FE Review Dynamics G. Mauer UNLV Mechanical Engineering.
-
date post
21-Dec-2015 -
Category
Documents
-
view
235 -
download
4
Transcript of FE Review Dynamics G. Mauer UNLV Mechanical Engineering.
FE Review
Dynamics
G. MauerUNLV
Mechanical Engineering
A (x0,y0)
B (d,h)v
0g
horiz.
distance = dx
yh
X-Y Coordinates
Point Mass Dynamics
A ball is thrown horizontally from A and passes through B(d=20,h = -20) meters.
The travel time t to Point B is
(A) t = 4 s(B) t = 1 s(C) t = 0.5 s(D) t = 2 sUse g = 10 m/s2
A (x0,y0)
B
v0 g
horiz. distance d = 20 m
h
x
y
Use g = 10 m/s2
A ball is thrown horizontally from A and passes through B(d=20,h = -20) meters.
The travel time t to Point B is
(A) t = 4 s(B) t = 1 s(C) t = 0.5 s(D) t = 2 sUse g = 10 m/s2
A (x0,y0)
B
v0 g
horiz. distance d = 20 m
h
x
y
Use g = 10 m/s2
st
tsmm
tgtvty
2
*/10*5.020
**5.0*)sin(*0)(22
2
A ball is thrown horizontally from A and passes through B(d=20,h = -20) meters at time t = 2s.
The start velocity v0 is
(A) v0 = 40 m/s(B) v0 = 20 m/s(C) v0 = 10 m/s(D) v0 = 5 m/s
A (x0,y0)
B
v0 g
horiz. distance d = 20 m
h
x
y
Use g = 10 m/s2
A ball is thrown horizontally from A and passes through B(d=20,h = -20) meters at time t = 2s.
The start velocity v0 is
(A) v0 = 40 m/s(B) v0 = 20 m/s(C) v0 = 10 m/s(D) v0 = 5 m/s
A (x0,y0)
B
v0 g
horiz. distance d = 20 m
h
x
y
Use g = 10 m/s2
smv
svm
tvtx
/100
2*1*020
*)cos(*0)(
12.7 Normal and Tangential Coordinatesut : unit tangent to the pathun : unit normal to the path
Normal and Tangential CoordinatesVelocity Page 53 tusv *
Normal and Tangential Coordinates
Fundamental Problem 12.27 ttn uau
va **
2
(A) constant•(B) 1 m/s2
•(C) 2 m/s2
•(D) not enough information•(E) 4 m/s2
The boat is traveling along the circular path with = 40m and a speed of v = 0.5*t2 , where t is in seconds. At t = 4s, the normal acceleration is:
Fundamental Problem 12.27 ttn uau
va **
2
(A) constant•(B) 1 m/s2
•(C) 2 m/s2
•(D) not enough information•(E) 4 m/s2
2/1*2_:2__
*5.0*2/
smastAt
tdtdva
t
t
The boat is traveling along the circular path with = 40m and a speed of v = 0.5*t2 , where t is in seconds. At t = 4s, the normal acceleration is:
Polar coordinates
Polar coordinates
Polar coordinates
Polar Coordinates
Point P moves on a counterclockwise circular path, with r =1m, dot(t) = 2 rad/s. The radial and tangential accelerations are:•(A) ar = 4m/s2 a = 2 m/s2
•(B) ar = -4m/s2 a = -2 m/s2
•(C) ar = -4m/s2 a = 0 m/s2
•(D) ar = 0 m/s2 a = 0 m/s2
Polar Coordinates
Point P moves on a counterclockwise circular path, with r =1m, dot(t) = 2 rad/s. The radial and tangential accelerations are:•(A) ar = 4m/s2 a = 2 m/s2
•(B) ar = -4m/s2 a = -2 m/s2
•(C) ar = -4m/s2 a = 0 m/s2
•(D) ar = 0 m/s2 a = 0 m/s2
e
Unit vectors
Cr
disk = 10 rad/s
er
B
Point B moves radially outward from center C, with r-dot =1m/s, dot(t) = 10 rad/s. At r=1m, the radial acceleration is:•(A) ar = 20 m/s2
•(B) ar = -20 m/s2
•(C) ar = 100 m/s2
•(D) ar = -100 m/s2
e
Unit vectors
Cr
disk = 10 rad/s
er
B
Point B moves radially outward from center C, with r-dot =1m/s, dot(t) = 10 rad/s. At r=1m, the radial acceleration is:•(A) ar = 20 m/s2
•(B) ar = -20 m/s2
•(C) ar = 100 m/s2
•(D) ar = -100 m/s2
Example cont’d: Problem 2.198 Sailboat tacking against Northern Wind
BoatWindBoatWind VVV /
2. Vector equation (1 scalar eqn. each in i- and j-direction)
500
150
i
Given:r(t) = 2+2*sin((t)), dot= constantThe radial velocity is
(A) 2+2*cos((t ))*-dot,(B) -2*cos((t))*-dot(C) 2*cos((t))*-dot(D) 2*cos((t))(E) 2* +2*cos((t ))*-dot
Given:r(t) = 2+2*sin((t)), dot= constantThe radial velocity is
(A) 2+2*cos((t ))*-dot,(B) -2*cos((t))*-dot(C) 2*cos((t))*-dot(D) 2*cos((t))(E) 2* +2*cos((t ))*-dot
2.9 Constrained Motion
L
B
A
i
J
vA = const
vA is given as shown.Find vB
Approach: Use rel. Velocity:vB = vA +vB/A
(transl. + rot.)
