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KINENAMITCS OF RIGID
BODY
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GENARAL
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PLANAR KINEMATICS OF A RIGID BODY
We will now start to study rigid body motion. The analysis
will be limited to planar motion.
For example, in the design of gears, cams, and links in
machinery or mechanisms, rotation of the body is an
important aspect in the analysis of motion.
There are cases where an object cannot be treated as a
particle. In these cases the size or shape of the body must be
considered. Also, rotation of the body about its center of
mass requires a different approach.
A body is said to undergo planar motion when all parts of
the body move along paths equidistant from a fixed plane.
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PLANAR RIGID BODY MOTION
Translation: Translation occurs if every line segment on the
body remains parallel to its original direction during the
motion. When all points move along straight lines, the motion
is called rectilinear translation. When the paths of motion are
curved lines, the motion is called curvilinear translation.
There are three types of planar rigid body motion.
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General plane motion. In this case, the
body undergoes both translation and
rotation. Translation occurs within a plane and rotation occurs about an axis
perpendicular to this plane.
Rotation about a fixed axis. In this case, all
the particles of the body, except those on
the axis of rotation, move along circular paths in planes perpendicular to the axis of
rotation.
PLANAR RIGID BODY MOTION (continued)
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PLANAR RIGID BODY MOTION (continued)
The piston (C) undergoes rectilinear translation since it is
constrained to slide in a straight line. The connecting rod (D)
undergoes curvilinear translation, since it will remain horizontal asit moves along a circular path.
The wheel and crank (A and B)
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.
An example of bodies undergoing
the three types of motion is shown
in this mechanism.
The connecting rod (E) undergoes general plane motion, as it will
both translate and rotate.
C
E
A
D
B
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TRANSLATION AND
ROTATION ABOUT A
FIXED AXIS
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RIGID-BODY MOTION: TRANSLATION
The positions of two points A and B
on a translating body can be related by
r B = r A + r B/Awhere r A & r B are the absolute position
vectors defined from the fixed x-y
coordinate system, and r B/A is the
relative-position vector between B and A.
Note, all points in a rigid body subjected to translation
move with the same velocity and acceleration.
The velocity at B is vB = vA+ dr B/A/dt .
Now dr B/A/dt = 0 since r B/A is constant. So, vB = vA, and by
following similar logic, aB = aA.
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RIGID-BODY MOTION: ROTATION ABOUT A FIXED AXIS
The change in angular position, d θ , is called the
angular displacement, with units of either
radians or revolutions. They are related by
1 revolution = 2π radians
When a body rotates about a fixed axis, any
point P in the body travels along a circular
path. The angular position of P is defined by θ .
Angular velocity, ω, is obtained by taking thetime derivative of angular displacement:
ω = d θ /dt (rad/s) +Similarly, angular acceleration is
α = d2θ/dt2 = dω/dt or α = ω(dω/dθ) + rad/s2
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If the angular acceleration of the body is
constant, α αα α = α αα α C, the equations for angular
velocity and acceleration can be integratedto yield the set of algebraic equations
below.
θO and ωO are the initial values of the body’sangular position and angular velocity. 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)
ω = ωO + αCt
θ = θO + ω Ot + 0.5 αCt2
ω2 = (ωO)2 + 2 αC (θ – θO)
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In the vector formulation, the magnitudeand direction of v can be determined
from the cross product of ω ωω ω and r p .
Here r p
is a vector from any point on the
axis of rotation to P.
RIGID-BODY ROTATION: VELOCITY OF POINT P
˙
˙̇̇̇
Using polar coordinates, radial and
transverse components of velocity of P
can be expresses as
vr = r = 0 and vθ = r θ ˙̇̇̇˙̇̇̇
As θ = ω ; therefore,˙̇̇̇
v = ωr
v = ω ωω ω x r p = ω ωω ω x r
The direction of v is determined by the
right-hand rule.
