Download ME 242 Chapter 13

Final Exam Review
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Review
Always work from first Principles!
Review
Always work from first Principles!
Kinetics:
Free-Body Analysis
Newton’s Law
Constraints
Review
Unit vectors
J

i
G
B
g
L
1. Free-Body
A
Unit vectors
J

i
G
B
B_x
g
A
mg
B_y
L
1. Free-Body
Unit vectors
J

i
G
B
B_x
g
A
mg
B_y
L
2. Newton
Moments about B: -mg*L/2 =
IB*a
with IB = m*L2/3
Unit vectors
J

i
G
B
B_x
g
mg
B_y
L
3. Constraint
aG
A
= a*L/2 = -g*3/4
J
R
R-h
i
mg
b
A_x
h= 0.05m
aCart,x = const
1. Free-Body
A
R= 0.8m
A_y
N
g
J
R
R-h
i
mg
b
2. Newton
A_x
h= 0.05m
A
R= 0.8m
A_y
aCart,x = const
Moments about Center of Cylinder:A_x
From triangle at left:
Ax*(R-h) –b*mg = 0
acart*(R-h) –b*g = 0
N
g
J
R
R-h
i
mg
g
b
Newton
A_x
h= 0.05m
aCart,x = const
A
R= 0.8m
A_y
N
N = 0 at impending rolling, thus Ay
= mg
Ax = m*acart
Kinematics (P. 16-126)
CTR
Kinematics (P. 16-126)
4r
-2r*i + 2r*j
CTR
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
Point Mass Dynamics
X-Y Coordinates
v
g
B (d,h)
0
y

A (x0,y0)
x
horiz.
distance = d
h
v0
g
A (x0,y0)
y
h
x
horiz. distance
d = 20 m
B
A ball is thrown horizontally
from A and passes
through B(d=20,h = -20)
meters.
The travel time t to Point B is
Use g = 10 m/s2 (A) t = 4 s
(B) t = 1 s
(C) t = 0.5 s
(D) t = 2 s
Use g = 10 m/s2
v0
g
A (x0,y0)
y
h
x
horiz. distance
d = 20 m
B
y (t ) Use
 v0 * sin(
t  0m/s
.5 * g *2t 2
g =) *10
 20m  0.5 *10m / s 2 * t 2
t  2s
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 s
Use g = 10 m/s2
v0
g
A (x0,y0)
y
h
x
horiz. distance
d = 20 m
B
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
v0
g
A (x0,y0)
y
h
x
horiz. distance
d = 20 m
B
x(t )  v0 * cos( ) * t
20m  v0 *1* 2s
v0  10m / s
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
12.7 Normal and Tangential Coordinates
ut : unit tangent to the path
un : unit normal to the path
Normal and Tangential
Coordinates
Velocity
Page 53

v  s * ut
Normal and Tangential Coordinates
Fundamental Problem 12.27
a
v
2

* un  at * ut
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:
(A) constant
•(B) 1 m/s2
•(C) 2 m/s2
•(D) not enough information
•(E) 4 m/s2
Fundamental Problem 12.27
a
v
2

* un  at * ut
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:
at  dv / dt  2 * 0.5 * t
At __ t  2s : _ at  2 *1m / s 2
(A) constant
•(B) 1 m/s2
•(C) 2 m/s2
•(D) not enough information
•(E) 4 m/s2
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
disk = 10 rad/s
B
C
r
e
er
Unit vectors
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
disk = 10 rad/s
B
C
r
e
er
Unit vectors
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
12.10 Relative (Constrained) Motion

 
VB  VA  VB / A
We Solve Graphically (Vector Addition)
vA
vB
vB/A
Example : Sailboat tacking against
Northern Wind



VWind  VBoat  VWind/ Boat
2. Vector equation (1 scalar eqn. each in
i- and j-direction)
500
150
i
Given:
r(t) = 2+2*sin((t)), _dot=
constant
The 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= constant
The 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
A
J
L
vA = const
vA is given
as shown.
Find vB

i
B
Approach:
Use rel.
Velocity:
vB = vA +vB/A
(transl. + rot.)
V
r = 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
6 m/s
(B) 40 rad/s
(C) -40 rad/s
(D) 4 rad/s
(E) none of the above
(A)
V
r = 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
6 m/s
(B) 40 rad/s
(C) -40 rad/s
(D) 4 rad/s
(E) none of the above
(A)
Omit all constants!
XB
xA
A
B
c
yE
Xc
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!
XB
xA
A
B
c
yE
Xc
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


