chapter 18 planar kinetics of a rigid body: work and energy (sections 18.1-18.4) objectives:...
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CHAPTER 18 PLANAR KINETICS OF A RIGID BODY:
WORK AND ENERGY (Sections 18.1-18.4)
Objectives:a) Define the various ways a
force and couple do work.b) Apply the principle of work
and energy to a rigid body.
APPLICATIONS
The work of the torque developed by the driving gears on the two motors on the mixer is transformed into the rotational kinetic energy of the mixing drum.
The work done by the compactor's engine is transformed into the translational kinetic energy of the frame and the translational and rotational kinetic energy of its roller and wheels
KINETIC ENERGY
The kinetic energy of a rigid body in general plane motion is given by
T = 1/2 m (vG)2 + 1/2 IG ω2
Several simplifications can occur.1. Pure Translation: the
rotational kinetic energy is zero (ω = 0): T = 0.5 m (vG)2
KINETIC ENERGY (continued)
If the rotation occurs about the mass center (pure rotation), G, then vG=0
2. Rotation: When a rigid body is rotating about a fixed axis passing through point O, the body has both translational and rotational kinetic energy:
T = 0.5m(vG)2 + 0.5IG2
Since
vG = rG, T = 0.5(IG + m(rG)2)2 = 0.5IO2
T = 0.5 IG 2
WORK OF A FORCE
The work done by a force is
UF = F•dr = (F cos θ) ds
Constant force : UFc = (Fc cos θ)s
s
Work of a weight: Uw = -WΔy.
Work of a spring force: Us = -0.5k[(s2)2 – (s1)2]
FORCES THAT DO NO WORK
1. Reactions at fixed supports because the displacement at their point of application is zero.
3. Internal forces because they always act in equal and opposite pairs.
2. Normal and frictional forces acting on bodies as they roll without slipping over a rough surface since there is no instantaneous displacement of the point in contact with ground.
THE WORK OF A COUPLE
A body subjected to a couple produces work only when it undergoes rotation.
constant M: UM = M (2 – 1)
If the body rotates through an angular displacement d, the work of the couple moment, M, is
2
1
M dUM
PRINCIPLE OF WORK AND ENERGY
The principle states: U1-2 = T12
i.e. the work done by all external forces and couple moments equals the change in body’s kinetic energy (translational and rotational).
This equation is a scalar equation. It can be applied to a system of rigid bodies by summing contributions from all bodies.
Find: Angular velocity of the disk when point G moves 0.5 m. The disk starts from rest and rolls without slipping. The spring is initially unstretched.
EXAMPLE
Given:The disk weighs 40 N, has a radius of gyration (kG) 0.6 m. A 15 N.m moment is applied. The spring constant is 10N/m.
EXAMPLE (solution)
Free body diagram of the disk:
Only the spring force and couple moment M do work.
Spring will stretch twice the amount of displacement of G, or 1 m. Why?
EXAMPLE (continued)
Work: U1-2 = -0.5k[(s2)2 – (s1)2] + M(2 – 1)
U1-2 = -0.5(10)(12 – 0) + 15(0.5/0.8) = 4.4N.m
Kinematic relation: vG = r = 0.8Kinetic energy: T1 = 0
T2 = 0.5m (vG)2 + 0.5 IG 2
T2 = 0.5(40/9.8)(0.8)2 + 0.5(40/9.8)(0.6)22
T2 = 2.0 2
Work and energy: U1-2 = T12
4.4 = 2.0 2
= 1.5 rad/s
GROUP PROBLEM SOLVING
Find: The angular velocity of the sphere when = 45° if the sphere rolls without slipping.
Given: A sphere weighing 10N rolls along a semicircular hoop. Its equals 0 when = 0.
GROUP PROBLEM SOLVING (continued)
Now calculate the _______________ :
Solution: Draw a FBD and calculate the vertical distance the mass center moves.
GROUP PROBLEM SOLVING (continued)
A kinematic equation for finding the velocity of the mass center is needed. It is
_________ energy:
Now apply the principle of work and energy equation:
= 13.1 rad/s
PLANAR KINETICS OF A RIGID BODY: CONSERVATION OF ENERGY (Section 18.5)
Objectives:a) Determine the potential
energy of conservative forces.
b) Apply the principle of conservation of energy.
APPLICATIONS
Torsional spring at the top of the door winds up as the door is lowered. When the door is raised, the spring potential energy is transferred into gravitational potential energy of the door’s weight, thus making it easy to open.
CONSERVATION OF ENERGY
The conservation of energy theorem is a “simpler” energy method for solving problems. Once again, the problem parameter of distance is a key indicator of when conservation of energy is a good method for solving the problem.
