under consideration. It shows that the determining parameters of
the real contact area are the size of the system, the indentation depth, and the surface gradient.
The main physical question which remains is: what
is the surface gradient? For fractal surfaces this is a
poorly defined property which depends on the accuracy
with which the surface topography is measured. On can
easily see this from the following simple estimation.
The average gradient of the surface profile ⟨∇z 2 ⟩ can
be expressed as an integral of the spectral power density
of the surface profile C(q) according to
⟨∇z 2 ⟩ ≈ ∫ C (q )q 3dq.
(8)
It is known that many natural surfaces have the property
of fractality, and for randomly self-affine, fractal surfaces the spectral power density is known to be a power
function of the wave vector q, that is
C2D (q) = const × (q q0 )−2 H − 2 ,
(9)
where H is the Hurst exponent and q0 is some reference
wave vector [44]. For typical Hurst exponents smaller
than one, the resulting integral (8) diverges at the upper
limit of integration. This means that for a true fractal
surface (without an upper cut-off wave vector), the
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WHAT DOES FRICTION REALLY DEPEND ON? ROBUST GOVERNING PARAMETERS
(a)
119
(b)
Fig. 1. Contact between an elastomer and a rigid conical indenter which is moved tangentially with the velocity L (a); rheological
model for a viscoelastic medium (b).
surface gradient is infinitely large. In practice, of course,
there is always some upper cut-off wave vector qmax
and the surface gradient is determined by one or two
orders of magnitude of wave vectors at and below qmax .
In other words, for typical fractal surfaces, the friction
force is determined by the roughness components with
the largest wave vectors (or the smallest scale of the
system). One can say that understanding friction is
equivalent to understanding the nature of this smallest
relevant scale. This statement is not limited to the
spectral representation of fractal surfaces. It is sufficient
that the characteristic surface gradient is increasing
toward the smaller space scales. The above said means
that the question about the coefficient of friction is
essentially replaced by the question of determining the
“relevant maximum wave vector”, which is not just a
geometrical but in most cases a physical problem.
Further essential parameter is the “typical size” a
of the “microcontacts”. This size determines the characteristic frequency of interaction between the bodies and
thus the effective complex modulus of their interaction.
In the approximation of Greenwood and Williamson
[45], this size is given by a ≈ Rh , where R is the
radius of curvature of asperities and h is the rms roughness. Using the simple estimation ∇z ≈ ( h R)1 2 (see
[2]), it can also be written in the following form:
h
a≈
.
(10)
∇z
The characteristic frequency ω can then be estimated
as
L L∇z
ω≈ ≈
.
(11)
a
h
Thus, we anticipate that the coefficient of friction
should be a function of dimensionless combinations of
the following parameters: indentation depth d and sliding velocity L as loading parameters, apparent contact
size L as system parameter, rms roughness h and the
relevant rms surface gradient ∇z as roughness parameters, relaxation time τ as material parameter.
PHYSICAL MESOMECHANICS
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4. EXAMPLES OF GENERALIZED LAWS
OF FRICTION
To get feeling how the parameters discussed above
determine the coefficient of friction, it is instructive to
analyze several particular examples of generalized laws
of friction.
As an example, we consider a rigid wedge-shaped
indenter of the form
z = g ( x) = c x ,
(12)
which is pressed into a viscoelastic foundation to a certain depth d and is moved tangentially with the velocity
L (Fig. 1a).
In this case, the coefficient of friction is evaluated
as follows [29]:
2(cLτ d ) − 1 2 (cLτ d )2
F
.
µ= x =c
(13)
Fn
1 + 1 2 (cLτ d )2
By taking into account that c = ∇z , and (cLτ) d ≈
Lτ a ≈ ωτ, this result can be rewritten in the form
 Lτ 
µ = ∇z Ψ   .