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

Vy/vx =cot 


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()
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.
F
=
F
=
m
*g
T
=
m
a2
net
2
2
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 T
m2 g - ( m1 g + m1 a ) = m2 a
( m2 - m1 ) g = ( m1 + m2 ) a
( m1 + m2 ) a = ( m2 - m1 ) g
a = ( m 2 - m 1 ) g / ( m1 + m 2 )
Rules
1. Free-Body Analysis, one for
each mass
2. Constraint equation(s):
Define connections.
You should have as many
equations as Unknowns.
COUNT!
3. Algebra:
Solve system of equations
for all unknowns
J
g
m
M*g*sin*i
i
0 = 30
M*g
0
-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.
J
g
m
M*g*sin*i
i
0 = 30
M*g
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
N
-M*g*cos*j
 (Forces _ x)  m * g * sin  * i  m * x
 (Forces _ y)  N - m * g * cos * j  0(static _ only)
Friction F = mk*N:
Another horizontal
reaction is added in
negative x-direction.
J
g
m
M*g*sin*i
i
0 = 30
M*g
0
mk*N
N
-M*g*cos*j
 (Forces _ x)  (m * g * sin   mk * N ) * i  m * x
 (Forces _ y)  N - m * g * cos * j  0(static _ only)
Mass m rests on the 30
deg. Incline as shown.
The free-body reaction
seen by the incline in jdirection is
J
g
m
i
0 = 30
0
(A) -mg*sin30o
(B) +mg*sin30o
(C) -mg*cos30o
(D) +mg*cos30o
(E) None of the above
Mass m rests on the 30
deg. Incline as shown.
The free-body reaction
seen by the incline in jdirection is
J
mg
g
m
i
0 = 30
mg*cos()
0
(A) -mg*sin30o
(B) +mg*sin30o
(C) -mg*cos30o
(D) +mg*cos30o
(E) None of the above
m
g
Mass m rests on the 30
deg. Incline as shown.
J
The static friction
required to keep the
mass from sliding in idirection is
i
0 = 30 0
(A) -mg*sin30o
(B) +mg*sin30o
(C) -mg*cos30o
(D) +mg*cos30o
(E) None of the above
m
0 = 30 0
Mass m rests on the 30
deg. Incline as shown.
J
The static friction
required to keep the
g
mass from sliding in idirection is
mg*sin() i
(A) -mg*sin30o
(B) +mg*sin30o
(C) -mg*cos30o
(D) +mg*cos30o
(E) None of the above
Newton applied to mass B
gives:
(A) SFu = 2T = mB*aB
(B) SFu = -2T + mB*g = 0
(C) SFu = mB*g-2T = mB*aB
(D) SFu = 2T- mB*g-2T = 0
Newton applied to mass B
gives:
(A) SFu = 2T = mB*aB
(B) SFu = -2T + mB*g = 0
(C) SFu = mB*g-2T = mB*aB
(D) SFu = 2T- mB*g-2T = 0
Newton applied to mass A
gives:
(A) SFx = T +F= mA*ax ; SFy = N - mA*g*cos(30o) = 0
(B) SFx = T-F= mA*ax SFy = N- mA*g*cos(30o) = mA*ay
(C) SFx = T = mA*ax ; SFy = N - mA*g*cos(30o) =0
(D) SFx = T-F = mA*ax ; SFy = N-mA*g*cos(30o) =0
Newton applied to mass A
gives:
(A) SFx = T +F= mA*ax ; SFy = N - mA*g*cos(30o) = 0
(B) SFx = T-F= mA*ax SFy = N- mA*g*cos(30o) = mA*ay
(C) SFx = T = mA*ax ; SFy = N - mA*g*cos(30o) =0
(D) SFx = T-F = mA*ax ; SFy = N-mA*g*cos(30o) =0
Energy Methods
 
dW  F  dr
Scalar _ Pr oduct
Only Force components in direction of
motion do WORK
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. mk =
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
A car is traveling at 20 m/s on a level
road, when the brakes are suddenly
applied and all four wheels lock. mk =
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
T 1 above
0  1 / 2 * m * v02   m k * m * g * s
(E) none Tof2 the
Solve _ for _ brake _ dist _ s :
s
v02
2 * mk * g

400
meters
10
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
y
(A) Wg <0, Wspr <0
(B) Wg >0, Wspr <0
(C) Wg <0, Wspr >0
(D) Wg >0, Wspr >0
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
y
 
