1. sports
2. football

ES 210
Dynamics
Week 15
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Reminder: The Final Exam is Thursday, 18 December 2014, 08:00 –
10:00 in Duckering 354.
o The Final is comprehensive only in that the material in this course
has built upon itself from the start of the semester. The topics
covered on the Final will be those covered since Exam #2.
This is rigid body kinematics, the Principle of Work & Energy
for rigid bodies, and the Principle of Impulse & Momentum
for rigid bogies.
This is the material covered in Weeks 10 thru 15, chapters
16, 17, 18, & 19 in the text, and the associated homework
assignments.
o The exam will be similar in format to the previous two exams.
However, this exam may be slightly longer since the final exam
period is 2 hours vs. the 1.5 hour period for the previous exams.
o The exam is closed book/closed notes. You will be allowed three
8½
x 11 crib sheets.
As usual, the crib sheets may contain
equations and conversion factors only.
They may not include
worked example problems nor explanations of procedures or other
narrative instructions..
Let’s look at a few more examples to demonstrate the Principle of Impulse
and Momentum for rigid bodies.
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ES 210
Dynamics
Week 15
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Problem:
Gear A has a mass of 10 kg and a radius of gyration of 200 mm, gear B has a
mass of 3 kg and a radius of gyration of 80 mm. The system is at rest when a 6
N-m moment is applied to gear B. (Note: this is the same gear system from one
of our previous examples.) Neglecting friction, determine (a) the time required
for the angular velocity of gear B to reach 600 rpm, (b) the tangential force which
gear B exerts on gear A.
Solution:
Note that this is the same problem I previously solved using the Principle of Work
and Energy.
I’ll apply the Principle of Angular Impulse & Momentum to each gear separately.
The forces and moments are constant so the impulses are these forces/moments
x the unknown time.. From the previous example using this gear train I had:
I A = 0.40kg ⋅ m 2
I B = 0.0192kg ⋅ m 2
(ω A )2 = 25.1 rad s
(ω B )2 = 62.8 rad s
Apply the Principle to gear A. Draw the diagram for the Principle:
If I take moments about the center (A), ccw+, the Principle can be stated
as:
− I A (ω A )1 − FtrA = − I A (ω A )2
week15.docx
ES 210
Dynamics
Week 15
(
⇒ 0 + Ft (0.25m ) = 0.40kg ⋅ m
2
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)(
)
 N ⋅ s2 
rad


25.1
s  kg ⋅ m 


⇒ Ft = 40.16 N ⋅ s
Next, I’ll apply the Principle to gear B taking moments about its center (ccw+):
− I B (ω B )1 + Mt − FtrB = I B (ω B )2
(
)(
⇒ 0 + (6 N ⋅ m )t − (40.16 N ⋅ s )(0.1m ) = 0.0192kg ⋅ m 2 62.8 rad
)s  kgN ⋅⋅sm 
2


⇒ t = 0.8703s
F (0.8703s ) = 40.16 N ⋅ s
⇒ F = 46.15 N
Note that this is the same answer I obtained analyzing this problem using
the Principle of Work & Energy during Week 13.
Problem:
A uniform sphere of mass, m, and radius, r, is projected along a rough surface
with a linear velocity of v1 and no angular velocity. Denoting by µk the coefficient
of kinetic friction between the sphere and the surface, determine (a) the time, t2,
at which the sphere will begin to roll without sliding, (b) the linear and angular
velocities at this time.
Solution:
While sliding on the surface, the sphere is subject to the normal force, N, the
friction force, F, and its own weight, W = mg. A diagram of this drawn in a
manner to aid the application of the Principle of Impulse & Momentum is:
week15.docx
ES 210
Dynamics
Week 15
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Note that the sphere will stop sliding and begin rolling when the velocity at C is
zero. At this time, C becomes the instantaneous zero center for rotation.
I’ll analyze the problem from time t = t1 when the sphere is placed in motion on
the table to time t = t2 when the sphere begins to roll without sliding.
From the above diagram:
∑ components
∑ components
∑M
Nt −Wt = 0
(1)
mv1 − Ft = mv2
(2)
Ftr = Iω 2
(3)
+↑
+→
G
cw+
From (1)
N = W = mg
While the sphere is sliding, at C I have F = µ k N = µ k mg . Substituting this
into (2):
mv1 − µ k mgt = mv 2
⇒ v 2 = v1 − µ k gt
Substituting the expression for F and noting that I =
(4)
2 2
mr for a sphere
5
into (3):
2 2
mr ω 2
5
5 µk g
⇒ ω2 =
t
2 r
µ k mgtr =
(5)
When the sphere is rolling without sliding v2 = rω 2 . Substituting from (4) &
(5), this gives me:
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ES 210
Dynamics
Week 15
v1 − µ k gt = r
⇒t =
5 of 11
5 µk g
t
2 r
2 v1
7 µk g
Now, back substituting for t, I obtain:
ω2 =
5 µ k g  2 v1  5 v1