Vr = 150 mm
The conveyor belt is moving to the left at v = 6 m/s. The angular velocity of the drum (Radius = 150 mm) is
(A) 6 m/s(B) 40 rad/s(C) -40 rad/s(D) 4 rad/s(E) none of the above
Vr = 150 mm
The conveyor belt is moving to the left at v = 6 m/s. The angular velocity of the drum (Radius = 150 mm) is
(A) 6 m/s(B) 40 rad/s(C) -40 rad/s(D) 4 rad/s(E) none of the above
yE
Xc
c
XBxA
A B
The rope length between points A and B is:
•(A) xA – xB + xc
•(B) xB – xA + 4xc
•(C) xA – xB + 4xc
•(D) xA + xB + 4xc
Omit all constants!
yE
Xc
c
XBxA
A B
The rope length between points A and B is:
•(A) xA – xB + xc
•(B) xB – xA + 4xc
•(C) xA – xB + 4xc
•(D) xA + xB + 4xc
Omit all constants!
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.
Dynamics
M1: up as positive:Fnet = T - m1*g = m1 a1
M2: down as positive.Fnet = F = m2*g - T = m2 a2
3. Constraint equation:a1 = a2 = a
Equations
From previous:T - m1*g = m1 a
T = m1 g + m1 a Previous for Mass 2:m2*g - T = m2 a
Insert above expr. for Tm2 g - ( m1 g + m1 a ) = m2 a
( m2 - m1 ) g = ( m1 + m2 ) a( m1 + m2 ) a = ( m2 - m1 ) g
a = ( m2 - m1 ) g / ( m1 + m2 )
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
-M*g*cosj
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
-M*g*cosj
xmxForces *i*sin*g*m)_(
)_(0j*cos*g*m-N )_( onlystaticyForces
N
Friction F = k*N:Another horizontal
reaction is added in negative x-direction.
0 = 30 0
g
i
J
m
M*g
M*g*sin
-M*g*cosj
xmNkxForces *i*)*sin*g*m()_(
)_(0j*cos*g*m-N )_( onlystaticyForces
N k*N
Problem 3.27 in Book:Find accel of Mass AStart with:
(A)Newton’s Law for A.(B)Newton’s Law for A
and B(C) Free-Body analysis of
A and B(D) Free-Body analysis of
A
Problem 3.27 in Book:Find accel of Mass AStart with:
(A)Newton’s Law for A.(B)Newton’s Law for A
and B(C) Free-Body analysis of
A and B(D) Free-Body analysis of
A
Problem 3.27 in Book cont’dNewton applied to mass B gives:
(A)Fu = 2T = mB*aB (B) Fu = -2T + mB*g = 0(C) Fu = mB*g-2T = mB*aB
(D)Fu = 2T- mB*g-2T = 0
Problem 3.27 in Book cont’dNewton applied to mass B gives:
(A)Fu = 2T = mB*aB (B) Fu = -2T + mB*g = 0(C) Fu = mB*g-2T = mB*aB
(D)Fu = 2T- mB*g-2T = 0
Problem 3.27 in Book cont’dNewton applied to mass A gives:
(A)Fx = T +F= mA*ax ; Fy = N - mA*g*cos(30o) = 0(B) Fx = T-F= mA*ax Fy = N- mA*g*cos(30o) = mA*ay
(C) Fx = T = mA*ax ; Fy = N - mA*g*cos(30o) =0
(D)Fx = T-F = mA*ax ; Fy = N-mA*g*cos(30o) =0
Problem 3.27 in Book cont’dNewton applied to mass A gives:
(A)Fx = T +F= mA*ax ; Fy = N - mA*g*cos(30o) = 0(B) Fx = T-F= mA*ax Fy = N- mA*g*cos(30o) = mA*ay
(C) Fx = T = mA*ax ; Fy = N - mA*g*cos(30o) =0
(D)Fx = T-F = mA*ax ; Fy = N-mA*g*cos(30o) =0
Energy Methods
Only Force components in direction of motion do WORK
oductScalar
rdFdW
Pr_
Work of
Gravity
Work of a
Spring
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.