W  F * ds
(A) Wg <0, Wspr <0
(B) Wg >0, Wspr <0
(C) Wg <0, Wspr >0
(D) Wg >0, Wspr >0
Potential Energy
• For any conservative force F we can
define a W   F gdr  U
potential energy function U in the following
way:
r2
r r
U  U 2  U1  W    F gdr
r2
U2
r1
–
The work done by a conservative force is
r1 Uin
1 the
equal and opposite to the change
potential energy function.
Hooke’s Law
• Force exerted to compress a spring is
proportional to the amount of
Fs  kx
compression.
1 2
PE s  kx
2


Conservative Forces:
Gravity is a conservative force:
GMm
Ug  
R
• Gravity near the Earth’s surface:
Ug  mgy
• A spring produces a conservative force: U  1 kx 2
s
2
h
d

(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
h
(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
d

PE of block released = KE of block = PE gained by spring
Height dropped, h  d sin   4 sin 30m  2m
Potential energy released, mgh  1  9.8m / s 2  2m  19.6J
Kinetic energy of the block =
v U
1 2
mv  19.6J
2
2
2
 19.6  6.3 m/s
m
1
(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
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!
Page 336:
 dr
v
 ωr
dt
an =  x (  x r)
at = a x r
Arm BD is rotating with constant dot= h
>0, while point D moves at vD*i. Seen
from D, the velocity vector at B is:
(A) vB = vD*i - BD*h*cosi  BDh*sin*j
(B) vB = vD*i - BD*h*cosi  BDh*sin*j
(C) vB = vD*i + BD*h*cosi BDh*sin*j
(D) vB = - BD*h*cosi  BDh*sin*j
(E) none of the above
J
B
B
i
BD
AB
A
 (t)
 (t),
h(t)
D
vD(t)
Arm BD is rotating with constant dot= h
>0, while point D moves at vD*i. Seen
from D, the velocity vector at B is:
(A) vB = vD*i - BD*h*cosi  BDh*sin*j
(B) vB = vD*i - BD*h*cosi  BDh*sin*j
(C) vB = vD*i + BD*h*cosi BDh*sin*j
(D) vB = - BD*h*cosi  BDh*sin*j
(E) none of the above
J
B
B
i
BD
AB
A
 (t)
 (t),
h(t)
D
vD(t)
Meriam Problem 5.71
Given are: BC wBC  2 (clockwise), Geometry: equilateral triangle
with l  0.12 meters. Angle   60

180
Collar slides rel. to bar AB.
Mathcad EXAMPLE
Guess
Values:
(outward
motion of
collar is
positive)
wOA  1
vcoll  1
Vector Analysis: OA  rA vCOLL  BC  rAC
Mathcad does not evaluate cross products symbolically, so the LEFT and
RIGHT sides of the above equation are listed below. Equaling the i- and jterms yields two equations for the unknowns OA and vCOLL
Mathcad Example
part 2:
Solving the vector
equations
Mathcad
Examples
OA X rOA
BC X rAC
part 3
Graphical Solution
B
C
ARM BC: VA
= BC X rAC
Right ARM OA:
VA =OA X rOA
Collar slides rel. to Arm BC
at velocity vColl. The angle
of vector vColl = 60o
vB = vA + vB/A
Given: Geometry and
VA
Find: vB and AB
vA +
AB
x
vA = const
J
Graphical Solution
Veloc. of B
vA = const
r
AB Counterclo
ckw.

B
AB x r
vB
vB = ?
vA is
given

iA
AB x r
r
vB = vA + vB/A
Given: Geometry and
VA
Find: vB and AB
vA +
AB
x
vA = const
J
iA
r
AB Counterclo
ckw.

B
AB x r
vB
vA is
given

vA = const
Solution:
vB = vA + AB X r
AB x r
r
vB = 3 ft/s down,  = 60o
and vA = vB/tan. The relative
velocity vA/B is found from
vector eq.
y
(A)vA = vB+ vA/B ,vA/B points
(B) vA = vB+ vA/B ,vA/B points
(C) vB = vA+ vA/B ,vA/B points
(D) VB = vB+ vA/B ,vA/B points
x
vB
vA
vB = 3 ft/s down,  = 60o
and vA = vB/tan. The relative
velocity vA/B is found from
vector eq.
y
(A)vA = vB+ vA/B ,vA/B points
(B) vA = vB+ vA/B ,vA/B points
(C) vB = vA+ vA/B ,vA/B points
(D) VB = vB+ vA/B ,vA/B points
x
vB
vA
vA
vB
vA/B
Rigid Body Acceleration
Stresses and Flow Patterns in a Steam Turbine
FEA Visualization (U of Stuttgart)
aB = aA + aB/A,centr+ aB/A,angular
Given: Geometry and
VA,aA, vB, AB
Find: aB and aAB
r* AB2 + r* a
J
Look at the Accel. of B relative to A:
iA
vA = const
r
AB
Counterclockw
.