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The acceleration of P is expressed in terms of
its normal (an) and tangential (at) components
The tangential component, at, represents thetime rate of change in the velocity's
magnitude. It is directed tangent to the path of
motion.
The normal component, an, represents the time
rate of change in the velocity’s direction. It is
directed toward the center of the circular path.
RIGID-BODY ROTATION: ACCELERATION OF POINT P
at = dv/dt and an =v2 / ρ
but ρ = r ; v = ω r and α = d ω /dt
Therefore, at = α r and an= ω 2r
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Using the vector formulation, the acceleration
of P can also be defined by differentiating the
velocity.
a = dv/dt = dω ωω ω /dt x r P + ω ωω ω x dr P/dt
= α αα α x r P + ω ωω ω x (ω ωω ω x r P)
It can be shown that this equation reduces to
RIGID-BODY ROTATION: ACCELERATION OF POINT P
(continued)
The magnitude of the acceleration vector is a = (at)2 + (an)
2
a = α αα α x r – ω 2r
= at + an
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ROTATION ABOUT A FIXED AXIS: PROCEDURE
• Establish a sign convention along the axis of rotation.
• Alternatively, the vector form of the equations can be used(with i , j , k components).
v = ω ωω ω x r P = ωωωω x r a = a
t
+ an
= α αα α x r P
+ ω ωω ω x (ω ωω ω x r P
) = α αα α x r – ω2r
• If α is constant, use the equations for constant angularacceleration.
• 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
• 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.
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EXAMPLE 1
Given: Starting from rest when s = 0, pulley
A (r A = 50 mm) is given a constant
angular acceleration, αA = 6 rad/s2.
Pulley C (r C = 150 mm) has an innerhub D (r D = 75 mm) which is fixed
to C and turns with it.
Find:The speed of block B when it has risen s = 6 m.
Plan: 1) The angular acceleration of pulley C (and hub D) can be
related to αA if it is assumed the belt is inextensible anddoes not slip.
2) The acceleration of block B can be determined by using
the equations for motion of a point on a rotating body.
3) The velocity of B can be found by using the constant
acceleration equations.
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EXAMPLE 1 (continued)
Solution:
Since C and D turn together, αD = αC = 2 rad/s2
1) Assuming the belt is inextensible and does not slip, it will have
the same speed and tangential component of acceleration as it
passes over the two pulleys (A and C). Thus,
at = αAr A = αCr C => (6)(50) = αC(150) => αC = 2 rad/s2
2) Assuming the cord attached to block B is inextensible and
does not slip, the speed and acceleration of B will be the same
as the speed and tangential component of acceleration along
the outer rim of hub D:
aB = (at)D = αDr D = (2)(0.075) = 0.15 m/s2
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3) Since αA is constant, αD and aB will be constant. The constantacceleration equation for rectilinear motion can be used to
determine the speed of block B when s = 6 m (so = vo = 0):
(vB)2 = (vo)
2 + 2aB(s – so) +
(vB)2 = 0 + 2(0.15)(6 – 0)
vB = 1.34 m/s
EXAMPLE 1 (continued)
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GENERAL PLANE MOTION
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Absolute Motion Analysis
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PROCEDURE FOR ANALYSIS
The absolute motion analysis method (also called the parametric method) relates the position of a point, P, on a
rigid body undergoing rectilinear motion to the angular
position, θ (parameter), of a line contained in the body.(Often this line is a link in a machine.) Once a relationship
in the form of sP = f(θ) is established, the velocity andacceleration of point P are obtained in terms of the angular
velocity, ω, and angular acceleration, α, of the rigid body bytaking the first and second time derivatives of the positionfunction. Usually the chain rule must be used when taking
the derivatives of the position coordinate equation.
Absolute Motion Analysis
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Given:Two slider blocks are connected
by a rod of length 2 m. Also,
vA = 8 m/s and aA = 0.
