Transcript Lecture Notes for Section 17.4 - North Dakota State University

EQUATIONS OF MOTION: ROTATION ABOUT A
FIXED AXIS (Section 17.4)
Today’s Objectives:
Students will be able to
analyze the planar kinetics of a
rigid body undergoing
rotational motion.
In-Class Activities:
• Check homework, if any
• Reading quiz
• Applications
• Rotation about an axis
• Equations of motion
• Concept quiz
• Group problem solving
• Attention quiz
READING QUIZ
1. In rotational motion, the normal component of acceleration
at the body’s center of gravity (G) is always
A) zero.
B) tangent to the path of motion of G.
C) directed from G toward the center of rotation.
D) directed from the center of rotation toward G.
2. If a rigid body rotates about point O, the sum of the
moments of the external forces acting on the body about
point O equals
A) IGa
B) IOa
C) m aG
D) m aO
APPLICATIONS
The crank on the oil-pump rig
undergoes rotation about a fixed axis,
caused by the driving torque M from a
motor.
As the crank turns, a dynamic reaction
is produced at the pin. This reaction is
a function of angular velocity, angular
acceleration, and the orientation of the
crank.
Pin at the center of
rotation.
If the motor exerts a constant torque M on
the crank, does the crank turn at a constant
angular velocity? Is this desirable for such
a machine?
APPLICATIONS (continued)
The “Catherine wheel” is a fireworks
display consisting of a coiled tube of
powder pinned at its center.
As the powder burns, the mass of
powder decreases as the exhaust gases
produce a force directed tangent to the
wheel. This force tends to rotate the
wheel.
If the powder burns at a constant rate, the exhaust gases
produce a constant thrust. Does this mean the angular
acceleration is also constant? Why or why not? What is
the resulting effect on the fireworks’ display?
EQUATIONS OF MOTION FOR PURE ROTATION
When a rigid body rotates about a fixed axis
perpendicular to the plane of the body at
point O, the body’s center of gravity G moves
in a circular path of radius rG. Thus, the
acceleration of point G can be represented by
a tangential component (aG)t = rG a and a
normal component (aG)n = rG w2.
Since the body experiences an angular acceleration, its inertia
creates a moment of magnitude IGa equal to the moment of
the external forces about point G. Thus, the scalar equations
of motion can be stated as:
 Fn = m (aG)n = m rG w2
 Ft = m (aG)t = m rG a
 MG = IG a
EQUATIONS OF MOTION (continued)
Note that the MG moment equation may be replaced by a
moment summation about any arbitrary point. Summing the
moment about the center of rotation O yields
MO = IGa + rG m (aG) t = (IG + m (rG)2 ) a
From the parallel axis theorem, IO = IG + m(rG)2, therefore
the term in parentheses represents IO. Consequently, we can
write the three equations of motion for the body as:
Fn = m (aG) n = m rG w2
Ft = m (aG) t = m rG a
MO = IO a
PROCEDURE FOR ANALYSIS
Problems involving the kinetics of a rigid body rotating about
a fixed axis can be solved using the following process.
1. Establish an inertial coordinate system and specify the sign and
direction of (aG)n and (aG)t.
2. Draw a free body diagram accounting for all external forces
and couples. Show the resulting inertia forces and couple
(typically on a separate kinetic diagram).
3. Compute the mass moment of inertia IG or IO.
4. Write the three equations of motion and identify the
unknowns. Solve for the unknowns.
5. Use kinematics if there are more than three unknowns (since
the equations of motion allow for only three unknowns).
EXAMPLE
Given:A rod with mass of 20
kg is rotating at 5 rad/s
at the instant shown. A
moment of 60 N·m is
applied to the rod.
Find: The angular acceleration a and the reaction at pin O when
the rod is in the horizontal position.
Plan: Since the mass center, G, moves in a circle of radius
1.5 m, it’s acceleration has a normal component toward O
and a tangential component acting downward and
perpendicular to rG. Apply the problem solving procedure.
EXAMPLE (continued)
Solution:
FBD & Kinetic Diagram
Equations of motion:
+ Fn = man = mrGw2
On = 20(1.5)(5)2 = 750 N
+ Ft = mat = mrGa
-Ot + 20(9.81) = 20(1.5)a
+ MO = IG a + m rG a (rG)
Using IG = (ml2)/12 and rG = (0.5)(l), we can write:
MO = a[(ml2/12) + (ml2/4)] = (ml2/3)a where (ml2/3) = IO.
After substituting:
60 + 20(9.81)(1.5) = 20(32/3)a
Solving: a = 5.9 rad/s2
Ot = 19 N
CONCEPT QUIZ
•
q
O
l
1. If a rigid bar of length l (above) is released from rest in the
horizontal position (q = 0), the magnitude of its angular
acceleration is at maximum when
A) q = 0
B) q = 90
C) q = 180
D) q = 0 and 180
2. In the above problem, when q = 90°, the horizontal
component of the reaction at pin O is
A) zero
B) m g
C) m (l/2) w2
D) None of the above.
GROUP PROBLEM SOLVING
Given: Wdisk = 15 lb,
Wrod = 10 lb,
w = 8 rad/s at
this instant.
Find: The horizontal and vertical components of the
reaction at pin O when the rod is horizontal.
Plan: Draw the free body diagram and kinetic diagram of
the rod and disk as one unit. Then apply the
equations of motion.
GROUP PROBLEM SOLVING (continued)
Solution:
2
m
(3.75)(8)
d
15 lb 10 lb
(IG)ra
(IG)da
mr(1.5)(8)2
Ox
=
Oy
md(3.75a) mr(1.5a)
Equations of motion:
Fx = m(aG)x: Ox = (15/32.2)(3.75)(8)2 + (10/32.2)(1.5)(8)2
Ox = 142 lb
Fy = m(aG)y: Oy –15 – 10 = -(15/32.2)(3.75a) – (10/32.2)(1.5a)
MO = Ioa: 15(3.75) + 10(1.5) = [0.5(15/32.2)(0.75)2
+ (15/32.2)(3.75)2 ]diska
+ [(1/12)(10/32.2)(3)2
Therefore,
+ (10/32.2)(1.5)2]roda
a = 9.36 rad/s2, Oy = 4.29 lb
(a)
m
ATTENTION QUIZ
(b)
m
T
10 lb
10 lb
1. A drum of mass m is set into motion in two ways: (a) by a
constant 10 lb force, and, (b) by a block of weight 10 lb. If aa
and ab represent the angular acceleration of the drum in each
case, select the true statement.
A) aa > ab
B) aa < ab
C) aa = ab
D) None of the above.
2. For the same problem, the tension T in the cable in case (b) is
A) T = 10 lb
B) T < 10 lb
C) T > 10 lb
D) None of the above.