B
vB
Given: Geometry and
VA,aA, vB, AB
aB = aA + aB/A,centr+ aB/A,angular
r* AB2 + r* a
Find: aB and aAB
J
Look at the Accel. of B relative to A:
iA
We know:
vA = const
r

B
Centrip.
r* AB 2
1. Centripetal: magnitude r2 and
direction (inward). If in doubt, compute
the vector product x(*r)
Given: Geometry and
VA,aA, vB, AB
aB = aA + aB/A,centr+ aB/A,angular
r* AB2 + r* a
Find: aB and aAB
J
Look at the Accel. of B relative to A:
iA
We know:
vA = const
r
Centrip. r*
AB 2

B
r* a
1. Centripetal: magnitude r2 and
direction (inward). If in doubt, compute
the vector product x(*r)
2. The DIRECTION of the angular accel
(normal to bar AB)
Given: Geometry and
VA,aA, vB, AB
aB = aA + aB/A,centr+ aB/A,angular
r* AB2 + r* a
Find: aB and aAB
J
Look at the Accel. of B relative to A:
iA
We know:
vA = const
r
Centrip. r*
AB 2

B
Angular r* a
aB
1. Centripetal: magnitude r2 and
direction (inward). If in doubt, compute
the vector product x(*r)
2. The DIRECTION of the angular accel
(normal to bar AB)
3. The DIRECTION of the accel of point B
(horizontal along the constraint)
We can add graphically:
Start with Centipetal
Given: Geometry and
VA,aA, vB, AB
aB = aA + aB/A,centr+ aB/A,angular
Find: aB and aAB
r is the vector from reference
point A to point B
J
i
A
r
vA = const
AB
Angular r* a
Centrip. r* AB 2

B
aB
Given: Geometry and
VA,aA, vB, AB
We can add graphically:
Start with Centipetal
Find: aB and aAB
r is the vector from
reference point A to point B
J
i
aB = aA + aB/A,centr+ aB/A,angular
Now
Complete the
Triangle:
r* a
r* AB2
aB
A
vA = const
r
AB
Centrip. r* AB2

B
Result:
a is < 0 (clockwise)
aB is negative (to the
left)
E
F
A
a (t)
B
H
AB
The instantaneous center
of Arm BD is located at
Point:
(A) F
(B) G
(C) B
(D) D
(E) H
BD
J
 (t)
O
i
G
D
vD(t)
The instantaneous
center of Arm BD is
located at Point:
(A) F
(B) G
(C) B
(D) D
(E) H
E
F
A
a (t)
B
H
AB
BD
J
 (t)
O
i
G
D
vD(t)
Plane Motion
3 equations:
S Forces_x
S Forces_y
S Moments about G
fig_06_002
Plane Motion
3 equations:
S Forces_x
S Forces_y
S Moments about G
Translatio n :  Fx  m * x
..................... Fy  m * y
Rotation : ..... M G  I G *a
fig_06_002
Parallel Axes Theorem
Pure rotation about fixed point P
I P  IG  m * d
fig_06_005
2
Constrained Motion:
The system no longer has all three
Degrees of freedom
Describe the constraint(s) with an
Equation
Given: I_G=m*k2=300*1.5^2 =
675 kgm^2. The angular
accel of the rocket is
(A) 0.102 rad/s2
(B) 0.31 rad/s2
(C) 3.1 rad/s2
(D) 5.9 rad/s2
F_y = m*a =
300*8.69N
Thrust T = 4 kN
6.78
Given: I_G= 675 kgm^2, m = 300
kg. The angular accel of the
rocket is
(A) 0.102 rad/s2
(B) 0.31 rad/s2
(C) 3.1 rad/s2
(D) 5.9 rad/s2
F_y = m*a =
300*8.69N
Answer: (sum of moments about G
= I_G*alpha)
4000N * sin(1deg)*3m = 675*alpha
alpha = 0.31 rad/s^2
Thrust T = 4 kN
6.78