Transcript Lecture Notes for Section 16.4 (Absolute Motion Analysis)

ABSOLUTE MOTION ANALYSIS
Today’s Objective:
Students will be able to:
1. Determine the velocity and acceleration of a rigid body undergoing
general plane motion using an absolute motion analysis.
In-Class Activities:
• General Plane Motion
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
APPLICATIONS
The position of the piston, x, can be defined as a function of
the angular position of the crank, q. By differentiating x with
respect to time, the velocity of the piston can be related to the
angular velocity, w, of the crank.
The stroke of the piston is defined as the total distance moved
by the piston as the crank angle varies from 0 to 180°. How
does the length of crank AB affect the stroke?
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
APPLICATIONS
(continued)
The rolling of a cylinder is an example of general plane motion.
During this motion, the cylinder rotates clockwise while it
translates to the right.
The position of the center, G, is related to the angular position, q, by,
sG = r q, if the cylinder rolls without slipping.
Can you relate the translational velocity of G and the angular
velocity of the cylinder?
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
ABSOLUTE MOTION ANALYSIS
(Section 16.4)
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, q
(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(q) is established, the velocity and acceleration of point P are
obtained in terms of the angular velocity, w, and angular
acceleration, a, 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 derivatives of the
position coordinate equation.
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
READING QUIZ
1. A body subjected to general plane motion undergoes a/an
A) translation.
B) rotation.
C) simultaneous translation and rotation.
D) out of plane movement.
2. In general plane motion, if the rigid body is represented by a
slab, the slab rotates
A) about an axis perpendicular to the plane.
B) about an axis parallel to the plane.
C) about an axis lying in the plane.
D) None of the above.
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
EXAMPLE
Given:Two slider blocks are
connected by a rod of
length 2 m. Also,
vA = 8 m/s and aA = 0.
Find: Angular velocity, w, and
angular acceleration, a, of
the rod when q = 60°.
Plan: Choose a fixed reference point and define the position of
the slider A in terms of the parameter q. Notice from the
position vector of A, positive angular position q is
measured clockwise.
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
EXAMPLE
(continued)
Solution:
By geometry, sA = 2 cos q
reference
q
A
By differentiating with respect to time,
vA = -2 w sin q
sA
Using q = 60° and vA = 8 m/s and solving for w:
w = 8/(-2 sin 60°) = - 4.62 rad/s
(The negative sign means the rod rotates counterclockwise as
point A goes to the right.) Differentiating vA and solving for a,
aA = -2a sin q – 2w2 cos q = 0
a = - w2/tan q = -12.32 rad/s2
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
EXAMPLE II
Given:Crank AB rotates
at a constant
velocity of w =
150 rad/s
Find: Velocity of P
when q = 30°
Plan: Define x as a function of q and differentiate with
respect to time.
Solution: xP = 0.2 cos q + (0.75)2 – (0.2 sin q)2
vP = -0.2w sin q + (0.5)[(0.75)2
– (0.2sin q)2]-0.5(-2)(0.2sin q)(0.2cos q)w
vP = -0.2w sin q – [0.5(0.2)2sin2q w] / (0.75)2 – (0.2 sin q)2
At q = 30°, w = 150 rad/s and vP = -18.5 ft/s = 18.5 ft/s
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
CONCEPT QUIZ
1. If the position, s, is given as a function of angular position,
q, by s = 10 sin 2q, the velocity, v, is
A) 20 cos 2q
B) 20 sin 2q
C) 20 w cos 2q
D) 20 w sin 2q
2. If s = 10 sin 2q, the acceleration, a, is
A) 20 a sin 2q
B) 20 a cos 2q - 40 w2 sin 2q
C) 20 a cos 2q
D) -40 a sin2 q
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
GROUP PROBLEM SOLVING
Given: The w and a of the disk and
the dimensions as shown.
Find: The velocity and acceleration
of cylinder B in terms of q.
Plan: Relate s, the length of cable
between A and C, to the
angular position, q. The
velocity of cylinder B is
equal to the time rate of
change of s.
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
GROUP PROBLEM SOLVING
(continued)
Solution:
Law of cosines:
s = (3)2 + (5)2 – 2(3)(5) cos q
vB = (0.5)[34 – 30 cosq]-0.5(30 sinq)w
vB = [15 sin q w]/
34 – 30 cos q
(15w2 cosq + 15a sinq) (-0.5)(15w sinq)(30w sinq)
+
aB =
34 - 30cosq
(34 - 30cosq)3/2
2cosq + asinq)
2sin2q
15(w
225w
aB =
(34 - 30 cosq)0.5
(34 - 30 cosq)3/2
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU
ATTENTION QUIZ
1. The sliders shown below are confined to move in the
horizontal and vertical slots. If vA=10 m/s, determine the
connecting bar’s angular velocity when q = 30.
A) 10 rad/s
B) 10 rad/s
C) 8.7 rad/s
D) 8.7 rad/s
2. If vA=10 m/s, determine the angular
acceleration, a, when q = 30.
A) 0 rad/s2
B) -50.2 rad/s2
C) -112 rad/s2
D) -173 rad/s2
vA=10 m/s
“Dynamics by Hibbeler,” Dr. S. Nasseri, MET Department, SPSU