(14)
 a
In a similar contact of a sphere of radius R1 with the
Kelvin body we have [29]
ξ[2 − 3ξ − 2ξ3 + 2(1 + ξ2 )3 2 ]
(15)
µ = ∇z
[1 − ξ3 + (1 + ξ 2 )3 2 ]4 3
with
Lτ
Lτ
ξ=
≈ .
12
a
(2R1d )
Thus, the coefficient of friction again has the general
form (14). In a contact of a cone with a Maxwell body,
the coefficient of friction is given by [29]
d ( cLτ) − 2(1 − e− d (cLτ ) ) + ln(2 − e− d (cLτ ) )
, (16)
µ=c
d ( cLτ) − ln(2 − e− d (cLτ) )
which is obviously again of the form (14).
Till now, we considered simple “one-asperity”
shapes of indenters. However, as the dependence (14)
contains only local geometrical parameters of asperities,
120
POPOV
which according to the general ideas of Archard [7],
do depend on the loading conditions only weakly, this
will be valid also for multicontact configuration, at least
qualitatively. We thus come to the conclusion that
form (14) may be very general.
Let us discuss the dimensionless parameter entering
Eq. (14). There are two parameters ∇z and a characterizing roughness and one further parameter, relaxation
time τ characterizing the rheology of the material. We
note that two parameters are the minimum with which
a rough surface can be characterized. For example, the
Greenwood and Williamsion model uses the following
two parameters: the rms roughness and the radius of
curvature of asperities. However, in the most cases, the
relevant curvatures and roughnesses “on the relevant
scale” cannon be measured straightforwardly, just because we often do not know exactly the physical processes determining the smallest “relevant scale”. They
therefore will be in most cases used as empirical fitting
parameters. However, in this case it is more sensible to
use as empirical fitting parameters the parameters ∇z
and a which, according to Eq. (14), straightforwardly
determine the coefficient of friction.
Of course, the assumed independence of the parameters ∇z and a on the loading parameters is valid only
approximately. Both these parameters depend on the
actual “contact configuration”. The latter, however, as
argued in Sect. 2, depends solely on the indentation
depth d. We thus can combine Eq. (14) with the argument of Sect. 2 in the following form:
 Lτ(T ) 
µ = ∇z ( d ) Ψ 
(17)
,
 a( d ) 
where we denoted explicitly ∇z(d) and a(d) as (relatively weak) function of only indentation depth, while
the relaxation time τ(T) is a material parameter which
does not depend on the contact configuration but may
depend on the temperature T. This form of the law of
friction allows to suggest a new generalized master
curve procedure. Equation (17) can be identically rewritten in the form
ln µ = ln ∇z ( d ) + ln Ψ
× [exp(ln L + ln τ(T ) − ln a( d ))],
(18)
or introducing a new function Φ (⋅) = ln Ψ (exp(⋅)), in
the form
ln µ − ln ∇z ( d ) = Φ(ln L + ln τ(T ) − ln a(d )). (19)
This equation states that if we plot the logarithm of the
coefficient of friction as function of the logarithm of
velocity, for different values of the indentation depth d
and the temperature T, all curves will “look the same”
and it is possible to move them by some vertical shift
factor ln∇z(d) depending solely on the indentation
depth as well as horizontally by the shift factor ln τ(T) –
ln a(d) which is the additive superposition of a function
of d and a function of T.
Of course, in practice, the controlling (governing)
parameter of the external loading is in most cases the
normal force and not the indentation depth. The indentation depth generally will depend on the normal force,
the size or the shape of the body, elastic properties,
and for viscoelastic bodies also on the sliding velocity.
However, all these influence factors are lying in a realm
of the macroscopic contact mechanics and mechanical
design and are principally solvable with standard techniques of contact mechanics. The whole “tribology”,
on the contrary, is completely contained in the phenomenological equation (19).
The suggested form (18) of the generalized law of
friction is, of course, subjected to some restrictions.
For example, we have assumed that the temperature is
an independent external parameter. This assumption is
only valid as long as the local “flash temperatures” are
of no importance. However, it was shown in [46] that
the general form (19) remains valid even under consideration of flash temperatures if a temperature dependent shift in the vertical direction is allowed.
