Transcript Lecture Notes for Section 16.5 (General Plane Motion

RELATIVE MOTION ANALYSIS: VELOCITY
Today’s Objectives:
Students will be able to:
1. Describe the velocity of a rigid body
in terms of translation and rotation
components.
2. Perform a relative-motion velocity
analysis of a point on the body.
In-Class Activities:
• Translation and
Rotation
Components of
Velocity
• Relative Velocity
Analysis
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
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, , be found?
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
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?
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
RELATIVE MOTION ANALYSIS (Section 16.5)
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
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
RELATIVE MOTION ANALYSIS: VELOCITY
=
+
The velocity at B is given as : (drB/dt) = (drA/dt) + (drB/A/dt) or
vB = vA + vB/A
Since the body is taken as rotating about A,
vB/A = drB/A/dt =  x rB/A
Here  will only have a k component since the axis of rotation
is perpendicular to the plane of translation.
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
RELATIVE MOTION ANALYSIS: VELOCITY
vB = vA +  x rB/A
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.
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
RELATIVE MOTION ANALYSIS: VELOCITY
(continued)
vB = vA +  x 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.
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
PROCEDURE FOR ANALYSIS
The relative velocity equation can be applied using either a
Cartesian vector analysis or by writing scalar x and y component
equations directly.
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 and by using the kinematic
diagram, underneath each term represent the vectors
graphically by showing their magnitudes and directions.
3. Write the scalar equations from the x and y components of
these graphical representations of the vectors. Solve for
the unknowns.
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
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 . If the magnitudes are unknown, the sense of
direction may be assumed.
2. Express the vectors in Cartesian vector form and substitute
into vB = vA +  x 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.
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
READING QUIZ
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)  x rB/A .
D) All of the above.
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
EXAMPLE
Given: Block A is moving down
at 2 m/s.
Find: The velocity of B at the
instant  = 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 +  x rB/A.
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
EXAMPLE
(continued)
Solution:
vB = vA + AB x rB/A
vB i = -2 j + ( k x (0.2 sin 45 i - 0.2 cos 45 j ))
vB i = -2 j + 0.2  sin 45 j + 0.2  cos 45 i
Equating the i and j components gives:
vB = 0.2  cos 45
0 = -2 + 0.2  sin 45
Solving:
 = 14.1 rad/s or AB = 14.1 rad/s k
vB = 2 m/s or vB = 2 m/s i
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
EXAMPLE II
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
+ CB x rB/C. (Why do CB first?) Then do a similar
process for link AB.
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
Solution:
EXAMPLE II
(continued)
Link CB. Write the relative-velocity
equation:
vB = vC + CB x rB/C
vB i = -2 j + CB k x (0.2 i - 0.2 j )
vB i = -2 j + 0.2 CB j + 0.2 CB i
By comparing the i, j components:
i: vB = 0.2 CB
=> vB = 2 m/s i
j: 0 = -2 + 0.2 CB
=> CB = 10 rad/s k
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
EXAMPLE II
(continued)
Link AB experiences only rotation
about A. Since vB is known, there is
only one equation with one unknown
to be found.
vB = AB x rB/A
2 i = AB k x (-0.2 j )
By comparing the i-components:
2 i = 0.2 AB i
2 = 0.2 AB
So, AB = 10 rad/s k
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
CONCEPT QUIZ

1. If the disk is moving with a velocity at point
O of 15 ft/s and  = 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 ft
V=15 ft/s
O
A
2. If the velocity at A is zero, then determine the angular
velocity, .
A) 30 rad/s
B) 0 rad/s
C) 7.5 rad/s
D) 15 rad/s
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
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.
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
GROUP PROBLEM SOLVING
(continued)
Solution:
1) First draw the kinematic diagram of link AB.
B
•
rB/A
AB
A
Link AB rotates about a fixed axis at
A. Since AB is ccw, vB will be
directed down, so vB = -vB j.
•
100 mm y
vB
x
Applying the relative velocity equation with vA = 0:
vB = vA + AB x rB/A
-vB j = (500 k) x (-0.1 i + 0 j)
-vB j = -50 j + 0 i
Equating j components: vB = 50
vB = -50 j m/s“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
GROUP PROBLEM SOLVING
(continued)
2) Now consider link BC.
C
Since point C is attached to the
vC
piston, vC must be directed up or
500 mm
down. It is assumed here to act
rC/B
BC
down, so vC = -vC j. The unknown
y
B
sense of BC is assumed here to be
ccw: BC = BC k.
vB
x
Applying the relative velocity equation:
vC = vB + BC x rC/B
-vC j = -50 j + (BC k) x (0.5 cos60 i + 0.5 sin60 j)
-vC j = -50 j + 0.25BC j – 0.433BC i
i: 0 = -0.433BC => BC = 0
j: -vC = -50 + 0.25BC => vC = 50
vC = -50 j m/s“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
ATTENTION QUIZ
vA
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 + gear x rB/A
B) vA = vC + gear x rA/C
C) vB = vC + gear x rC/B
D) vA = vC + gear x rC/A
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  = 60º?
A) A
B) B
C) C
D) No difference.
B
4m
A

C
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU