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Final Exam Review
Please Return Loan Clickers to the MEG office after Class!
Today!
Always work from first Principles!
Review
Always work from first Principles!Kinetics:
Free-Body AnalysisNewton’s Law
Constraints
Review
g
A
i
JUnit vectors
B
L
G
1. Free-Body
Review
g
A
i
JUnit vectors
B
L
G
1. Free-Body
B_x
B_y
mg
g
A
i
JUnit vectors
B
L
G
2. Newton
B_x
B_y
mg
Moments about B: -mg*L/2 = IB*a with IB = m*L2/3
g
A
i
JUnit vectors
B
L
G
3. Constraint
B_x
B_y
mg
aG = *L/2 = -g*3/4
1. Free-Body
aCart,x = const
A
i
J
g
R= 0.8m
h= 0.05m
b
R R-hmg
A_yA_x
N
aCart,x = const
A
i
J
g
R= 0.8m
h= 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= 0.8m
h= 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
Kinematics (P. 16-126)
CTR
4r
-2r*i + 2r*j
Feedback
Overall, when comparing traditional Homework formats with Mastering, I
prefer
(A) Paper submission of Homework (B) Electronic Submission
Feedback
For me, the most useful benefit of Mastering is
(A) Hints while developing the solution to a problem
(B) Instant grading of results(C) Practice Exams
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
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
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
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!
Given: v0 = const.The vertical velocity
component of point A (in y-direction) is
(A)vA,y = v0*tan()(B) vA,y = v0*cot()(C) vA,y = v0*cos((D)vA,y = 2*v0
Given: v0 = const.The velocity of point A
in vertical y-direction is
(A)vA,y = v0*tan()(B) vA,y = v0*cot()(C) vA,y = v0*cos((D)vA,y = 2*v0
(E) vA,y = v0/cos()
Vy/vx =cot
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
Mass m rests on the 30 deg. Incline as shown. The free-body reaction seen by the incline in j-direction is
(A) -mg*sin30o
(B) +mg*sin30o (C) -mg*cos30o
(D) +mg*cos30o
(E) None of the above
0 = 30 0
g
i
J
m
Mass m rests on the 30 deg. Incline as shown. The free-body reaction seen by the incline in j-direction is
(A) -mg*sin30o
(B) +mg*sin30o (C) -mg*cos30o
(D) +mg*cos30o
(E) None of the above
0 = 30 0
g
i
J
m
mg
mg*cos()
Mass m rests on the 30 deg. Incline as shown. The static friction required to keep the mass from sliding in i-direction is
(A) -mg*sin30o
(B) +mg*sin30o (C) -mg*cos30o
(D) +mg*cos30o
(E) None of the above
0 = 30 0
g
i
J
m
Mass m rests on the 30 deg. Incline as shown. The static friction required to keep the mass from sliding in i-direction is
(A) -mg*sin30o
(B) +mg*sin30o (C) -mg*cos30o
(D) +mg*cos30o
(E) None of the above
0 = 30 0
g
i
J
m
mg*sin()
Newton applied to mass B gives:
(AFu = 2T = mB*aB (B) Fu = -2T + mB*g = 0(C) Fu = mB*g-2T = mB*aB
(DFu = 2T- mB*g-2T = 0
(AFu = 2T = mB*aB (B) Fu = -2T + mB*g = 0(C) Fu = mB*g-2T = mB*aB
(DFu = 2T- mB*g-2T = 0
Newton applied to mass B gives:
(AFx = 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
(DFx = T-F = mA*ax ; Fy = N-mA*g*cos(30o) =0
Newton applied to mass A gives:
(AFx = 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
(DFx = T-F = mA*ax ; Fy = N-mA*g*cos(30o) =0
Newton applied to mass A gives:
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(B20 m(C) 80 m(D) 10 m(E) none of the above
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(B20 m(C) 80 m(D) 10 m(E) none of the above
metersg
vs
sdistbrakeforSolve
sgmvmTT
k
k
10
400
**2
:____
*****2/1012
20
20
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
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
sdFW
*
Potential Energy
• For any conservative force F we can define a potential energy function U in the following way:
– The work done by a conservative force is equal and opposite to the change in the potential energy function.
