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Final Exam Review. Please Return Loan Clickers to the MEG office after Class! Today!. Final Exam Wed. Wed. May 14 8 – 10 a.m. Review. Always work from first Principles!. Review. Always work from first Principles! Kinetics: Free-Body Analysis Newton’s Law Constraints. Review. - PowerPoint PPT Presentation
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Final Exam Review
Please Return Loan Clickers to the MEG office after Class!
Today!
Final ExamWed. Wed. May 14
8 – 10 a.m.
Always work from first Principles!
Review
Always work from first Principles!Kinetics:
Free-Body AnalysisNewton’s Law
Constraints
Review
gA
i
JUnit vectors
B
L
G
1. Free-Body
Review
gA
i
JUnit vectors
B
L
G
1. Free-Body
B_x
B_ymg
gA
i
JUnit vectors
B
L
G
2. Newton
B_x
B_ymg
Moments about B: -mg*L/2 = IB*a with IB = m*L1/3
gA
i
JUnit vectors
B
L
G
3. Constraint
B_x
B_ymg
aG = a*L/2 = -3g/(2L) * L/2 = -3g/4
1. Free-Body
aCart,x = const
A
i
J
g
R= 0.8mh= 0.05m
b
R R-hmg
A_yA_x
N
aCart,x = const
A
i
J
g
R= 0.8mh= 0.05m
b
R R-hmg
A_yA_x
N
2. Newton
Moments about Center of Cylinder:A_xFrom triangle at left:
Ax*(R-h) –b*mg = 0acart*(R-h) –b*g = 0
aCart,x = const
A
i
J
g
R= 0.8mh= 0.05m
b
R R-hmg
A_yA_x
N
2. Newton
N = 0 at impending rolling, thus Ay = mg
Ax = m*acart
Kinematics (P. 16-126)
CTR
4r
-2r*i + 2r*j
A (x0,y0)
B (d,h)v0
g
horiz.
distance = dx
yh
X-Y Coordinates
Point Mass Dynamics
Normal and Tangential CoordinatesVelocity Page 53 tusv *
Normal and Tangential Coordinates
Polar coordinates
Polar coordinates
Polar coordinates
ABAB VVV /
We Solve Graphically (Vector Addition)
12.10 Relative (Constrained) Motion
vB
vA
vB/A
Example : Sailboat tacking against Northern Wind
BoatWindBoatWind VVV /
2. Vector equation (1 scalar eqn. each in i- and j-direction)
500
150
i
Constrained Motion
LB
A
i
JvA = const
vA is given as shown.Find vB
Approach: Use rel. Velocity:vB = vA +vB/A
(transl. + rot.)
NEWTON'S LAW OF INERTIA A body, not acted on by any force, remains in
uniform motion. NEWTON'S LAW OF MOTION
Moving an object with twice the mass will require twice the force.
Force is proportional to the mass of an object and to the acceleration (the change in velocity).
F=ma.
Rules1. Free-Body Analysis, one for each mass
3. Algebra:Solve system of equations for all unknowns
2. Constraint equation(s): Define connections.You should have as many equations as Unknowns.COUNT!
0 = 30 0
g
i
J
m
M*g
M*g*sin*i
-M*g*cos*j
Mass m rests on the 30 deg. Incline as shown. Step 1: Free-Body Analysis. Best approach: use coordinates tangential and normal to the path of motion as shown.
Mass m rests on the 30 deg. Incline as shown. Step 1: Free-Body Analysis.
Step 2: Apply Newton’s Law in each Direction:
0 = 30 0
g
i
J
m
M*g
M*g*sin*i
-M*g*cos*j
xmxForces *i*sin*g*m)_(
)_(0j*cos*g*m-N )_( onlystaticyForces
N
Friction F = mk*N:Another horizontal
reaction is added in negative x-direction.
0 = 30 0
g
i
J
m
M*g
M*g*sin*i
-M*g*cos*j
xmNkxForces *i*)*sin*g*m()_( m
)_(0j*cos*g*m-N )_( onlystaticyForces
N mk*N
Energy Methods
Only Force components in direction of motion do WORK
oductScalarrdFdW
Pr_
The potential energy V is defined as:
The work is defined as
dr*F- W - V
𝑾= 𝑭 ∗𝒅𝒓 𝒐𝒓𝑾=𝑻∗𝒅𝝑
The work-energy relation: The relation between the work done on a particle by the forces which are applied on it and how its kinetic energy changes follows from Newton’s second law.
