Transcript Kinetics of Rigid Bodies - Energy

Tenth Edition
CHAPTER
17
VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Ferdinand P. Beer
E. Russell Johnston, Jr.
Phillip J. Cornwell
Lecture Notes:
Brian P. Self
California State Polytechnic University
Plane
MotionofofRigid
Rigid
Plane Motion
Bodies:Energy
Bodies:
Energy and Momentum
Methods
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
Introduction
To predict the launch from a catapult, you
must apply the principle of work-energy.
To determine the forces acting on the stopper pin when the
catapult reaches its final position, angular impulse momentum
equations are used.
17 - 2
Introduction
• Method of work and energy and the method of impulse and
momentum will be used to analyze the plane motion of rigid
bodies and systems of rigid bodies.
• Principle of work and energy is well suited to the solution of
problems involving displacements and velocities.
T1  U12  T2
• Principle of impulse and momentum is appropriate for
problems involving velocities and time.
t2 
t2 




H O 1    M O dt  H O 2
L1    Fdt  L2
t1
t1
• Problems involving eccentric impact are solved by supplementing
the principle of impulse and momentum with the application of
the coefficient of restitution.
17 - 3
Introduction
Approaches to Rigid Body Kinetics Problems
Forces and
Accelerations
Velocities and
Displacements
Velocities and
Time
Newton’s Second
Law (last chapter)
Work-Energy
ImpulseMomentum
å F = ma
åM = H
t2
G
G
2-4
G
T1  U12  T2
mv1   F dt  mv2
t1
t2
I G1   M G dt  I G2
t1
Principle of Work and Energy for a Rigid Body
•Work and kinetic energy are scalar quantities.
• Assume that the rigid body is made of a large
number of particles.
T1  U12  T2
T1 , T2  initial and final total kinetic energy of
particles forming body
U12  total work of internal and external
forces acting on particles of body.
• Internal forces between particles A
and B are equal and opposite.
• Therefore, the net work of internal
forces is zero.
17 - 5
Work of Forces Acting on a Rigid Body
• Work of a force during a displacement of its
point of application,
A2 
s
 2
U12   F  dr   F cos ds
A1
s1


• Consider the net work of two forces
F and  F

forming a couple of moment M during a
displacement of their points of application.
     
dU  F  dr1  F  dr1  F  dr2
 F ds2  Fr d
 M d
2
U12   M d
1
 M  2  1  if M is constant.
17 - 6
Kinetic Energy of a Rigid Body in Plane Motion
• Consider a rigid body of mass m in plane motion consisting of individual
particles i. The kinetic energy of the body can then be expressed as:
T  12 mv 2  12  Δmi vi2
 12 mv 2  12   ri2 Δmi   2
 12 mv 2  12 I  2
• Kinetic energy of a rigid body can be
separated into:
- the kinetic energy associated with the
motion of the mass center G and
- the kinetic energy associated with the
rotation of the body about G.
T  12 mv 2 
17 - 7
1
2
I 2
Translation + Rotation
Kinetic Energy of a Rigid Body in Plane Motion
• Consider a rigid body rotating about a fixed axis through O.
T
1
2
 Δm v 
2
i i
1
2
2
2
1
Δ
m
r


r
Δ
m





 i i
i
2  i
2
 12 I O 2
• This is equivalent to using:
T  12 mv 2 
1
2
I 2
• Remember to only use
T  12 I O 2
when O is a fixed axis of rotation
17 - 8
Systems of Rigid Bodies
• For problems involving systems consisting of several rigid bodies, the
principle of work and energy can be applied to each body.
• We may also apply the principle of work and energy to the entire system,
T1  U12  T2
T1 ,T2 = arithmetic sum of the kinetic energies
of all bodies forming the system
U12 = work of all forces acting on the various
bodies, whether these forces are
internal or external to the system as a
whole.
T
T
17 - 9
W = 120 g
Systems of Rigid Bodies
• For problems involving pin connected members, blocks and pulleys
connected by inextensible cords, and meshed gears,
- internal forces occur in pairs of equal and opposite forces
- points of application of each pair move through equal distances
- net work of the internal forces is zero
- work on the system reduces to the work of the external forces
17 - 10
Conservation of Energy
• Expressing the work of conservative forces as a
change in potential energy, the principle of work
and energy becomes
T1  V1  T2  V2
• Consider the slender rod of mass m.
T1  0, V1  0
T2  12 mv22  12 I  22
 
 12 m 12 l
2
 12


2
1
ml
2
1 ml  

12
2 3
2
2
V2   12 Wl sin    12 mgl sin 
T1  V1  T2  V2
• mass m
• released with zero velocity
• determine  at 
17 - 11
1 ml 2 2 1
0
  mgl sin 
2 3
2
 3g
   sin  
 l

Power
• Power = rate at which work is done


• For a body acted upon by force F and moving with velocity v,
dU  
Power 
 F v
dt