=
2 r  7 µ k g  7 r
5 v  5
v 2 = r  1  = v1
7 r  7
Now let’s look at a couple of examples of using the Principle of Impulse &
Momentum to solve eccentric impact problems.
Problem:
A 0.05 lbf bullet, B, is fired with a horizontal velocity of 1500 ft/s into the side of a
20 lbf square panel suspended from a hinge at A. If the panel is initially at rest,
determine (a) the angular velocity of the panel immediately after the bullet
becomes embedded, (b) the impulsive reaction at A assuming that the bullet
becomes embedded in 0.0006s.
Solution:
I’ll treat the panel and bullet as a single system. Since the time interval is very
short, I’ll neglect all non-impulsive external forces. This leaves me with just the
reaction impulses at the hinge.
Start by drawing a diagram representing the Principle of Impulse & Momentum:
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ES 210
Dynamics
Week 15
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Next, sum the components:
∑M
∑ components
∑ components
A
ccw +
 14 
9 
mB v B 
ft  + 0 = m p v P 
ft  + I P ω 2
 12 
 12 
m B v B + Ax t = m P v 2
(1)
(2)
+→
0 + Ay t = 0
(3)
+↑
I need the moment of inertia of the panel:

1
1  20lbf
2
I P = mP b = 
6
6  32.2 ft
s2


 18  2
2
 ft  = 0.2329lbf ⋅ ft ⋅ s
 12 

9 
Since the panel pivots about A: v 2 = rω 2 =  ft ω 2 . Substituting these last
 12 
two facts and the given data into (1):

 0.05lbf

 32.2 ft 2
s



ft  14   20  9 ω  9  + 0.2329lbf ⋅ ft ⋅ s 2 ω

 
1500 s  ft  = 
2
 12   32.2  12 2  12 


⇒ ω 2 = 4.667 rad
s
(
(
)
)
9 
⇒ v2 = 
ft  4.667 rad = 3.50 ft
s
s
 12 
Substituting into (2):


