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inematics
a
acceleration
gravitational
acceleration
position
radius
distance
time
velocity
ft/sec
2
m/s2
ft/sec
2
m/s2
ft
rn
ft m
ft
m
sec
s
ft/sec m/s
9
r
v
S y m b o l s
0:
e
rad/sec''
rad
tad/see
rad/s
2
rad
rad/s
angular acceleration
angular position
angular velocity
deals only with relationships among position, velocity,
acceleration, and time.
A body in motion can be considered a particle if rota-
tion of the body is absent or insignificant. A particle
does not possess rotational kinetic energy. All parts of
a particle have the same instantaneous displacement,
velocity, and acceleration.
A
rigid body
does not deform when loaded and can be
considered a combination of two or more particles that
remain at a fixed, finite distance from each other. At
any given instant, the parts (particles) of a rigid body
can have different displacements, velocities, and acceler-
ations if the body has rotational as well as translational
motion.
If r is the position vector of a particle, the instantaneous
velocity and acceleration are
r
[position] 1 4 . 1
dr
[velocity]
1 4 . 2
=-
dt
dv d
2
r
[acceleration]
1 4 .3
=-=-
dt dt
2
R E C T I L IN E A R M O T I O N
A rectilinear system is one in which particles move only
in straight lines. (Another name is
linear system.)
The
relationships among position, velocity, and acceleration
for a linear system are given by Eqs. 14.4 through 14.6.
S u b s c r i p t s
s(t) = v(t)dt =
J
a(t)de
1 4 . 4
0
initial
f
final
v(t)
=
d:;t)
=
ja(t)dt
n
normal
1 4 . 5
r
radial
( ) _ dv(t) _ d
2
s(t)
t
tangential
a t - dt - dt
2
1 4 . 6
e
transverse
I N T R O O O C T I O N T o K I N E M A T I C S
Dynamics
isthe study ofmoving objects. The subject is
. divided into kinematics and kinetics.
Kinematics
is the
study of a body's motion independent of the forces on
the body. It is a study of the geometry of motion with-
out consideration of the causes of motion. Kinematics
R e c ta n g u la r C o o r d in a te s
The position of a particle is specified with reference to
a coordinate. system. Three coordinates are necessary
to identify the position in three-dimensional space; in
two dimensions, two coordinates are necessary. A coor-
dinate can represent a linear position, as in the rectan-
gular coordinate 'system, or it can represent an angular
position, as in the polar system.
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ P r o f e s s i o n a l P ub lu ut io n s, ln e .
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out of the integrals in Eqs. 14.4 and 14.5. The ini
distance from the origin is
so;
the initial velocity
constant, va; and a constant acceleration is denoted
Consider the particle shown in Fig. 14.1. Its position,
as well as its velocity and acceleration, can be speci-
fied in three primary forms: vector form, rectangular
coordinate form, and unit vector form.
F i g u r e
1 4 .1
R e c t a n g u l a r C o o r d i n a t e s
~ath of particle
y
(x,
y.
z
x
k
z
The vector form of the particle's position is
r,
where
the vector
r
has both magnitude and direction. The
rectangular coordinate form is (x, y, z). The unit vector
form
is
0\
r
=
xi
+
yj
+
zk
1 4 .7
The velocity and acceleration are the first two deriva-
tives of the position vector, as shown in Eqs. 14.8
and 14.9.
dr
v=-
dt
= xi + yj + zk
dv d
2
r
a=-=-
dt dt
2
=xi+jjj+ik
1 4 .9
1 4 .8
C o n s t a n t A c c e le r a ti on
Acceleration is a constant in many cases, such as a free-
.falling
body with constant acceleration g. If the accel-
eration is constant, the acceleration term can be taken
a(t) =
o
v(t)
=
ao
J
dt
=
va
+
aot
s(t) =o J J dt
2
1 4 . 1
] 4 .1 ]
aot2
=S o
+
vat
+
-2-
v
2
(t)
=
6
+
2ao(s - so)
1 4 .1
1 4 .1
C U R V I L I N E A R M O T I O N
Curvilinear motion describes the motion of a part
o
along a path that isnot a straight line. Special exam
of curvilinear motion include plane circular motion
projectile motion. For particles traveling along cu
linear paths, the position, velocity, and accelerat
may be specified in rectangular coordinates as they
w
for rectilinear motion, or it may be more convenien
express the kinematic variables in terms of other c
dinate systems (e.g., polar coordinates).