Find: Angular velocity, ω, andangular acceleration, α, of therod when θ = 60°.
Plan: Choose a fixed reference point and define the position of
the slider A in terms of the parameter θ. Notice from the position vector of A, positive angular position θ ismeasured clockwise.
EXAMPLE 2
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Relative motion analysis
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RELATIVE MOTION ANALYSIS: DISPLACEMENT
When a body is subjected to general plane motion, it undergoes acombination of translation and rotation.
=
dr B = dr A + dr B/A
Disp. due to translation and rotation
Disp. due to translation
Disp. due to 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:
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The velocity at B is given as : (dr B/dt) = (dr A/dt) + (dr B/A/dt) or
vB = vA + vB/A
RELATIVE MOTION ANALYSIS: VELOCITY
= +
Since the body is taken as rotating about A,vB/A = dr B/A/dt = ω ωω ω x r B/A
Here ω ωω ω will only have a k component since the axis of rotation
is perpendicular to the plane of translation.
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When using the relative velocity equation, points A and B
should generally be points on the body with a known motion.
Often these points are pin connections in linkages.Here both points A and B have
circular motion since the disk and
link BC move in circular paths.
The directions of vA and vB are
known since they are always
tangent to the circular path of
motion.
vB = vA + ω ωω ω x r B/A
RELATIVE MOTION ANALYSIS: VELOCITY (continued)
(16-16)
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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.
vB = vA + ω ωω ω x r B/A
RELATIVE MOTION ANALYSIS: VELOCITY (continued)
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.
ROLLING WITHOUT SLIPPING
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PROCEDURE FOR ANALYSIS
The relative velocity equation can be applied using either aCartesian vector analysis or by writing scalar x and y component
equations directly.
Vector Analysis:
3. If the solution yields a negative answer, the sense of
direction of the vector is opposite to that assumed.
2. Express the vectors in Cartesian vector form and substitute
into vB = vA + ω ωω ω x r B/A. Evaluate the cross product and
equate respective i and j components to obtain two scalar
equations.
1. Establish the fixed x-y coordinate directions and draw the
kinematic diagram of the body, showing the vectors vA, vB,
r B/A and ω ωω ω . If the magnitudes are unknown, the sense of
direction may be assumed.
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PROCEDURE FOR ANALYSIS (continued)
3. Write the scalar equations from the x and y components of
these graphical representations of the vectors. Solve forthe unknowns.
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.
Scalar Analysis:
2. Write the equation vB = vA + vB/A and by using the kinematic
diagram, underneath each term represent the vectorsgraphically by showing their magnitudes and directions.
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Example 3: Determining Velocities
and Angular Velocities
• Question
Bar AB in the figure below rotates with aclockwise angular velocity of 10 rad/s.
Determine the angular velocity of bar BC
and the velocity of point C.
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Example 3: Determining Velocities
and Angular Velocities
• Strategy
Bar AB rotates about the fixed point A witha known angular velocity, so we can
determine the velocity of B. Then, by
expressing the horizontal velocity of C in
terms of the velocity of B and the angular
velocity of bar BC, we can obtain twoequations in terms of the velocity of C and
the angular velocity of bar BC.
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Example 3: Determining Velocities
and Angular Velocities• Solution
Because the angular velocity of bar AB isgiven and point A is stationary, we canuse Eq. 16-16 to determine the velocity ofpoint B. From Fig a, the position vector ofB relative to A is r B/A = 0.4i + 0.4j (m).
Fig a
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Example 3: Determining Velocities
and Angular Velocities
The angular velocity vector of bar AB is ω AB
= -10k (rad/s), so the velocity of B is
m/s j4i4
04.04.0
10000
r vv /
−=
−+=
×+=
k ji
A B A B ω
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Example 3: Determining Velocities
and Angular VelocitiesWe can use Eq. 16-16 to express the velocity of
point C in terms of the velocity of point B. Let ωBCbe the counterclockwise angular velocity of bar
BC (Fig b), so that the angular velocity vector of
bar BC is ωBC = ωBC k. Because point C is
moving horizontally, we can write its velocity asvC = vC i.