If it is appropriate, conservation of energy is easier to use than the principle of work and energy.
CONSERVATIVE FORCES
A force F is conservative if the work done by the force is independent of the path.
Typical conservative forces encountered in dynamics are gravitational forces (i.e., weight) and elastic forces (i.e., springs).
What is a common force that is not conservative?
CONSERVATION OF ENERGY
When a rigid body is acted upon by a system of conservative forces, the work done by these forces is conserved. Thus, the sum of kinetic energy and potential energy remains constant:
T1 + V1 = T2 + V2 = Constant
i.e. as a rigid body moves from one position to another when acted upon by only conservative forces, kinetic energy is converted to potential energy and vice versa.
GRAVITATIONAL POTENTIAL ENERGY
The gravitational potential energy is a function of the height of the body’s center above or below a datum.
Gravitational potential energy is +ve when yG is +ve, since the weight has the ability to do +ve work when the body is moved back to the datum.
The gravitational potential energy is found by:
Vg = W yG
ELASTIC POTENTIAL ENERGY
Spring forces are also conservative forces.
Notice that the elastic potential energy is always +ve
The potential energy of a spring force (F = ks) is found by the equation
Ve = ½ ks2
Find: The angular velocity of rod AB at = 0° if the rod is released from rest when = 30°.
EXAMPLE 1
Given:The rod AB has a mass of 10 kg. Piston B is attached to a spring of constant k = 800 N/m. The spring is un-stretched when = 0°. Neglect the mass of the pistons.
EXAMPLE 1 (solution)
Potential Energy: put datum in line with rod when =0°. Gravitational and elastic energy=0 at 2. => V2 = 0
Gravitational energy at 1: - (10)( 9.81) ½ (0.4 sin 30)Elastic energy at 1: ½ (800) (0.4 sin 30)2
So V1 = - 9.81J + 16.0 J = 6.19 J
Initial Position Final Position
EXAMPLE 1 (continued)
The rod released from rest (vG1=0, 1=0). Thus, T1 = 0. At 2, the angular velocity is 2 and the velocity is vG2 .
Kinetic Energy:
Initial Position Final Position
EXAMPLE 1 (continued)
Thus, T2 = ½ (10)(vG2)2 + ½ (1/12)(10)(0.42)(2)2
At 2, vA=0. Hence, vG2 = r = 0.2 2 .
Then, T2 = 0.2 22 + 0.067 2
2 = 0.267 22
conservation of energy:
T1 + V1 = T2 + V2
0 + 6.19 = 0.26722 + 0 => 2 = 4.82 rad/s
EXAMPLE 2
Given: The weight of the disk is 30 N and its kG equals 0.6 m. The spring has a stiffness of 2 N/m and an unstretched length of 1m.
Find: The angular velocity at the instant G moves 3 m to the left. The disk is released from rest in the position shown and rolls without slipping.
EXAMPLE 2 (solution)
Potential Energy: No changes in the gravitational potential energy
The elastic potential energy at 1:
V1 = 0.5 k (s1)2 where s1 = 4 m.
Thus, V1 = ½ 2 (4)2 = 16 N.m.
The elastic potential energy at 2 is
V2 = ½ 2 (3)2 = 9 N.m.
EXAMPLE 2 (continued)
Kinetic Energy:
Released from rest: vG1=1=0, T1=0
At 2, 2 and vG2:
T2 = ½ m (vG2)2 + ½ IG (2) 2
= ½ (30/9.8) (vG2)2 + ½ (30/9.8) 0.62 (2) 2
Disk is rolling without slipping: vG2 = (0.75 2)
T2 = ½(30/9.8)(0.75 2)2 + ½(30/9.8) 0.62 (2)2 = 1.41 (2)2
EXAMPLE 2 (continued)
conservation of energy:
T1 + V1 = T2 + V2
0 +16.0 = 1.41 22 + 9
Solving , 2 = 2.23 rad/s
GROUP PROBLEM SOLVING
Find: The angle (measured down from the horizontal) to which the bar rotates before it stops its initial downward movement.
Given:A 50 N bar is rotating downward at 2 rad/s. The spring has an unstretched length of 2m and a spring constant of 12 N/m
GROUP PROBLEM SOLVING (solution)
Potential Energy:Put the datum in line with the rod when = 0.
Gravitational potential energy at 2:
Elastic potential energy at 2:
So, V2 =
CHAPTER 19PLANAR KINETICS: IMPULSE AND
MOMENTUM (Sections 19.1-19.2)
Today’s Objectives:a) Develop formulations for
the linear and angular momentum of a body.
b) Apply the principle of linear and angular impulse and momentum.