Finally let us mention that the coefficient of friction
is a kinetic property which under nonstationary conditions can be described with differential equations containing additional state variables [18–20]. However, as
it was argued in [39] and [23, 47] both the characteristic
slip length and the relaxation time are governed by the
indentation depth and the dimensionless parameter
Lτ a just as the stationary properties.
5. CONCLUSION
In the present paper we discussed the question of
whether it is possible to find such “robust governing
parameters” which determine the friction most directly
while remaining in the framework of a simple analytical
structure. We argued that the property which most directly and robustly determines the contact configuration
is the indentation depth, whereas the parameters of the
contact configuration which are relevant for friction
are the surface gradient and the characteristic size of
micro-asperities. Both parameters depend only on the
indentation depth, though this dependence is relatively
weak. Another parameter which is of importance is the
characteristic relaxation time of the medium which only
depends on temperature. The suggested structure of the
coefficient of friction allows to apply the phenomenoPHYSICAL MESOMECHANICS
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WHAT DOES FRICTION REALLY DEPEND ON? ROBUST GOVERNING PARAMETERS
logical “generalized master curve procedure” by measuring the coefficient of friction at different velocities,
temperatures and indentation depths and applying the
shifting procedure. As the indentation depth is not a
parameter which normally can be controlled in experiment, the suggested frictional law has to be completed
by a contact mechanical derivation connecting the indentation depth with the normal force, shape of the bodies and their materials parameters.
ACKNOWLEDGMENTS
The author acknowledges a valuable discussion with
I. Argatov. This work was supported by Deutsche Forschungsgemeinschaft (DFG).
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PHYSICAL MESOMECHANICS
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QUICK SUMMARY OF PRINCIPLE OF IMPULSE AND MOMENTUM
; L = mvG
CONSERVATION OF MOMENTUM
• Understand the conditions for CONSERVATION of linear
and angular momentum.
• Use the condition of conservation of linear/ angular
momentum.
CONSERVATION OF MOMENTUM
If there are no external impulses acting on a body:
Both linear momentum and angular momentum are
conserved
CONSERVATION OF LINEAR MOMENTUM
Recall that the linear impulse and momentum relationship is
t2
0
t2
0
L1 + ∑ ∫ F dt = L2 or (m vG)1 + ∑ ∫ F dt = (m vG)2
t1
t1
If the sum of all the linear impulses acting on the rigid
body (or a system of rigid bodies) is zero, all the impulse
terms are zero. Thus, the linear momentum for a rigid
body (or system) is constant, or conserved.
So (m vG)1 = (m vG)2
This equation is referred to as the conservation of linear
momentum.
CONSERVATION OF LINEAR MOMENTUM
t2
̴0
t2
̴0
L1 + ∑ ∫ F dt = L2 or (m vG)1 + ∑ ∫ F dt = (m vG)2
t1
t1
The conservation of linear momentum equation can be
used ALSO if the linear impulses are small or nonimpulsive in a specified direction. Then the integral terms
are negligible e.g. when small forces act over very short
periods of time e.g. force of a slightly deformed spring or
in some cases weight of the body etc
CONSERVATION OF ANGULAR MOMENTUM
Similarly, if the sum of all the angular impulses due to
external forces acting on the rigid body (or a system of rigid
bodies) is zero, all the impulse terms are zero. Thus, angular
momentum is conserved
t2
0 or ̴ 0
t2
0 or ̴ 0
(HG)1 + ∑ ∫ MG dt = (HG)2 or IGω1 +∑ ∫ MG dt = IGω 2
t1
t1
The resulting equation is referred to as the conservation of
angular momentum or (HG)1 = (HG)2 .
If the initial condition of the rigid body (or system) is known,
conservation of momentum is often used to determine the final
linear or angular velocity of a body just after an event occurs.
APPLICATIONS
When the skater executes a spin, the angular impulse acting on her is
very close to zero since there is relatively little friction between her
skates and the ice. Also, the friction is exerted very close to the pivot
point/axis of rotation.