Fig b
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Example 3: Determining Velocities
and Angular Velocities
( ) ( ) j8.04i4.04
04.08.0
00
k ji
j4i4iv
r vv /
BC C
BC C
BC BC BC
ω ω
ω
ω
−−+=
−+−=
×+=
Therefore, we have
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Example 3: Determining Velocities
and Angular Velocities
Equating the i and j components in this
equation, we obtain
BC
BC C v
ω
ω
8.040
,4.04
−=
+=
Solving it, we have vC = 6 m/s and ωBC = 5 rad/s.
the angular velocity of bar BC is ωBC
= 5k (rad/s),
and the velocity of point C is vC = 6i m/s.
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INSTANTANEOUS CENTER(IC) OF ZERO VELOCITY
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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 (if it were rigidly connected to the body).
This point is called the instantaneous center of zero velocity,
or IC. It may or may not lie on the 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.
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LOCATION OF THE INSTANTANEOUS CENTER
To locate the IC, we can 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 ωω ω
of the body are known.
In this case, the IC is located along the line
drawn perpendicular to vA at A, a distance
r A/IC = vA/ω from A. Note that the IC lies upand to the right of A since vA must cause a
clockwise angular velocity ω ωω ω about the IC.
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A second case is when the
lines of action of two non-
parallel 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)
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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 proportionaltriangles. 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.
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VELOCITY ANALYSIS
The velocity of any point on a body undergoing general plane
motion can be determined easily 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. Note the velocity has a sense ofdirection which tends to move the point in
a manner consistent with the angular
rotation direction.
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Example 4: Linkage Analysis by
Instantaneous Centers
• Question
Bar AB in the Figure rotates with acounterclockwise angular velocity of 10
rad/s. what are the angular velocities of
bars BC and CD?
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Example 4: Linkage Analysis by
Instantaneous Centers
• Strategy
Because bars AB and CD rotate aboutfixed axes, we know the directions of
motion of points B & C so we can locate
the instantaneous center of bar BC.
Beginning with bar AB, we can use the
instantaneous centers of the bars todetermine both the velocities of the points
where they are connected and their
angular velocities.
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Example 4: Linkage Analysis by
Instantaneous Centers
• Solution
The velocity of B due to the rotation of bar AB about A (Fig a) is
( )( ) m/s20rad/s10m2 == B
v
Fig a
E l 4 Li k A l i b
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Example 4: Linkage Analysis by
Instantaneous Centers
Drawing lines perpendicular to the
directions of motion of B & C, we locate theinstantaneous center of bar BC (Fig b). The
velocity of B is equal to the product of its
distance from the instantaneous center ofbar BC and the angular velocity ωBC:
Hence, ωBC = 10 rad/s with bar BC rotating
in the clockwise direction.
( ) C Bv ω m2m/s20 ==
E l 4 Li k A l i b
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Example 4: Linkage Analysis by
Instantaneous Centers
Fig b
E l 4 Li k A l i b
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Example 4: Linkage Analysis by
Instantaneous Centers
Using the instantaneous center of bar BC
and its angular velocity ωBC, we candetermine the velocity of point C:
m/s810m8 == C C v ω
Lastly is to use the velocity of C to determine the
angular velocity of bar CD about point D (Fig c).
We have
ckwisecounterclorad/s10
m8m/s810
=∴==
CD
CDC v
ω
ω
E l 4 Li k A l i b
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Example 4: Linkage Analysis by
Instantaneous Centers
Fig c
ATTENTION QUIZ
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ATTENTION QUIZ
2. Point A on the rod has a velocity of 8 m/s to the right.
Where is the IC for the rod?
A) Point A.
B) Point B.
C) Point C.D) Point D.