• This can be written as: r1
r2 U2
U1
W Fgdr U
U U2 U1 W
rFgd
rr
r1
r2
Hooke’s Law• Force exerted to compress a spring is proportional to the
amount of compression.
Fs kx
PEs 1
2kx 2
Conservative Forces:
Gravity is a conservative force:
• Gravity near the Earth’s surface:
• A spring produces a conservative force: Us 1
2kx2
Ug mgy
Ug GMm
R
(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
(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
PE of block released = KE of block = PE gained by spring
Height dropped, h d sin 4sin 30m 2m
Potential energy released, mgh 19.8m / s2 2m 19.6J
Kinetic energy of the block = 1
2mv2 19.6J
v U2
m 19.6
2
16.3 m/s
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 change of potential energy is
(A) -882 Nm(B) 882 Nm(C) 1470 Nm(D) -1470 Nm(E) None of the
above
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 change of potential energy is
Change of PE = mg(hfinal-h0)
= 30*9.8*(0-3) = -882 Nm (A) -882 Nm(B) 882 Nm(C) 1470 Nm(D) -1470 Nm(E) None of the
above
Rot. about Fixed Axis Memorize!
rωr
dt
dv
Page 336:
at = x r
an = x ( x r)
Arm BD is rotating with constant dot= , while point D moves at vD*i. Seen from D, the velocity vector at B is:
(A) vB = vD*i - BD**cosiBD*sinj(B) vB = vD*i - BD**cosi BD*sinj(C) vB = vD*i + BD**cosiBD*sinj(D) vB = - BD**cosiBD*sinj(E) none of the above
i
J
BD
(t)
B
AB
B
D
(t),(t)
A vD(t)
Arm BD is rotating with constant dot= , while point D moves at vD*i. Seen from D, the velocity vector at B is:
(A) vB = vD*i - BD**cosiBD*sinj(B) vB = vD*i - BD**cosi BD*sinj(C) vB = vD*i + BD**cosiBD*sinj(D) vB = - BD**cosiBD*sinj(E) none of the above
i
J
BD
(t)
B
AB
B
D
(t),(t)
A vD(t)
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
Mathcad Examples
part 3Graphical Solution
BC
BC X rAC
OA X rOA
ARM BC: VA= BC X rAC
Right ARM OA:VA =OA X rOA
Collar slides rel. to Arm BCat velocity vColl. The angle
of vector vColl = 60o
Find: vB and AB
Graphical Solution Veloc. of Bi
J
B
AvA = const
ABCounterclockw.
vB
Given: Geometry andVA
vB = vA + vB/A
AB rxvA +vA = const
vA isgiven
vB = ?
AB rx
AB rx
Find: vB and AB
i
J
B
AvA = const
ABCounterclockw.
vB
Given: Geometry andVA
vB = vA + vB/A
AB rxvA +vA = const
vA isgiven
Solution:vB = vA + AB X r
AB rx
AB rx
vB = 3 ft/s down, = 60o
and vA = vB/tanThe relative velocity vA/B is found from vector eq.
(A)vA = vB+ vA/B ,vA/B points
(BvA = 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 = 3 ft/s down, = 60o
and vA = vB/tanThe relative velocity vA/B is found from vector eq.
(A)vA = vB+ vA/B ,vA/B points
(BvA = 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
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
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
6.78
Given: I_G=m*k2=300*1.5^2 = 675 kgm^2. The angular accel of the rocket is
Thrust T = 4 kN
F_y = m*a = 300*8.69N
(A) 0.102 rad/s2
(B) 0.31 rad/s2
(C) 3.1 rad/s2
(D) 5.9 rad/s2
6.78
Given: I_G= 675 kgm^2, m = 300 kg. The angular accel of the rocket is
Thrust T = 4 kN
F_y = m*a = 300*8.69N
(A) 0.102 rad/s2
(B) 0.31 rad/s2
(C) 3.1 rad/s2
(D) 5.9 rad/s2
Answer: (sum of moments about G = I_G*alpha)4000N * sin(1deg)*3m = 675*alpha
alpha = 0.31 rad/s^2