• For a rigid body rotating with an
angular
velocity
and acted

upon by a couple of moment M parallel to the axis of
rotation, dU M d
Power 

 M
dt
dt
17 - 12
Sample Problem 17.1
SOLUTION:
• Consider the system of the
flywheel and block. The work
done by the internal forces
exerted by the cable cancels.
• Note that the velocity of the block
and the angular velocity of the
drum and flywheel are related by
v  r
For the drum and flywheel, I = 16kg ×m2 .
The bearing friction is equivalent to a
couple of 90 N ×m. At the instant shown,
the block is moving downward at 2 m/s.
Determine the velocity of the block after
it has moved 1.25 m downward.
17 - 13
• Apply the principle of work and
kinetic energy to develop an
expression for the final velocity.
Sample Problem 17.1
SOLUTION:
• Consider the system of the flywheel and block. The work
done by the internal forces exerted by the cable cancels.
• Note that the velocity of the block and the angular velocity of
the drum and flywheel are related by
v1 2m s
v2
v
=
= 5rad s
= 2
ωw 2 =
r 0.4m
r 0.4m
• Apply the principle of work and kinetic energy to develop an
expression for the final velocity.
v = rrω
w
w1 =
ω
T1 = 12 mv12 + 12 I w12
1
1
2
2
120kg
2m
s
+
16kg ×m 2 (5rad s )
(
)(
)
2
2
= 440J
=
(
)
T2 = 12 mv22 + 12 I w 22
2
æ
ö
v
1
1
= (120)v22 + (16) ç 2 ÷ = 110v22
2
2
è 0.4 ø
17 - 14
Sample Problem 17.1
T1 = 12 mv12 + 12 I w12 = 440J
T2 = 12 mv22 + 12 I w22 = 110v22
• Note that the block displacement and pulley
rotation are related by
qT22 =
s2 1.25m
=
= 3.125rad
r
0.4m
Then,
U1®1 - 22 = W (s2 - s1 ) - M (q2 - q1 )
(
)
= (120kg ) 9.81m/s 2 (1.2m) - (90 N ×m )(3.125rad )
= 1190J
• Principle of work and energy:
T1 + U1®1 -22 = T2
1.25 m
1.25 m
440J + 1190J = 110 v22
v2 = 3.85m s
17 - 15
v2 = 3.85m s
Sample Problem 17.2
SOLUTION:
• Consider a system consisting of the two
gears. Noting that the gear rotational
speeds are related, evaluate the final
kinetic energy of the system.
• Apply the principle of work and energy.
Calculate the number of revolutions
m A  10 kg k A  200 mm
required for the work of the applied
mB  3 kg k B  80 mm
moment to equal the final kinetic energy
of the system.
The system is at rest when a moment
• Apply the principle of work and energy to
of M  6 N  m is applied to gear B.
a system consisting of gear A. With the
Neglecting friction, a) determine the
final kinetic energy and number of
number of revolutions of gear B
revolutions known, calculate the moment
before its angular velocity reaches 600 and tangential force required for the
rpm, and b) tangential force exerted
indicated work.
by gear B on gear A.
17 - 16
Sample Problem 17.2
SOLUTION:
• Consider a system consisting of the two gears. Noting
that the gear rotational speeds are related, evaluate
the final kinetic energy of the system.
B 
600 rpm 2 rad rev  62.8 rad s
60 s min
r
0.100
 A   B B  62.8
 25.1rad s
rA
0.250
I A  m Ak A2  10kg 0.200m 2  0.400 kg  m 2
I B  mB k B2  3kg 0.080m 2  0.0192 kg  m 2
T2  12 I A A2  12 I B B2
 12 0.400 25.1 2  12 0.019262.82
 163.9 J
17 - 17
Sample Problem 17.2
• Apply the principle of work and energy. Calculate the
number of revolutions required for the work.
T1  U12  T2
0  6 B J  163.9J
 B  27.32 rad
B 
27.32
 4.35 rev
2
• Apply the principle of work and energy to a system
consisting of gear A. Calculate the moment and
tangential force required for the indicated work.
r
0.100
 A   B B  27.32
 10.93 rad
rA
0.250
T2  12 I A A2  12 0.40025.1 2  126.0 J
T1  U1 2  T2
0  M A 10.93 rad   126.0J
M A  rA F  11.52 N  m
17 - 18
F
11.52
 46.2 N
0.250
Sample Problem 17.3
SOLUTION:
• The work done by the weight of the
bodies is the same. From the principle
of work and energy, it follows that each
body will have the same kinetic energy
after the change of elevation.
A sphere, cylinder, and hoop, each
having the same mass and radius, are
released from rest on an incline.
Determine the velocity of each body
after it has rolled through a distance
corresponding to a change of elevation
h.
17 - 19
• Because each of the bodies has a
different centroidal moment of inertia,
the distribution of the total kinetic
energy between the linear and
rotational components will be different
as well.
Sample Problem 17.3
SOLUTION:
• The work done by the weight of the bodies is the
same. From the principle of work and energy, it
follows that each body will have the same kinetic
energy after the change of elevation.
v
With  
r
v 
T2  12 mv  12 I   12 mv  12 I  
r
I 