• C
D •
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RELATIVE MOTION ANALYSIS:ACCELERATION
RELATIVE MOTION ANALYSIS ACCELERATION
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RELATIVE MOTION ANALYSIS: ACCELERATION
The equation relating the accelerations of two points on the
body is determined by differentiating the velocity equation
with respect to time.
The result is aB = aA + (aB/A)t + (aB/A)n
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.
It will develop tangentialand normal components.
/+dt
dv AB
dt
dvA
dt
dvB =
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The relative normal acceleration component (aB/A)n is (-ω2 r B/A)
and the direction is always from B towards A.
RELATIVE MOTION ANALYSIS: ACCELERATION
= +
Graphically: aB = aA + (aB/A)t + (aB/A)n
The relative tangential acceleration component (aB/A
)t
is (α αα α x r B/A
)
and perpendicular to r B/A.
RELATIVE MOTION ANALYSIS: ACCELERATION
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Since the relative acceleration components can be expressed
as (aB/A)t =α ×
r B/A and (aB/A)n = - ω2
r B/A the relativeacceleration equation becomes
aB = aA + αααα × r B/A - ω2 r B/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, r B/A.
RELATIVE MOTION ANALYSIS: ACCELERATION
(continued)
(16-18)
APPLICATION OF RELATIVE ACCELERATION
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APPLICATION OF RELATIVE ACCELERATION
EQUATIONIn 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.
Point C, connecting link BC and the piston, moves along a straight-line
path. Hence, aC is directed horizontally.
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.
PROCEDURE FOR ANALYSIS: RELATIVE
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PROCEDURE FOR ANALYSIS: RELATIVE
ACCELERATION ANALYSIS1. Establish a fixed coordinate system.
2. Draw the kinematic diagram of the body.
3. Indicate on it aA, aB, ω ωω ω , α αα α , and r B/A. If the points A and Bmove 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 × r B/A - ω2 r B/A
5. If the solution yields a negative answer for an unknown
magnitude, it indicates the sense of direction of the vector
is opposite to that shown on the diagram.
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Example 5: Acceleration of a Point• Question
The rolling disk in the Figure below has
counterclockwise angular velocity ω and
counterclockwise angular acceleration α.
What is the acceleration of point A?
E l 5 A l ti f P i t
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Example 5: Acceleration of a Point
• Strategy
We know that the magnitude of the
acceleration of the center of the disk is theproduct of the radius and the angular
acceleration. Therefore, we can express
the acceleration of A as the sum of the
acceleration of the center of the disk and
the acceleration of A relative to the center.
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Example 5: Acceleration of a Point• Solution
First Method. In terms of the coordinate
system in Fig a, the acceleration of thecenter B is aB = -αRi. A’s motion in a
circular path of radius R relative to B
results in the tangential and normal
components of relative acceleration shown
in fig b:
ia 2/ R R B α ω +−=
E l 5 A l ti f P i t
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Example 5: Acceleration of a Point
The acceleration of A is
ji jiiaaa
2
2
/
R R R R R R B A B A
α ω α α ω α
+−−=+−−=+=
Fig a Fig b
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Example 5: Acceleration of a PointSecond Method. The angular acceleration vector
of the disk is α = αk, and the position of A relative
to B is r A/B = Ri (Fig c).
Fig c
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Example 5: Acceleration of a Point
From equation of acceleration, the
acceleration of A is
( ) ( ) ( )
jR i
iik i
r r aa
2
2
/2
/
α ω α
ω α α
ω α
+−−=
−×+−=
−×+=
R R
R R R
B A B A B A
Example 6: Angular Accelerations of
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p g
Members of a Linkage• Question
Bar AB in the figure below has a
counterclockwise angular velocity of 10 rad/sand a clockwise angular acceleration of 300rad/s2. What are the angular accelerations ofbars BC and CD?
