Transcript Lecture Notes for Section 16.4 (Absolute Motion Analysis)

ABSOLUTE MOTION ANALYSIS (Section 16.4)
Today’s Objective:
Students will be able to
determine the velocity and
acceleration of a rigid body
undergoing general plane
motion using an absolute
motion analysis.
In-Class Activities:
• Check homework, if any
• Reading quiz
• Applications
• General Plane Motion
• Concept quiz
• Group problem solving
• Attention quiz
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.
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?
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?
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.
EXAMPLE 1
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.
EXAMPLE 1 (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
EXAMPLE 2
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
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
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.
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
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