A car is traveling at 20 m/s on a level road, when the brakes are suddenly applied and all four wheels lock. k = 0.5. The total distance traveled to a full stop is (use Energy Method, g = 10 m/s2)
(A) 40 m(B)20 m(C) 80 m(D) 10 m(E) none of the above
Collar A is compressing the spring after dropping vertically from A. Using the y-reference as shown, the work done by gravity (Wg) and the work done by the compression spring (Wspr) are
(A) Wg <0, Wspr <0(B) Wg >0, Wspr <0(C) Wg <0, Wspr >0(D) Wg >0, Wspr >0
y
Conservative Forces
A conservative force is one for which the work done is independent of the path taken
Another way to state it:The work depends only on the initial
and final positions,not on the route taken.
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
A child of mass 30 kg is sliding downhill while the opposing friction force is 50 N along the 5m long incline (3m vertical drop). The work done by friction is
(A) -150 Nm(B) 150 Nm(C) 250 Nm(D) -250 Nm(E) 500 Nm
(Use Energy Conservation) A 1 kg block slides d=4 m down a frictionless plane inclined at =30 degrees to the horizontal. The speed of the block at the bottom of the inclined plane is
(A) 1.6 m/s(B) 2.2 m/s(C) 4.4 m/s(D) 6.3 m/s(E) none of the
above
dh
vmrHo
Angular Momentum
Linear Momentum
vmG
Rot. about Fixed Axis Memorize!
rωr
dt
dv
Page 336:
at = x r
an = x ( x r)
Meriam Problem 5.71Given are: BC wBC 2 (clockwise), Geometry: equilateral triangle
with l 0.12 meters. Angle 60
180 Collar slides rel. to bar AB.
GuessValues:(outwardmotion ofcollar ispositive)
wOA 1
vcoll 1
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
Enter vectors:
Mathcad EXAMPLE
Mathcad Example
part 2:Solving the vector equations
16.4 Motion Analysis
http://gtrebaol.free.fr/doc/flash/four_bar/doc/
Approach
1. Geometry: DefinitionsConstants
Variables
Make a sketch
2. Analysis: Derivatives (velocity and acceleration)
3. Equations of Motion
4. Solve the Set of Equations. Use Computer Tools.
Example
Bar BC rotates at constant BC. Find the angular Veloc. of arm BC.
Step 1: Define the Geometry
Example
Bar BC rotates at constant BC. 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((t))
All angles and distance AC(t) are time-variant
A
i
JB
C
(t) (t)
vA(t)
O
Velocities: = -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
Analysis: Solve the rel. Veloc. Vector equation conceptually
Seen from O: vA = x OA
A
i
JB
C
(t)
vA(t)
O (t)
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
Analysis: Solve the rel. Veloc. Vector equation
Seen from O: vA = x OA
A
i
JB
C
(t)
vA(t)
O (t)
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
Analysis: Solve the rel. Veloc. Vector equation
Seen from C: vCollar + BC x AC(t) A
i
JB
C
(t)
O (t)
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
BC x AC(t)
vA,rel
Analysis: Solve the rel. Veloc. Vector equation numerically
A
i
JB
C
(t)
vA(t)
O (t)
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
Enter vectors:
OA
0
0
wOA
rA
l cos ( )
l sin ( )
0
BC
0
0
wBC
rAC
l cos ( )
l sin ( )
0
Analysis: Solve the rel. Veloc. Vector equation numerically
A
i
JB
C
(t)
vA(t)
O (t)
Vector Analysis: OA rA vCOLL BC rAC
LEFT_i l wOA sin ( ) RIGHT_i l wBC sin ( ) vcoll cos ( )
LEFT_j l wOA cos ( ) RIGHT_j l wBC sin ( ) vcoll sin ( )
Here: BC is given as -2 rad/s (clockwise). Find OA
Analysis: Solve the rel. Veloc. Vector equation numericallyA
i
JB
C
(t)
vA(t)
O (t)
Solve the two vector (i and j) equations :
Given
l wOA sin ( ) l wBC sin ( ) vcoll cos ( )
l wOA cos ( ) l wBC sin ( ) vcoll sin ( )
vec Find wOA vcoll( ) vec4.732
0.568
A
i
JB
C
(t)
vA(t)
O (t)
Recap: The analysis is becoming more complex.
