EGR 232 Engineering Dynamics
Lecture 11: Review for Test #2
Today:
Homework Questions
Review for Test #2
Applications of Newton's 2nd Law
Linear and Rotational Momentum
New Homework: Problems from Chap 12: 34, 46, 51, 59
Next Class:
Test #2
-- closed book
-- closed computer
-- closed notes
-- equation sheet provided
Format:
-- complete 3 homework style problems.
Fall 2014
EGR 232 Eng. Dynamics:
Equation Sheet for Test 2:
from Chapter 11:
Average Velocity
Average acceleration:
x
v
vave 
aave 
t
t
Instantaneous kinematic expressions:
dx
dv
v
a
a dx  v dv
dt
dt
Uniform Rectilinear Motion Equations:
x  x0  vt
Uniform Accelereated Rectilinear Motion Equations:
1
v  v0  at
x  x0  v0t  at 2
2
1
v 2  v 02  2a (x  x 0 )
x  x0  (v  v0 )t
2
Relative Motion:
xB  x A  xB / A
vB  v A  vB / A
aB  aA  aB / A
Graphical Representation of Kinematics:
x-t plot: slope = instantaneous velocity
v-t plot: slope = instantaneous acceleration
area under curve = change of position
a-t plot: slope = instantaneous jerk
area under curve = change of velocity
Rectangular Motion Kinematic Equations:
ˆ
r  xˆ
i y ˆ
jzk
Displacement:
Velocity:
dr
dt
ˆ
vxˆ
i y ˆ
jzk
v v ˆ
i v ˆ
j  v kˆ
v
x
Acceleration:
y
z
dv
dt
ˆ
a  x Aˆ
i  yAˆ
j  z Ak
aaˆ
i a ˆ
j  a kˆ
a
x
y
z
Constrained Motion:
L  Ax1  Bx2
d
L  Ax1  Bx2
dt
d
0  Av1  Bv2
dt


0  Av1  Bv2
0  Aa 1  Ba 2
Tangential and Normal Components:
v2
ˆ
ˆn  at e
ˆt
v  v et
a
e

Radial and Transverse Components:
ˆr  r e
ˆ
ˆr
vr e
rr e
Cylindrical Coordinates
r  r eˆr  zkˆ
v  r eˆr  r  eˆ  zkˆ




ˆr  r  2r e
ˆ
a  r  r 2 e




a  r  r  2 eˆr  r   2r  eˆ  zkˆ
-----------------------------------------------------------------------------------------------------------from Chapter 12:
Newton's 2nd Law:
Linear Momentum:
F  ma
Rotational Momentum:
HO  r  mv
L  mv
or
HO  mr 2
Universal Law of Gravitation:
mm
where G = 34.4 x 10-9 ft4/lb-s4 = 66.73 x 10-12 m3/kg-s2
FG  G 1 2 2
r
Problem 1)
EGR 232 Test #2
What is the fastest speed at which a car may travel on a flat surface while
maintaining a circular path of diameter 200 feet without sliding if the coefficient of
friction between the surface and the tires is 0.4? Make sure you provide the FBD
and AD of the vehicle and give the final answer in mph.
1 mile = 5280 ft
200 ft
v
Problem 2)
EGR 232 Test #2
An unknown force acts on a 2 kg particle such that it moves in a vertical plane as
given in radial-transverse coordinates which are unknown functions of time. At one
point of the path, the transverse motion terms are known to be
and
  5 rad / s
  1 rad / s2
   rad
2
and the radial position of the particle at the at same point is r  2 cos  where r is
in meters.
a) Draw the FBD and the AD of the particle in terms of radial-transverse
coordinates. (hint: don't forget to include the action of the weight on the particle.)
b) Determine the velocity (in radial-transverse coordinates) at this position and time.
c) Determine the acceleration (in radial-transverse coord) at this position and time
d) What magnitude and direction of force are needed to cause the motion
described.
g
eθ
er
v
a
m
r
θ
Problem 3)
EGR 232 Test #2
The two-mass system shown below is released from rest.
a) Determine the acceleration of block A.
b) Determine the tension in the pulley cord.
c) Determine the velocity of block B after 3 seconds.
d) Determine the relative acceleration of block B with respect to block A.
100 lb
A
μs = 0.20
μk = 0.20
y
x
B
75 lb
Prob blem 1)
EGR 232 Test #2 Review
What is the fastest speed at which a car may travel on a flat surface while
maintaining a circular path of diameter 200 feet without sliding if the coefficient of
friction between the surface and the tires is 0.4? Make sure you provide the FBD
and AD of the vehicle and give the final answer in mph.
1 mile = 5280 ft
Solution:
200 ft
given: radius of curve = 200 feet
coef of friction = 0.4
FBD
AD
W
aN
v
Ff
N
Apply Newton's 2nd Law:
F  ma
Normal Direction: Fn  man
Vertical Direction: Fy  may
Ff  man
W  N  0
Friction Equation:
Ff  s N  0.4N
solve for N, Ff and an:
man  s N
and
so
man  s mg  an  s g
Normal acceleration equation:
v2
an 
N  mg

so
v
s g 
v2


v 2  s  g  0.4(100ft )(32.2ft / s2 )
s  g  0.4(100ft )(32.2ft / s2 )  35.9ft / s
v  35.9ft / s 
1mi
3600s