 12  m  2 v 2

r 
2
2
2
T1  U1 2  T2
I 

0  Wh  12  m  2 v 2

r 
2Wh
2 gh
v2 

m  I r 2 1  I mr 2
17 - 20
2
Sample Problem 17.3
• Because each of the bodies has a different
centroidal moment of inertia, the distribution of
the total kinetic energy between the linear and
rotational components will be different as well.
2 gh
v2 
1  I mr 2
I  52 mr 2
v  0.845 2 gh
Cylinder : I  12 mr 2
v  0.816 2 gh
Sphere :
Hoop :
I  mr 2
v  0.707 2 gh
NOTE:
• For a frictionless block sliding through the same
distance,   0, v  2 gh
17 - 21
• The velocity of the body is independent of its mass
and radius.
• The velocity of the body does depend on
k2
I

mr 2
r2
Sample Problem 17.4
SOLUTION:
• The weight and spring forces are
conservative. The principle of work and
energy can be expressed as
T1  V1  T2  V2
A 15-kg slender rod pivots about the
point O. The other end is pressed
against a spring (k = 300 kN/m) until
the spring is compressed 40 mm and
the rod is in a horizontal position.
• Evaluate the initial and final potential
energy.
• Express the final kinetic energy in terms
of the final angular velocity of the rod.
• Based on the free-body-diagram
equation, solve for the reactions at the
pivot.
If the rod is released from this position,
determine its angular velocity and the
reaction at the pivot as the rod passes
through a vertical position.
17 - 22
Sample Problem 17.4
SOLUTION:
• The weight and spring forces are conservative. The
principle of work and energy can be expressed as
T1  V1  T2  V2
• Evaluate the initial and final potential energy.
2
V1 = Vg + Ve = 0 + 12 kx12 = 12 (300,000 N m )(0.04m )
= 240J
V2 = Vg + Ve = Wh + 0 = (147.15 N )(0.75m )
I = 121 ml 2
1
(15kg )(2.5m)2
12
= 7.81kg ×m 2
=
= 110.4 J
• Express the final kinetic energy in terms of the angular
velocity of the rod.
2
T2 = 12 mv22 + 12 I w22 = 12 m (rw2 ) + 12 I w22
=
17 - 23
1
(15 )(0.75w2 )2 + 12 (7.81)w22 = 8.12w22
2
Sample Problem 17.4
From the principle of work and energy,
T1 + V1 = T2 + V2
0 + 240 J = 8.12w 22 + 110.4 J
w2 = 3.995rad s
• Based on the free-body-diagram equation, solve for the
reactions at the pivot.
2
2
an = r w22 = (0.75m )(3.995rad s ) = 11.97 m s 2 an = 11.97 m s
at = ra
 M O   M O eff
 Fx   Fx eff
 Fy   Fy eff
at = ra
0  I  mr  r
 0
Rx  mr  
Rx  0
R y - 147.15 N = - man
(
= - (15kg ) 11.97 m s 2
R y = - 32.4 N
R = 32.4 N
17 - 24
)
Sample Problem 17.5
SOLUTION:
• Consider a system consisting of the two
rods. With the conservative weight force,
T1  V1  T2  V2
• Evaluate the initial and final potential
energy.
• Express the final kinetic energy of the
Each of the two slender rods has a
system in terms of the angular velocities of
mass of 6 kg. The system is released the rods.
from rest with b = 60o.
• Solve the energy equation for the angular
Determine a) the angular velocity of
velocity, then evaluate the velocity of the
o
rod AB when b = 20 , and b) the
point D.
velocity of the point D at the same
instant.
17 - 25
Sample Problem 17.5
SOLUTION:
• Consider a system consisting of the two rods. With
the conservative weight force,
T1  V1  T2  V2
• Evaluate the initial and final potential energy.
V1  2Wy1  258.86 N 0.325 m 
 38.26 J
V2  2Wy2  258.86 N 0.1283 m 
 15.10 J

W  mg  6 kg  9.81m s 2
 58.86 N
17 - 26

Sample Problem 17.5
• Express the final kinetic energy of the system in terms
of the angular velocities of the rods.
vAB   0.375m 


Since vB is perpendicular to AB and vDis horizontal,
the instantaneous center of rotation for rod BD is C.
CD  20.75 msin 20  0.513 m
BC  0.75 m
and applying the law of cosines to CDE, EC = 0.522 m
Consider the velocity of point B

vB   AB   BC  AB
 BD  

vBD  0.522 m
For the final kinetic energy,
1 ml 2  1 6 kg 0.75 m 2  0.281kg  m 2
I AB  I BD  12
12
1 mv 2  1 I  2  1 mv 2  1 I  2
T2  12
AB 2 AB AB 12
BD 2 BD BD
1 6 0.375 2  1 0.281 2  1 6 0.522 2  1 0.281 2
 12
2
12
2
17 - 27
 1.520 2
Sample Problem 17.5
• Solve the energy equation for the angular velocity,
then evaluate the velocity of the point D.
T1  V1  T2  V2
0  38.26 J  1.520 2  15.10 J
  3.90 rad s

 AB  3.90 rad s
vD  CD 
 0.513 m 3.90 rad s 
 2.00 m s

vD  2.00 m s
17 - 28
Exercise 1
Exercise 2
THE END