•To succeed: TryClear Organization from the start
•Mathcad
•Vector Equation = 2 simultaneous equations, solve simultaneously!
fig_05_011
16.6 Relative Velocity
vA = vB + vA/B
Relative Velocity
vA = vB + vA/B
= VB (transl)+ vRot
vRot = x r
Seen from O:vB = x rSeen from A:
vB = vA + AB x rB/A
Seen from O:vB = x rSeen from A:
vB = vA + AB x rB/A
Rigid Body Acceleration
Stresses and Flow Patterns in a Steam TurbineFEA Visualization (U of Stuttgart)
Find: a B and AB
Look at the Accel. o f B re la tive to A :i
J
B
AvA = const
ABC o u n te rc lo ck w
.
vB
G iven: G eom etry andV A ,aA , vB , A B
r
aB = aA + aB/A ,centr+ aB/A ,angular
r* AB2 r* +
r
Find: a B and AB
Look at the Accel. o f B re la tive to A :
W e know:
1. Centripetal: m agnitude r2 anddirection (inward). If in doubt, com putethe vector product x(*r)
i
J
B
AvA = const
C entrip .r* AB
2
G iven: G eom etry andV A ,aA , vB , A B
aB = aA + aB/A ,centr+ aB/A ,angular
r* AB2 r* +
r
Find: a B and AB
Look at the Accel. o f B re la tive to A :
W e know:
1. Centripetal: m agnitude r2 anddirection (inward). If in doubt, com putethe vector product x(*r)
2. The DIRECTION o f the angular accel(norm al to bar AB)
i
J
B
AvA = const
Centrip. r* AB 2
r*
G iven: G eom etry andV A ,aA , vB , A B
aB = aA + aB/A ,centr+ aB/A ,angular
r* AB2 r* +
Find: a B and AB
Look at the Accel. o f B re la tive to A :
W e know:
1. Centripetal: m agnitude r2 anddirection (inward). If in doubt, com putethe vector product x(*r)
2. The DIRECTION o f the angular accel(norm al to bar AB)
3. The DIRECTION o f the accel o f po int B(horizonta l a long the constra int)
i
J
B
AvA = const
Centrip. r* AB 2
Angular r*
G iven: G eom etry andV A ,aA , vB , A B
aB = aA + aB/A ,centr+ aB/A ,angular
r* AB2 r* +
aB
r
B
A
vA = const
AB
C entrip . r* AB 2
aB
Angular r*
r is the vector from reference
point A to point B
r
i
J
W e can add graphically:S tart w ith C entipeta l
aB = aA + aB/A ,centr+ aB/A ,angular
G iven: G eom etry andV A ,aA , vB , A B
Find: a B and AB
W e can add graphically:S tart w ith C entipeta l
aB = aA + aB/A ,centr+ aB/A ,angular
aB
r* r* AB
2
Result: is < 0 (c lockwise)
aB is negative (to theleft)
B
AvA = const
AB
C entrip . r* A B2
r is the vector from
reference point A to point B
r
i
J
N owC om plete the
Triangle:
G iven: G eom etry andV A ,aA , vB , A B
Find: a B and AB
The instantaneous center of Arm BD is located at Point:
(A) F(B) G(C) B(D) D(E) H
AAB
B
BD
D (t)
(t)
vD(t)i
J
E
O G
F
H
fig_06_002
Plane Motion3 equations: Forces_x Forces_y Moments about G
fig_06_002
Plane Motion3 equations: Forces_x Forces_y Moments about G
*.....:
*.....................
*:
GG
y
x
IMRotation
ymF
xmFnTranslatio
fig_06_005
Parallel Axes TheoremPure rotation about fixed point P
2*dmII GP
Describe the constraint(s) with an Equation
Constrained Motion: The system no longer has all three
Degrees of freedom