Work of
Gravity
Conservative Forces:
Gravity is a conservative force:
• Gravity near the Earth’s surface:
• A spring produces a conservative force:U s
12
kx 2
U g mgy
U g GMm
R
Rot. about Fixed Axis Memorize!
rωr
dtdv
Page 336:
at = a x r
an = w x ( w x r)
fig_05_012Meriam Problem 5.71Given are: wBC wBC 2 (clockwise), Geometry: equilateral trianglewith l 0.12 meters. Angle 60
180 Collar slides rel. to bar AB.
GuessValues:(outwardmotion ofcollar ispositive)
wOA 1
vcoll 1
Vector Analysis: wOA rA vCOLL wBC rAC Mathcad does not evaluate cross products symbolically, so the LEFT andRIGHT sides of the above equation are listed below. Equaling the i- and j-terms yields two equations for the unknowns wOA and vCOLL
Enter vectors:
Mathcad EXAMPLE
fig_05_012Mathcad Example
part 2:Solving the vector equations
fig_05_012Mathcad Examples
part 3Graphical Solution
wBC
wBC X rAC wOA X rOA
ARM BC: VA= wBC X rAC
Right ARM OA:VA =wOA X rOA
Collar slides rel. to Arm BCat velocity vColl. The angle
of vector vColl = 60o
Find: vB and wAB
Graphical Solution Veloc. of Bi
J
B
AvA = const
w Counterclockw.
vB
Given: Geometry andVA
vB = vA + vB/A
wAB rxvA +vA = const
vA isgiven
vB = ?
wAB rx
wAB rx
Find: vB and wAB
i
J
B
AvA = const
w Counterclockw.
vB
Given: Geometry andVA
vB = vA + vB/A
wAB rxvA +vA = const
vA isgiven
Solution:vB = vA + wAB X r
wAB rx
wAB rx
vB = 3 ft/s down, = 60o
and vA = vB/tan. The relative velocity vA/B is found from the vector eq.
(A)vA = vB+ vA/B ,vA/B points
() vA = vB+ vA/B ,vA/B points
(C) vB = vA+ vA/B ,vA/B points
(D) VB = vB+ vA/B ,vA/B points
vB vA
x
y
vB
vA
vA/B
The instantaneous center of Arm BD is located at Point:
(A) B(B) D(C) F(D) G(E) H
AAB
B
BD
D (t)
a (t)
vD(t)i
J
E
O G
F
H
Rigid Body Acceleration
Stresses and Flow Patterns in a Steam TurbineFEA Visualization (U of Stuttgart)
Find: a B and aAB
Look at the Accel. o f B re lative to A :i
J
B
AvA = const
wC o u n te rc lo ck w
.
vB
G iven: G eom etry andV A ,aA , vB , wAB
r
aB = aA + aB/A ,centr+ aB/A ,angular
r* wAB2 r* a+
r
Find: a B and aAB
Look at the Accel. o f B re la tive to A :W e know:1. Centripetal: m agnitude rw2 anddirection (inward). If in doubt, com putethe vector product wx(w*r)
i
J
B
AvA = const
C entrip .r* wAB
2
G iven: G eom etry andV A ,aA , vB , wAB
aB = aA + aB/A ,centr+ aB/A ,angular
r* wAB2 r* a+
r
Find: a B and aAB
Look at the Accel. o f B re la tive to A :W e know:1. Centripetal: m agnitude rw2 anddirection (inward). If in doubt, com putethe vector product wx(w*r)
2. T he DIRECTION o f the angular accel(norm al to bar AB )
i
J
B
AvA = const
Centrip . r* wAB 2
r* a
G iven: G eom etry andV A ,aA , vB , wAB
aB = aA + aB/A ,centr+ aB/A ,angular
r* wAB2 r* a+
Find: a B and aAB
Look at the Accel. o f B re la tive to A :W e know:1. Centripetal: m agnitude rw2 anddirection (inward). If in doubt, com putethe vector product wx(w*r)
2. The DIRECTION o f the angular accel(norm al to bar AB)
3. The DIRECTION o f the accel of point B(horizonta l a long the constra int)
i
J
B
AvA = const
Centrip. r* wAB 2
Angular r* a
G iven: G eom etry andV A ,aA , vB , wAB
aB = aA + aB/A ,centr+ aB/A ,angular
r* wAB2 r* a+
aB
r
B
A
vA = constw
C entrip . r* wAB 2
aB
Angular r* a
r is the vector from referencepoint A to po int B
r
i
J
W e can add graphically:S tart w ith C entipetal
aB = aA + aB/A ,centr+ aB/A ,angular
G iven: G eom etry andV A ,aA , vB , wAB
Find: a B and aAB
W e can add graphically:S tart w ith C entipetal
aB = aA + a B/A ,centr+ aB/A ,angular
aB
r* a r* wAB2
Result:a is < 0 (c lockwise)
aB is negative (to theleft)
B
AvA = const
w
Centrip. r* wAB2
r is the vector fromreference point A to po int B
r
i
J
N owC om plete the
Triangle:
G iven: G eom etry andV A ,aA , vB , wAB
Find: a B and aAB
At =90o, arad = v2/r or v2 = 4.8*0.3 thus v = 1.2
The accelerating Flywheel has R=300 mm. At =90o, the accel of point P is -1.8i -4.8j m/s2. The velocity of point P is
(A) 0.6 m/s(B) 1.2 m/s(C) 2.4 m/s(D) 12 m/s(E) 24 m/s
At =90o, v = 1.2 w = v/r = 1.2/0.3 = 4 rad/s
The accelerating Flywheel has R=300 mm. At =90o, the velocity of point P is 1.2 m/s. The wheel’s angular velocity w is
(A) 8 rad/s(B) 2 rad/s(C) 4 rad/s(D) 12 rad/s(E) 36 rad/s
Accelerating Flywheel has R=300 mm. At =90o, the accel is -1.8i -4.8j m/s2. The magnitude of the angular acceleration a is
(A) 3 rad/s2
(B) 6 rad/s2
(C) 8 rad/s2
(D) 12 rad/s2
(E) 24 rad/s2
At =90o, atangential = a*R or a = -1.8/0.3 = -6 rad/s2
fig_06_002
fig_06_002Plane Motion3 equations:S Forces_xS Forces_yS Moments about G
a*.....:
*.....................