 0.05lbf 
 20 
ft
(3.50 )
1500 s  + Ax (0.0006 s ) = 



 32.2 
 32.2 ft 
s 

⇒ Ax = −258.8lbf = 258.8lbf ←
week15.docx
Week 15
ES 210
Dynamics
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From (3):
Ay = 0
Problem:
A 2 kg sphere moving horizontally to the right with an initial velocity of 5 m/s
strikes the lower end of an 8 kg rigid rod, AB. The rod is suspended from a hinge
at A and is initially at rest. If the coefficient of restitution between the rod and the
sphere is 0.80, determine the angular velocity of the rod and the velocity of the
sphere immediately after impact.
Solution:
i’ll again apply the Principle of Impulse & Momentum. i’ll treat the rod and sphere
as a single system. This means that the only external impulsive force(s) for the
system are the reaction forces at the support, A.
Start by diagramming the Principle:
Take moments about A:
m S v S (1.2m ) + 0 = m S′ v S (1.2m ) + m R v ′R (0.6m ) + Iω ′
(1)
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ES 210
Dynamics
Week 15
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Since the rob rotates about A: v′R = rω ′ = 0.6mω ′ . The moment of inertia for
the rod is:
I=
1
1
2
ml 2 = (8kg )(1.2m ) = 0.960kg ⋅ m 2
12
12
Substituting these facts and the given data into (1):
(2kg )(5 m s )(1.2m ) = (2kg )v′S (1.2m ) + (8kg )(0.6mω ′)(0.6m ) + (0.96kg ⋅ m 2 )ω ′
2
12 kg ⋅ m
s
(
)
= (2.4kg ⋅ m )v ′S + 3.84kg ⋅ m ω ′
2
(2)
From the coefficient of restitution:
v ′B − v ′S
vS − vB
v ′ − v ′S
0 .8 = B
5m − 0
s
⇒ v ′B − v ′S = 4.0 m
e=
( )
(3)
s
Since the rod rotates about A: v′B = 1.2mω ′
(4)
From (2) & (4):
 v′ 
= (2.4kg ⋅ m )v ′S + 3.84kg ⋅ m 2  B 
 1 .2 m 
⇒ 12 = 2.4v S′ + 3.2v ′B
12 kg ⋅ m
(
2
s
)
(5)
Solving (3) & (5), we obtain:
v ′B = 3.857 m
s
v ′S = −0.143 m
s
⇒ ω ′ = 3.214 rad
s
week15.docx
ES 210
Dynamics
Week 15
9 of 11
Problem:
The garage door CD has a mass of 50 kg and can be treated as a thin
plate. Determine the required unstretched length of each of the two side
springs when the door is in the open position, so that when the door falls
freely from the open position it comes to rest when it reaches the fully
closed position, i.e., when AC rotates 180°. Each of the two side springs
has a stiffness of k = 350 N-m. Neglect the mass of the side bars AC.
Solution:
Since I’m not given any information about rolling resistance or frictional
forces, the only force involve is the conservative force due to gravity.
Therefore, I’ll use Conservation of Energy to solve this problem.
(ke )1 + ( pe )1 = (ke)2 + ( pe )2
Note that the door starts from rest and ends at rest. Therefore, the initial
and final kinetic energy are zero. I’ll use the center of mass of the door as
the point through which the weight acts for calculating the potential energy
terms.
W = mg
(
)
(
)
(
)
 N ⋅ s2 
1

1
2

0 + 2  350 N x 2  = 0 + 2  350 N ( x + 1)  − (50kg ) 9.81 m 2 (1m )
m 
m
s
2
2

 kg ⋅ m 
⇒ x = 0.201m
So, the unstretched length of the spring is:
x Spr = 0.5m − 0.201m = 0.299m = 299mm
week15.docx
ES 210
Dynamics
Week 15
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Problem:
If the motor, M, exerts a constant force of on the cable wrapped around
the reel’s outer rim, determine the velocity of the 50-kg cylinder after it has
traveled a distance of 2m. Initially, the system is at rest. The reel has a
mass of 25 kg, and the radius of gyration about its center of mass A is kA =
125 mm.
Solution:
I’ll use the Principle of Work & Energy to solve this problem. Consider the
FBD of the mechanism:
The reel rotates about its fixed central axis:
⇒ vC = ω r rC
⇒ ωr =
vC
vC
=
= 13.33vC
rC 0.075m
I A = m r k A2 = (25kg )(0.125m ) = 390.6 × 10 −3 kg ⋅ m 2
2
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ES 210
Dynamics
Week 15
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The system is initially at rest. Therefore, its initial kinetic energy is zero.
Its final kinetic energy is:
(ke)2 = (ke )r ,2 + (ke)C ,2
(ke)2
(ke)2
1
1
I Aω r2 + mC vC2
2
2
1
1
2
= 390.6 × 10 −3 kg ⋅ m 2 (13.33vC ) + (50kg )vC2 = 59.72vC2
2
2
=
(
)
Next consider the work done. From the FBD, we note that Ax, Ay, & Wr act
thru the center of rotation which does not translate. Therefore, they do no
work. Forces P and WC do work (they move thru a distance) counter to
each other. I’ll establish a sign convention such that P does positive work
and W C does negative work since the cylinder is moving upward (against
gravity).
I need to relate the linear travel of the cable to the rotation of the reel:
θ=
sC
2m
=
= 26.67 rad
rC 0.075m
⇒ s P = rPθ = (0.15m )(26.67 ) = 4.0m
The work done by these forces is:
U P = Ps P = (300 N )(4m ) = 1200 N ⋅ m
(
U WC = −WC s C = −(50kg ) 9.81 m
 N ⋅s 
 = −981.0 N ⋅ m
)
(
2m )
s
 kg ⋅ m 
2
2
Applying the Principle of Work & Energy:
(ke )1 + ∑U = (ke)2
1→ 2
0 + (1200 − 981.0 )N ⋅ m = 59.72vC2
⇒ vC = 1.92 m ↑
s
week15.docx