APPLICATIONS
As the pendulum swings downward, its angular momentum and linear momentum both increase. By calculating its momenta in the vertical position, we can calculate the impulse the pendulum exerts when it hits the test specimen.
The space shuttle has several engines that exert thrust on the shuttle when they are fired. By firing different engines, the pilot can control the motion and direction of the shuttle.
LINEAR AND ANGULAR MOMENTUM
The linear momentum of a rigid body is: L = m vG
The angular momentum of a rigid body is: HG = IG
LINEAR AND ANGULAR MOMENTUM (continued)
Translation.When a rigid body undergoes rectilinear or curvilinear translation, its angular momentum is zero because ω = 0.
Therefore:
L = m vG
HG = 0
Rotation about a fixed axis.
HO = ( rG x mvG) + IG = IO
The body’s linear momentum and angular momentum are:
L = m vG
HG = IG
LINEAR AND ANGULAR MOMENTUM (continued)
General plane motion. The linear and angular
momentum computed
about G are required.
L = m vG
HG = IGω
The angular momentum about point A is HA = IGω + (d)mvG
LINEAR AND ANGULAR MOMENTUM (continued)
PRINCIPLE OF IMPULSE AND MOMENTUM
The principle is developed by combining the equation of motion with kinematics. The resulting equations allow a direct solution to problems involving force, velocity, and time.
Linear impulse-linear momentum equation:
L1 + F dt = L2 or (mvG)1 + F dt = (mvG)2
t2
t1
t2
t1
Angular impulse-angular momentum equation:
(HG)1 + MG dt = (HG)2 or IG1 + MG dt = IG2
t2
t1
t2
t1
PRINCIPLE OF IMPULSE AND MOMENTUM (continued)
The previous relations can be represented graphically by drawing the impulse-momentum diagram.
+ =
Find: The angular velocity of the disk after 4 seconds if it starts from rest and rolls without slipping.
EXAMPLE
Given: A disk weighing 50 N has a rope wrapped around it. The rope is pulled with a force P equaling 2 N.
EXAMPLE (solution)
Impulse-momentum diagram:
Kinematics: (vG)2 = r 2t2
t1
Impulse & Momentum: (HA)1 + MA dt = (HA)2
(P t) 2 r = (mvG)2 r + IG2 = m r2 2 + 0.5 m r2 2 = 1.5 m r2 2
3.5 rad/s3(50/9.81)(0.6)
4(2)(4)
3mr
4 Pt2
(m vG)1 (m vG)2G G
IG1 IG 2
P t
F t
N t
W t
r+ =A
y
x
GROUP PROBLEM SOLVING
Find: The angular velocity of gear A after 5 seconds if the gears start turning from rest.
Plan: Time is a parameter, thus
Given: A gear set with:WA = 15 NWB = 10 NkA = 0.5 mkB = 0.35 mM = 2(1 – e-0.5t) N.m
GROUP PROBLEM SOLVING (continued)
Angular impulse & momentum relation for gear A about point A yields:
For gear B:
Kinematics:
CONSERVATION OF MOMENTUM (Section 19.3)
Objectives:a) Understand the conditions
for conservation of linear and angular momentum
b) Use the condition of conservation of linear/ angular momentum
APPLICATIONS
A skater spends a lot of time either spinning on the ice or rotating through the air. To spin fast, or for a long time, the skater must develop a large amount of angular momentum.
If the skater’s angular momentum is constant, can the skater vary his rotational speed? How?
The skater spins faster when the arms are drawn in and slower when the arms are extended. Why?
CONSERVATION OF LINEAR MOMENTUM
Recall that the linear impulse and momentum relationship is
L1 + F dt = L2 or (m vG)1 + F dt = (m vG)2
t2
t1
t2
t1
If the sum of all the linear impulses acting on the rigid body (or a system of rigid bodies) is zero, all the impulse terms are zero. Thus, the linear momentum for a rigid body (or system) is constant, or conserved. So L1 = L2.
00
This equation is referred to as the conservation of linear momentum. The conservation of linear momentum equation can be used if the linear impulses are small or non-impulsive.
CONSERVATION OF ANGULAR MOMENTUM
The angular impulse-angular momentum relationship is:
(HG)1 + MG dt = (HG)2 or IG1 + MG dt = IG2
t2
t1
t2
t1
00
Similarly, if the sum of all the angular impulses due to external forces acting on the rigid body (or a system of rigid bodies) is zero, all the impulse terms are zero. Thus, angular momentum is conserved . The resulting equation is referred to as the conservation of angular momentum or (HG)1 = (HG)2 .