 24.4mi / hr
5280ft
1hr
Problem 2)
EGR 232 Test #2 Review
An unknown force acts on a 2 kg particle such that it moves in a vertical plane as
given in radial-transverse coordinates which are unknown functions of time. At one
point of the path, the transverse motion terms are known to be
and
  5 rad / s
  1 rad / s2
   rad
2
and the radial position of the particle at the at same point is r  2 cos  where r is
in meters.
a) Draw the FBD and the AD of the particle in terms of radial-transverse
coordinates. (hint: don't forget to include the action of the weight on the particle.)
b) Determine the velocity (in radial-transverse coordinates) at this position and time.
c) Determine the acceleration (in radial-transverse coord) at this position and time
d) What magnitude and direction of force are needed to cause the motion
described.
Solution: based on radial-transverse coordinates
g
eθ
er
nd
First find the kinematics then apply Newton's 2 Law
v
a
Radial and Transverse Components:
ˆr  r e
ˆ
ˆr
vr e
rr e



m
r

ˆr  r  2r e
ˆ
a  r  r 2 e
θ
where
   2 rad
and
r  2 cos 
and
  5 rad / s
and
  1 rad / s
2
d( 2 cos  )
d
 2 sin 
 2 sin 
dt
dt
d( 2 sin  )
d(sin  )
d(  )
r
 2
 2 sin 
dt
dt
dt
r
eθ
er
FBD
Fr
Fθ
W
θ
 2 cos   2  sin
2
then
r  2 cos   2 cos(  / 2 )  0
r  2 sin  2( 5 )sin(  / 2 )  10
r  2 cos   2  sin  2( 5 ) cos(  / 2 )  2 ( 1 )sin(  / 2 )  2
Therefore:
v  r eˆr  r  eˆ  10 eˆr  0(5) eˆ  10 eˆr  0 eˆ m/s
2

eθ
AD
er
2




aθ
m
ar

a  r  r  2 eˆr  r   2r  eˆ  2  0(5)2 eˆr  0(1)  2(10)(5) eˆ  2 eˆr  100 eˆ m/s2
Forces:
F  ma

(Fr  W sin  ) eˆr  (F  W cos  )eˆ  m 2 eˆr  100 eˆ 
 Fr  m (2  g sin  )  (2kg )(2  9.81sin( / 2)  23.6m / s 2
2
F  mg cos   100m  F  m (g cos   100)  2(9.81cos( / 2)  100)  200m / s
Fr  mg sin   2m
Problem 3)
EGR 232 Test #2 Review
The two-mass system shown below is released from rest.
a) Determine the acceleration of block A.
100 lb
b) Determine the tension in the pulley cord.
c) Determine the velocity of block B after 3 seconds.
A
d) Determine the relative acceleration of block B
μ
=
0.20
s
with respect to block A.
μk = 0.20
WA =100 lb
for Body A:
aA
Friction Equation:
T
Ff  k N
y
mA
Ff
x
N
nd
Newton's 2 Law:
B
Fx  max
 T  Ff  mAaA
Fy  may  N  WA  0
2T
Body B:
Newton's 2nd Law:
Fy  may

mB
WB  2T  mB aB
WB=75 lb
Constrained motion:
aB
aA  2aB
Put everything together:
N  WA  mAg
and
becomes
and
T  k N  mAaA
: T  k mAg  mAaA 
WB  2T  mB aB

T  k mAg  mAaA
T  0.5mB g  0.5mB aB
k mAg  mAaA  0.5mB g  0.5mB aB
Then
aB 
0.5mB  k mA
0.5WB  kWA
0.5(75)  0.2(100)
g 
g 
32.2  2.37ft / s 2
2mA  0.5mB
2WA  0.5WB
2(100)  0.5(75)
aA  2aB  2(2.37)ft / s 2  4.74ft / s 2
WAaA
100(4.74)
 0.2(100) 
 34.7lb
g
32.2
 aBt  0  2.37(3)  7.11ft / s
T  k m A g  m AaA  kWA 
vB  vB0
aB / A  aB  aA  ()  (4.74iˆ)  (4.74iˆ  2.37 jˆ) ft / s 2
75 lb