*:
GG
y
x
IMRotation
ymF
xmFnTranslatio
fig_06_005
fig_06_005Parallel Axes TheoremPure rotation about fixed point P
2*dmII GP
Rigid Body Energy MethodsChapter 18 in
Hibbeler, Dynamics
Stresses and Flow Patterns in a Steam TurbineFEA Visualization (U of Stuttgart)
PROCEDURE FOR ANALYSIS
Problems involving velocity, displacement and conservative force systems can be solved using the conservation of energy equation.
• Potential energy: Draw two diagrams: one with the body located at its initial position and one at the final position. Compute the potential energy at each position using
V = Vg + Ve, where Vg= W yG and Ve = 1/2 k s2.
• Apply the conservation of energy equation.
• Kinetic energy: Compute the kinetic energy of the rigid body at each location. Kinetic energy has two components: translational kinetic energy, 1/2m(vG)2, and rotational kinetic energy,1/2 IGw2.
Impulse and Momentum
This equation represents the principle of linear impulse and momentum. It relates the particle’s final velocity (v2) and initial velocity (v1) and the forces acting on the particle as a function of time.
The principle of linear impulse and momentum is obtained by integrating the equation of motion with respect to time. The equation of motion can be written
F = m a = m (dv/dt)Separating variables and integrating between the limits v = v1 at t = t1 and v = v2 at t = t2 results in
mv2 – mv1dvmF dtv2
v1
t2
t1
==
PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM(continued)
IMPACT (Section 15.4)
Impact occurs when two bodies collide during a very short time period, causing large impulsive forces to be exerted between the bodies. Common examples of impact are a hammer striking a nail or a bat striking a ball. The line of impact is a line through the mass centers of the colliding particles. In general, there are two types of impact:
Central impact occurs when the directions of motion of the two colliding particles are along the line of impact.Oblique impact occurs when the direction of motion of one or both of the particles is at an angle to the line of impact.
CENTRAL IMPACT
There are two primary equations used when solving impact problems. The textbook provides extensive detail on their derivation.
Central impact happens when the velocities of the two objects are along the line of impact (recall that the line of impact is a line through the particles’ mass centers).
vA vB
Line of impact
Once the particles contact, they may deform if they are non-rigid. In any case, energy is transferred between the two particles.
CENTRAL IMPACT (continued)
In most problems, the initial velocities of the particles, (vA)1 and (vB)1, are known, and it is necessary to determine the final velocities, (vA)2 and (vB)2. So the first equation used is the conservation of linear momentum, applied along the line of impact.
(mA vA)1 + (mB vB)1 = (mA vA)2 + (mB vB)2
This provides one equation, but there are usually two unknowns, (vA)2 and (vB)2. So another equation is needed. The principle of impulse and momentum is used to develop this equation, which involves the coefficient of restitution, or e.
The coefficient of restitution, e, is the ratio of the particles’ relative separation velocity after impact, (vB)2 – (vA)2, to the particles’ relative approach velocity before impact, (vA)1 – (vB)1. The coefficient of restitution is also an indicator of the energy lost during the impact.
The equation defining the coefficient of restitution, e, is
(vA)1 - (vB)1
(vB)2 – (vA)2e =
If a value for e is specified, this relation provides the second equation necessary to solve for (vA)2 and (vB)2.
CENTRAL IMPACT (continued)
OBLIQUE IMPACT
Momentum of each particle is conserved in the direction perpendicular to the line of impact (y-axis):
mA(vAy)1 = mA(vAy)2 and mB(vBy)1 = mB(vBy)2
In an oblique impact, one or both of the particles’ motion is at an angle to the line of impact. Typically, there will be four unknowns: the magnitudes and directions of the final velocities.
Conservation of momentum and the coefficient of restitution equation are applied along the line of impact (x-axis):
mA(vAx)1 + mB(vBx)1 = mA(vAx)2 + mB(vBx)2
e = [(vBx)2 – (vAx)2]/[(vAx)1 – (vBx)1]
The four equations required to solve for the unknowns are:
End of Review