So the skater’s angular momentum is (approx) conserved.
(IG)1 ω 1 = (IG)2 ω2
The skater spins faster when the arms are drawn in and slower when
the arms are extended. Why?
PROCEDURE FOR ANALYSIS
• Establish the x, y, z inertial frame of reference and draw FBDs.
• Write the conservation of linear momentum equation.
• Write the conservation of angular momentum equation about a
fixed point or at the mass center G.
• Solve the conservation of linear or angular momentum
equations in the appropriate directions.
• If the motion is complicated, use of the kinematic equations
relating the velocity of the mass center G and the angular
velocity ω may be necessary.
Example: Merry-go-Round
Given: Two children(mA= mB= 30 kg)
sit at the edge of the merry-goround, which has a mass of 180
kg and a radius of gyration of
kz = 0.6 m.
Find: The angular velocity of the merry-go-round if A jumps
off horizontally in the +t direction with a speed of 2 m/s,
measured relative to the merry-go-round.
Example: Acrobat stunt.
An acrobat of mass m falls without rotation from a height of h above the bar. He grabs the bar
with a firm but slippery grip and rotates about the bar. What is h so that the acrobat ends up in a
stationary handstand? Assume that center of mass of the acrobat is located in the middle and the
moment of inertia about the center of mass is IG .
29
30
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u
M A- �AJ 1 -1- Me, ��\
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•
The t 0-lb slender rod is suspended from the pin at A. If 11 2-lb ball B is lhrown at the rod and
strikes its center with a velocity of30 ft/s, determine the angular velocity of the rod just after
impact The coefficient of restitution is e = 0.4 and the contact between the ball and the rod is
frictionless.
-,
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A
/,.·. � t'\\�
s :
1.s r1
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t.itv co f- lMM
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EXAMPLE
A solid ball with a mass m is thrown on the ground such that at the instant of contact it has an
angular velocity a'i,and velocity components (vr. )..1 and (vr, ) 11 as shown. If the ground is
frictionless, detennine the velocity of the contact point just after impact. The coefficient of
restitution is e.
'
Luv
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of
IMRAcr
l � Lf - D \ U, f.
get rid of this
term..
unknown
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SECTION 6
VIBRATIONS
6.1 UNDAMPED FREE VIBRATIONS
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: � - b_(__�t
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Example: Bent rod
The bent rod hos a negligible mass and supports a 5-kg collar at its end. If the rod is in the
equilibrium position shown, determine the natural period of vibration for the system.
100mm
I
Tl' QrL-� 0-t"-«s c-e. vJo��I
•
�,
(A_
s:�JJ
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5. (20%)A uniform thin bar of mass m and length l is pinned at its base at O and
supported by a horizontal spring with stiffness k at its other end as shown in .the
diagram below. The support for the spring is such that it remains horizontal for all
motion of the bar.
Draw the Free Body Diagram and Kinetic Diagram of the bar and derive the equation of
motion of this system for small oscillations. What is the natural frequency of these small
oscillations?
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'
13
. 0
S )._s 2·6s7- 6.2 Energy Methods
U '\-kl- \AA._ '\-k �" { undamped free vibration
-\k__ �Jo�I/J) � � hy (JtalVWi,W)� 01
dCA.b�C Y\U �� � f �
- tom a)lAJ!��
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s () ::-
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Example
The ends of the uniform slender bar of mass m and length L move freely in the
vertical and horizontal slots under the action of the two precompressed springs
each of stiffness k as shown. If the bar is in static equilibrium when θ = 0,
determine the natural frequency wn of small oscillations.
6.3 Undamped Forced Vibration
• T"- � (JYJifl1
{ °" �'t�\vM.
C
/
ci' /
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1 / 1 // / /
r' I I
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.'
ru Sui\.� 1c - �X. = MY-
-------·-0
i_ �!- � = �S\�wot
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Example 6..7
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