If the initial condition of the rigid body (or system) is known, conservation of momentum is often used to determine the final linear or angular velocity of a body just after an event occurs.
Given: A 10 kg wheel(IG = 0.156 kg·m2) rolls
without slipping and does not bounce at A.
Find: The minimum velocity, vG, of the wheel to just roll over
the obstruction at A.
EXAMPLE
Note: Since no slipping or bouncing occurs, the wheel pivots about point A. The force at A is much greater than the weight, and since the time of impact is very short, the weight can be considered non-impulsive. The reaction force at A, since we don’t know either its direction or magnitude, can be eliminated by applying the conservation of angular momentum equation about A.
Conservation of angular momentum: (HA)1 = (HA)2
r ' (mvG)1 + IG 1 = r (mvG)2 + IG 2
(0.2 - 0.03)(10)(vG)1 + 0.156 1 = 0.2(10)(vG)2 + 0.156 2
Kinematics: no slip, = vG / r = 5 vG .Substituting and
solving: (vG)2 = 0.892 (vG)1
Impulse-momentum diagram:
+ =
x
y
EXAMPLE (solution)
EXAMPLE (continued)
T2 + V2 = T3 + V3
{½(10)(vG)22 + ½(0.156)2
2 } + 0 = 0 + 98.1 (0.03)
Substituting 2 = 5 (vG)2 and (vG)2 = 0.892 (vG)1 and
solving yields: (vG)1 = 0.729 m/s, (vG)2 = 0.65 m/s
To complete the solution, conservation of energy can be used. Since it cannot be used for the impact (why?), it is applied just after the impact. In order to roll overthe bump, the wheel must go to position 3 from 2. When (vG)2 is a minimum, (vG)3 is zero. Why?
GROUP PROBLEM SOLVING
Find: The angular velocity of the pendulum just after impact.
Given: A slender rod (Wr = 5 N) has a wood block (Ww = 10 N) attached. A bullet (Wb = 0.2 N) is fired into the center of the block at 1000 m/s. Assume the pendulum is initially at rest and the bullet embeds itself into the block.
GROUP PROBLEM SOLVING (continued)
To use conservation of angular momentum,
IA = ?
IA = 7.2kg.m2
Solution:
First draw a FBD.
GROUP PROBLEM SOLVING (continued)
Solving yields:2 = 7.0 rad/s
Apply the conservation of angular momentum equation:
CHAPTER 22VIBRATIONS (section 22.1)
Objectives:
Discuss Undamped single degree of freedom (SDOF) vibration of a rigid body
Vibration definition
Vibration is the periodic motion of a body or system of connected bodies displaced from a position of equilibrium
Types of vibrations Free vibrations: motion is maintained by
conservative forces (like gravitational or elastic restoring forces).
Forced vibrations: motion is caused by external periodic or intermittent forces.
Undamped free response for SDOF systems
If you displace mass m a distance u from equilibrium position and release it, vibration occurs
Undamped free response for SDOF systems (cont):
Equilibrium eq: A linear, constant
coefficients, homogeneous, second order ordinary differential equation
2
2
2
20
uKu m
t
um Ku
t
2
2
22
2
0
0
cos sin
n
n n
um ku
t
uu
tu A t t
A and B are found from initial conditions
Tn =natural period of vibration
ωn =natural circular frequency of vibration
n
k
m
2n
n
T
fn =1/Tn= natural cyclic frequency of vibration in hertz (cycles per second)
Undamped free response for SDOF systems:
Solving initial conditions
0
. .
0
.
0
0
( 0)
( 0)
cos sin
n
n nn
u t u A
u t u B
uu u t t
cos sinn nu A t t
Solving initial conditions
.
0
0
max
0, 0 0
0, 0 /
(1 cos )
2 / 2
n
static
t u B
t u A P k
Pu t
Ku P K u
cos sin /u A t t P k
90
Viscous Damped free vibration :
xmxckxF
0 kxxcxm
02 2 xxx nn
k
c
m
x0x
)(dashpot
nm
czeta
2)(
c : viscous damping constant (in N s/m)
Define: viscous damping factor (no unit)
02 22 nn Use a trial solution x = A et :
Obtain two possible ’s :
)1( 21 n )1( 2
2 nand
tt eAeAx 2121
te 1λ te 2λ are two solutions. General solution is a linear combination
and
91
Case 1 : > 1 (overdamping) : 1 and 2 < 0
tt eAeAx 21
2
1
x decays to zero without oscillation
t
x
1,
overdamped1Critical
damped
1 underdamping
Three cases:
92
Case 2: = 1 (critical damping) : 1 = 2 = -n
A1 exp(-nt) is a solution
A2 t exp(-nt) is also a solution (prove !)
