Lecture16 5

Today’s Objectives:
Students will be able to:
a) Describe the velocity of a rigid
body in terms of translation
and rotation components.
b) Perform a relative-motion
velocity analysis of a point on
the body.
RELATIVE MOTION ANALYSIS: VELOCITY (Section 16.5)
In-Class Activities:
• Check homework, if any
• Reading quiz
• Applications
• Translation and rotation
components of velocity
• Relative velocity
analysis
• Concept quiz
• Group problem solving
• Attention quiz

READING QUIZ
1. When a relative-motion analysis involving two sets of
coordinate axes is used, the x’ - y’ coordinate system will
A) be attached to the selected point for analysis.
B) rotate with the body.
C) not be allowed to translate with respect to the fixed frame.
D) None of the above.
2. In the relative velocity equation, vB/A is
A) the relative velocity of B with respect to A.
B) due to the rotational motion.
C) ω x rB/A .
D) All of the above.

APPLICATIONS
As the slider block A moves horizontally to the left with vA,
it causes the link CB to rotate counterclockwise. Thus vB is
directed tangent to its circular path.
Which link is undergoing general plane motion?
How can its angular velocity, ω, be found?

APPLICATIONS (continued)
Planetary gear systems are used
in many automobile automatic
transmissions. By locking or
releasing different gears, this
system can operate the car at
different speeds.
How can we relate the angular velocities of the
various gears in the system?

RELATIVE MOTION ANALYSIS: DISPLACEMENT
When a body is subjected to general plane motion, it undergoes a
combination of translation and rotation.
=
drB = drA + drB/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:

The velocity at B is given as : (drB/dt) = (drA/dt) + (drB/A/dt)or
vB = vA + vB/A
RELATIVE MOTION ANALYSIS: VELOCITY
= +
Since the body is taken as rotating about A,
vB/A = drB/A/dt = ω x rB/A
Here ω will only have a k component since the axis of rotation
is perpendicular to the plane of translation.

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 x rB/A
RELATIVE MOTION ANALYSIS: VELOCITY (continued)

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 x rB/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.

PROCEDURE FOR ANALYSIS
3. Write the scalar equations from the x and y components of
these graphical representations of the vectors. Solve for
the 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 vectors
graphically by showing their magnitudes and directions.
The relative velocity equation can be applied using either a
Cartesian 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 x rB/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,
rB/A and ω. If the magnitudes are unknown, the sense of
direction may be assumed.
PROCEDURE FOR ANALYSIS (continued)

Given: Block A is moving down
at 2 m/s.
Find: The velocity of B at the
instant θ = 45°.
Plan: 1. Establish the fixed x-y directions and draw a
kinematic diagram.
2. Express each of the velocity vectors in terms of
their i, j, k components and solve vB = vA + ω x rB/A.
EXAMPLE 1

EXAMPLE 1 (continued)
Equating the i and j components gives:
vB = 0.2 ω cos 45
0 = -2 + 0.2 ω sin 45
vB = vA + ωAB x rB/A
vB i= -2 j + (ω k x (0.2 sin 45 i - 0.2 cos 45 j ))
vB i= -2 j + 0.2 ω sin 45 j + 0.2 ω cos 45 i
Solution:
Solving:
ω = 14.1 rad/s or ωAB = 14.1 rad/s k
vB = 2 m/s or vB = 2 m/s i

Given:Collar C is moving
downward with a velocity of
2 m/s.
Find: The angular velocities of CB
and AB at this instant.
Plan: Notice that the downward motion of C causes B to move to
the right. Also, CB and AB both rotate counterclockwise.
First, draw a kinematic diagram of link CB and use vB = vC +
ωCB x rB/C. (Why do CB first?) Then do a similar process
for link AB.
EXAMPLE 2

Link CB. Write the relative-velocity
equation:
By comparing the i, j components:
i: vB = 0.2 ωCB => vB = 2 m/s i
j: 0 = -2 + 0.2 ωCB => ωCB = 10 rad/s k
vB = vC + ωCB x rB/C
vB i = -2 j + ωCB k x (0.2 i - 0.2 j )
vB i = -2 j + 0.2 ωCB j + 0.2 ωCB i
Solution:

Link AB experiences only rotation about
A. Since vB is known, there is only one
equation with one unknown to be found.
By comparing the i-components:
2 = 0.2 ωAB
So, ωAB = 10 rad/s k
x
y
vB = ωAB x rB/A
2 i = ωAB k x (-0.2 j )
2 i = 0.2 ωAB i

CONCEPT QUIZ
1. If the disk is moving with a velocity at point
O of 15 ft/s and ω = 2 rad/s, determine the
velocity at A.
A) 0 ft/s B) 4 ft/s
C) 15 ft/s D) 11 ft/s
2. If the velocity at A is zero, then determine the angular
velocity, ω.
A) 30 rad/s B) 0 rad/s
C) 7.5 rad/s D) 15 rad/s
2 ft
V=15 ft/s
ω
A
O

GROUP PROBLEM SOLVING
Given: The crankshaft AB is rotating at
500 rad/s about a fixed axis
passing through A.
Find: The speed of the piston P at the
instant it is in the position shown.
Plan: 1) Draw the kinematic diagram of each link showing
all pertinent velocity and position vectors.
2) Since the motion of link AB is known, apply the
relative velocity equation first to this link, then link BC.

GROUP PROBLEM SOLVING (continued)
Solution:
-vB j = (500 k) x (-0.1 i + 0 j)
-vB j = -50 j + 0 i
1) First draw the kinematic diagram of link AB.
Link AB rotates about a fixed axis at
A. Since ωAB is ccw, vB will be
directed down, so vB = -vB j.
Equating j components: vB = 50
vB = -50 j m/s
Applying the relative velocity equation with vA = 0:
vB = vA + ωAB x rB/A
ωAB
y
x
100 mm
rB/A
vB
B A
• •

GROUP PROBLEM SOLVING (continued)
-vC j = -50 j + (ωBC k) x (0.5 cos60 i + 0.5 sin60 j)
-vC j = -50 j + 0.25ωBC j – 0.433ωBC i
2) Now consider link BC.
Since point C is attached to the
piston, vC must be directed up or
down. It is assumed here to act
down, so vC = -vC j. The unknown
sense of ωBC is assumed here to be
ccw: ωBC = ωBC k.
i: 0 = -0.433ωBC => ωBC = 0
j: -vC = -50 + 0.25ωBC => vC = 50
vC = -50 j m/s
Applying the relative velocity equation:
vC = vB + ωBC x rC/B
y
x
500 mm
rC/B
vB
ωBC
B
C
vC

ATTENTION QUIZ
1. Which equation could be used to find
the velocity of the center of the gear, C,
if the velocity vA is known?
A) vB = vA + ωgear x rB/A B) vA = vC + ωgear x rA/C
C) vB = vC + ωgear x rC/B D) vA = vC + ωgear x rC/A
vA
2. If the bar’s velocity at A is 3 m/s, what
“base” point (first term on the RHS of the
velocity equation) would be best used to
simplify finding the bar’s angular
velocity when θ = 60º?
A) A B) B
C) C D) No difference.
A
4 m
B
θ
C

Lecture16 5

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