principles learn the method. principles basics should be automatic memorize and practice!
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Principles
Learn The Method
Principles
Basics should be automaticMemorize and Practice!
The angular velocity of the disk (r = 1m) in rad/s is
(A) 6 k(B) -6 k(C) 6 i(D) - 3k(E) 3 k
5.129
i
J
wDisk = vo / r = -3/1 = -3 rad/s
The angular accel. of the disk (r = 2 m) in rad/s2 is
(A) 5 k(B) - 5 k(C) - 10 k(D) 10k(E) 2.5 k
5.129
i
J
aDisk = ao / r = 5/2 = 2.5 rad/s2
A
i
J
B
D (t) vD(t)
velangular_
Arm AB with Length L moves the slotted rod BD in x-direction only.
How do we determine the velocity vD of point D as a function of angle q(t)?
L
X(q) =L*cosq
A
i
J
B
D (t) vD(t)
velangular_
Arm AB with Length L moves the slotted rod BD in x-direction only.
How do we determine the velocity vD of point D as a function of angle q(t)?
L
X(q) =L*cosq
X-dot(q) =-L*sin * _q q dot
Chapter 14
Energy Methods
Only Force components in direction of motion do WORK
oductScalar
rdFdW
Pr_
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.
2. If the pendulum is released from the horizontal position, the velocity of its bob in the vertical position is _____
A) 3.8 m/s. B) 6.9 m/s.
C) 14.7 m/s. D) 21 m/s.
The work done is mgh½*m*v2 = mgh orV = sqrt(2gh) = sqrt(14.7)
Power
P E
t
dE
dt
Units of power:
J/sec = N-m/sec = Watts
1 hp = 746 W
The potential energy V is defined as:
dr*F- W - V
Conservative Forces
T1 + V1 = T2 + V2
Potential Energy
Potential energy is energy which results from position or configuration. An object may have the capacity for doing work as a result of its position in a gravitational field. It may have elastic potential energy as a result of a stretched spring or other elastic deformation.
Potential Energy
elastic potential energy as a result of a stretched spring or other elastic deformation.
Potential Energy
Potential Energy
y
Procedure
1. Frame, start and end points
1.2.xc
3. Apply the Energy Principle (only ext. Forces!)
2. Constraint equations?
xA
T1 = 0, start from rest0 + F*xc = 0.5*m*(vA)2
Ex. Problem 14.26
2 𝑥𝑐+𝑥𝐴=𝐿
Chapter 16
Rigid Body Kinematics
16.1
16.3 Rot. about Fixed Axis Memorize!
Vector Product is NOT commutative!
Cross Product
xyyxzxxzyzzy
zyx
zyx
babababababa
bbb
aaa
ba
kji
ba
Derivative of a Rotating Vector
• vector r is rotating around the origin, maintaining a fixed distance
• At any instant, it has an angular velocity of ω
rωr
dt
d
rω
rω
rωr
dt
dv
Page 317:
at = a x r
an = w x ( w x r)
General Motion = Translation + RotationVector sum vA = vB + vA/B
fig_05_006
fig_05_007
16.4 Motion Analysis
Approach1. Geometry: Definitions
Constants
Variables
Make a sketch
2a. Analysis (16.4) Derivatives (velocity and acceleration)
3. Equations of Motion
4. Solve the Set of Equations. Use Computer Tools.
2b. Rel. Motion (16.5)
Example
Bar BC rotates at constant wBC. Find the angular Veloc. of arm BC.
Step 1: Define the Geometry
Example
Bar BC rotates at constant wBC. Find the ang. Veloc. of arm BC.Step 1: Define the Geometry
A
i
JB
C
(t) (t)
vA(t)
O
Geometry: Compute all lengths and angles as f(q(t))
All angles and distance AC(t) are time-variant
A
i
JB
C
(t) (t)
vA(t)
O
Velocities: w = g-dot is given.
Vector Analysis: OA rA vCOLL BC 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 OA and vCOLL
Rigid Body AccelerationChapter 16.7
Stresses and Flow Patterns in a Steam TurbineFEA Visualization (U of Stuttgart)
A
i
JAB
B
D
(t) = 45deg
(t)
vD(t)= const
General Procedure
1.Compute all velocities and angular velocities.
2.Start with centripetal acceleration: It is ALWAYS oriented inward towards the center.
A
i
JAB
B
D
(t) = 45deg
(t)
vD(t)= const
General procedure
1.Compute all velocities and angular velocities.
2.Start with centripetal acceleration: It is ALWAYS oriented inward towards the center.
3. The angular accel is NORMAL to theCentripetal acceleration.
A
i
JAB
B
D
(t) = 45deg
(t)
vD(t)= const
General Procedure
1.Compute all velocities and angular velocities.
2.Start with centripetal acceleration: It is ALWAYS oriented inward towards the center.
3. The angular accel is NORMAL to theCentripetal acceleration. The direction of the angular acceleration is found from the mathematical analysis.
Example HIBBELER 16-1251. Find all vi and wi (Ch. 16.5)
2. View from A: aB = aABXrB – wAB
2*rB
3. View from D: aB = aC + aBCXrB/C – wBC2*rB/C
wAB = -11.55k
wBC = -5k
HIB 16-125Centripetal Terms: We know magnitudes and directions
aABXrB – wAB2*rB = aC + aBCXrB/C – wBC
2*rB/C
– wBC2*rB/C
– wAB2*rB
aD
We now can solve two simultaneous vector equations for wAB and wBC
HIB 16-125
aABXrB – wAB2*rB = aC + aBCXrB/C – wBC
2*rB/C
fig_05_11
fig_05_01116.8 Relative Motion
aA = aB + aA/B,centr+ aA/B,angular + aA,RELATIVE
fig_05_01116.8 Relative Motion
aA = aB + aA/B,centr+ aA/B,angular + aA,RELATIVE
fig_05_01116.8 Relative Motion
aP = *ur + ωx(ωxr) + (α x r) + 2 (ω x )
radial
tangential
From Ch. 12.8
End of Review Chapters 14 and 16