The general solution is x = (A1 + A2 t )exp(-nt)
x approaches zero quickly without oscillation.
93
Case 3: < 1 (underdamping) :
ttiti nnn eeAeAx )(22 1
21
121 ndDefine damped natural frequency
d
1x 2x
1t 2t
x
t
)sin()(
]e2
2 and e2
1Let
conjucate.complex are 2 and 1 that so real, is [
)( 21
tdtnCex
iCAiCA
AAx
ttiti ndd eeAeAx
Period d = 2/d
94
22
21
1
2
12
/decrement clogarithmi Define
n
ndn
xxn
It is because = c/(2mn), measurement of determines the viscous damping constant c.
dn
dn
n
eCe
Cexx
t
t
)(2
1
1
1
cycles successive twoof ntsdisplaceme theof ratio The
95
xmtFxckxF o sin
B. Viscous Damped Forced vibration
The equation of motion :
tFkxxcxm o sin t
mFxxx o
nn sin2 2
kcm
x0x
)(dashpot
tFF O sin
)(spring
96
2/1222
2
2/1222
/2/1/1/
factor ion magnificator ratio amplitude Define
/1
/2tan
/2/1
/
nno
n
n
nn
o
kF
CM
kFC
)sin( and )cos( then),sin( Let
2
tCxtCxtCx
pp
p
tmFxxx o
nn sin2 2
97
M
n /
1
1
01.0
2.01
Maximum M occurs at: 0 })2
(])(1{[d
d 222 nn
nres 21 2The resonance frequency is
98
n /1)(resonance
02.0
1
0
2/
Consider the following regions:
(1) is small, tan > 0, 0+, xp in phase with the driving force(2) is large, tan < 0, 0-, = , xp lags the driving force by 90o
(3) n-, tan +, /2(-)
n+, tan -, /2(+)
tan = 2n
n
)(
99
If the driving force is not applied to the mass, but is applied to the base of the system:
tsin2 22nn bxxx
If b2 is replaced by Fo/m:
)( Bxxmkxxc
k
c
x0x
tbxB sin
m
tsin2 2nn
m
Fxxx o
This can be used as a device to detect earthquake.
Bxmkxxcxm
100
Determine : (a) steady-state displacement (b) max. force transmitted to the base.
k c
m
pA
m =45 kg, k = 35 kN/m, c = 1250 N.s/m, p = 4000 sin (30 t) Pa, A= 50 x 10-3 m2.
Example
)(498.0)9.27452/(1250)2/(/9.2745/1035/ 3
dunderdampemcsradmk
n
n
101
The amplitude of the steady-state vibration is:
m
nn
koFX
00528.02/12
79.2/30498.0222
9.27/301
31035/
310504000
2/12
/222
/1
/
rad
n
n716.12
9.27/301
9.27/30498.021tan2
/1
/21tan
#)716.130cos(00528.0)cos( mttXxP
102
The force transmitted to the base is :
)(cos)(sin
tXctkXxckxF pptr
For max Ftr :
kct
tddFtr
1tan0
o
3
31 9.46
10351030351250tan
t
#
o
o3
max
2719.46sin00528.0301250
9.46cos00528.01035
N
Ftr
Group Working ProblemsProblem 1: Chapters 17, 18 and 19
Given: The 30-kg disk is pin connected at its center. If it starts from rest,
Find:
the number of revolutions it must make and the time needed to attain an angular velocity of 20rad/s?
Problem 1: Newton Laws
Find angular acceleration: (11.7rad/s2)
Find angle (17.1 radians)
Find number of revolutions: (2.73)
Find time (1.71 sec)
Problem 1: Energy Theorems
Find angle (17.1 radians)
Find number of revolutions: (2.73)
Find angular acceleration: (11.7rad/s2)
Find time (1.71 sec)
Problem 1: Momentum Principle
Find time (1.71 sec)
Find angular acceleration: (11.7rad/s2)
Find angle (17.1 radians)
Find number of revolutions: (2.73)
Problem 2: Chapter 22 in view of Chapters 17, 18 and 19
Given: a 10-kg block suspended from a cord wrapped around a 5kg disk.
Find: natural period of vibration
Problem 2: Energy Theorems: use conservation of energy Write conservation of energy equation
Differentiate equation
Find period (1.57s)
Problem 2: Momentum Principle: use M=d(HG)/dt
Write M=d(HG)/dt for a displacement θ from equilibrium position
Differentiate equation
Find period (1.57s)