Example 6: Angular Accelerations of
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p g
Members of a Linkage
• Strategy
Since we know the angular velocity of bar AB, we can determine the velocity of point
B. We can apply Eq. 16-16 to points C & D
to obtain an equation for vC in terms of theangular velocity of bar CD. We can repeat
the procedure to points B & C to obtain an
equation for vC in terms of the angular
velocity of bar BC.
Example 6: Angular Accelerations of
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p g
Members of a Linkage
By equating the two expressions for vC, we
will obtain a vector equation in twounknowns: the angular velocities of bars
BC and CD. Then, by following the same
sequence of steps, but using formula for
acceleration (Eq. 16-18) we can obtain the
angular accelerations of bars BC and CD.
Example 6: Angular Accelerations of
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p g
Members of a Linkage• Solution
The velocity of B is
( ) ( )
( )m/si20
j2k 100
r vv /
−=
×+=×+= A B AB A B ω
Fig a
Example 6: Angular Accelerations of
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p g
Members of a Linkage
Fig b
Example 6: Angular Accelerations of
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g
Members of a LinkageLet ωCD be the unknown angular velocity of
bar CD (Fig b). The velocity of C in terms ofthe velocity of D is
j2i2
022
00
k ji
0
r vv /
CDCD
CD
DC CD DC
ω ω
ω
ω
−−=
−
+=
×+=
Example 17.6: Angular Accelerations
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of Members of a Linkage
Fig c
Example 6: Angular Accelerations of
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Members of a LinkageDefining the angular velocity of bar BC by
ωBC
(Fig c), we obtain the velocity of C in
terms of the velocity of B:
( ) ( ) j2i20
i2k i20
r vv /
C
BC
BC BC BC
ω ω
ω
+−=
×+−=
×+=
Equating both eqns for vC yields
j2i20 j2i2 BC CDCD ω ω ω +−=−−
Example 6: Angular Accelerations of
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Members of a LinkageEquating the i and j components, we obtain
ωCD = 10 rad/s and ωBC = -10 rad/s.Using the same sequence to determine the
angular accelerations, the acceleration of B is
(Fig a) is
( ) ( ) ( ) ( )(m/s) j200i600
j210 j2k 3000
r r aa
2
/2
/
−=−×−+=
−×+= A B AB A B AB A B ω α
Example 6: Angular Accelerations of
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Members of a LinkageThe acceleration of C in terms of the
acceleration of D is (Fig b)
( ) ( )
( ) ( ) (m/s) j2200i2200
j2i210
022
00k ji
0
r r aa
2
/2
/
CDCD
CD
DC CD DC CD DC
α α
α
ω α
+−−=
+−−
−
+=
−×+=
Example 6: Angular Accelerations of
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Members of a LinkageThe acceleration of C in terms of the
acceleration of B is (Fig c)
( ) ( ) ( ) ( )( ) (m/s) j2200i400
i210i2k j200i600
r r aa
2
/2
/
BC
BC
BC BC BC BC BC
α
α
ω α
−−=−−×+−=
−×+=
Example 6: Angular Accelerations of
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Members of a LinkageEquating the expressions for aC, we obtain
( ) ( ) ( ) j2200i400 j2200i2200 C CDCD α α α −−=+−−
Equating i and j components, we obtain theangular accelerations αBC = 100 rad/s
2 and αCD =
-100 rad/s2 .
ATTENTION QUIZ
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1. Two bodies contact one another withoutslipping. If the tangential component of
the acceleration of point A on gear B is
100 cm/sec2, determine the tangential
component of the acceleration of point
A’ on gear C.
A) 50 cm/sec2 B) 100 cm/sec2
C) 200 cm/sec2 D) None of above.
2. If the tangential component of the acceleration of point A on
gear B is 100 cm/sec2, determine the angular acceleration of
gear B.
A) 50 rad/sec2 B) 100 rad/sec2
C) 200 rad/sec2 D) None of above.
2 m1 m
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