Download Today`s Objectives: Students will be able to: a) Describe the velocity

RELATIVE MOTION ANALYSIS: VELOCITY (Section 16.5)
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.
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.
=
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 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.
RELATIVE MOTION ANALYSIS: VELOCITY (continued)
vB = vA + ω 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.
RELATIVE MOTION ANALYSIS: VELOCITY (continued)
vB = vA + ω 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.
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.
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 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.
3. If the solution yields a negative answer, the sense of
direction of the vector is opposite to that assumed.
EXAMPLE 1
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 (continued)
Solution:
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
Equating the i and j components gives:
vB = 0.2 ω cos 45
0 = -2 + 0.2 ω sin 45
Solving:
ω = 14.1 rad/s or ωAB = 14.1 rad/s k
vB = 2 m/s or vB = 2 m/s i
EXAMPLE 2
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 (continued)
Solution:
Link CB. Write the relative-velocity
equation:
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
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
EXAMPLE 2 (continued)
y
x
Link AB experiences only rotation about
A. Since vB is known, there is only one
equation with one unknown to be found.
vB = ωAB x rB/A
2 i = ωAB k x (-0.2 j )
2 i = 0.2 ωAB i
By comparing the i-components:
2 = 0.2 ωAB
So, ωAB = 10 rad/s k
CONCEPT QUIZ
ω
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 ft
O
A
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
V=15 ft/s
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:
1) First draw the kinematic diagram of link AB.
B
•
rB/A
ωAB
A
Link AB rotates about a fixed axis at
A. Since ωAB 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 + ωAB 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
GROUP PROBLEM SOLVING (continued)
2) Now consider link BC.
C
vC
500 mm
ωBC
rC/B
y
B
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.
vB
x
Applying the relative velocity equation:
vC = vB + ωBC x rC/B
-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
i: 0 = -0.433ωBC => ωBC = 0
j: -vC = -50 + 0.25ωBC => vC = 50
vC = -50 j m/s