T r an s v e r se a n d R a d ia l C o m p o n en t s
In polar coordinates, the position of a particle is
scribed by a radius,
r,
and an angle, B . The posit
may also be expressed as a vector of magnitude r
direction specified by unit vector e
r
Since the velo
of a particle is not usually directed radially out from
center of the coordinate system, it can be divided
two components, called
radial
and
transverse,
wh
are parallel and perpendicular, respectively, to the
radial vector. Figure 14.2 illustrates the radial
transverse components of velocity ina polar coordin
system, and the unit radial and unit transverse vect
e
r
and
ee,
used in the vector forms of the motion eq
tions.
[position]
1 4 .1
v=;e;
+
Veee
= fer + reee
[velocity)
]4. ]
.. B 0 2 )
=
r-r e
r
+ r B +
2 rB )ee
[acceleration] / 4 . / 6
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K in e m a t i ( s 14~3
F i g u r e i 4 . 2 R a d i a l
; ; d
T r ~ ~ 5 v e ~ 5 e C o o r d i ~ ~ t e ~ .
y
r
path of
particle
\ 8
eQ~ _ _e ~r _
~l _
x
T a n g e n t i a l a n d N o r m a l C o m p o n e n t s
Aparticle moving in a curvilinear path will have instan-
taneous linear velocity and linear acceleration. These
linear variables will be directed tangentially to the path,
and, therefore, are known as
tangential velocity, Vt,
and
tangential acceleration, at,
respectively. The force that
constrains the particle to the curved path will generally
be directed toward the center of rotation, and the par-
ticle will experience an inward acceleration perpendic-
ular to the tangential velocity and .acceleration, known
a s
the
normal acceleration, an.
The resultant accelera-
tion, a, is the vector sum of the tangential and normal
accelerations. Normal and tangential components of ac-
o celeration are illustrated in Fig. 14.3. The vectors en
and
et
are normal and tangential to the path, respec-
tively. p is the principal radius of curvature.
1 4 . 1 7
1 4 . 1 8
F i g u r e 1 4 . 3 T a n g e n t i a l a n d N o r m a l C o o r d i n a te s
instantaneous
center of rotation
y
x
P la n e C ir tu la r M o tio n
Plane circular motion (also known as rotational particle
motion, angular motion, or circular motion) is motion
of a particle around a fixed circular path. The behav-
ior of a rotating particle is defined by its angular po-
sition,
B ,
angular velocity, w, and angular acceleration,
a. These variables are analogous to the s, v, and a
variables for linear systems .. Angular variables can be
substituted one-for-one in place of linear variables in
most equations.
B
[angular position]
1 4 . 1 9
dB
[angular velocity]
1 4 . 2 0=-
dt
dJ.JJ
a=-
dt
d
2
e
[angular acceleration] 1 4 . 2 1
dt
2
R E L A T I O N S H IP S B E T W E E N f i N E A if A N D
R O T A T IO N A L V A R IA B L E S
..
- _
_ _ .
. . . . . . . . . . . . . . . . . .. . . . . . . . . . . . .
s
=
rB
1 4 . 2 2
1 4 .2 3
1 4 . 2 4
Vt =rw
dVt
at = ra = dt
v r 2
an=-=rw
r
1 4 . 2 5
P R O J E C T IL E M O T I O N
A projectile is placed into motion by an initial impulse.
(Kinematics deals only with dynamics during the flight.
The force acting on the projectile during the launch
phase is covered in kinetics.) Neglecting air drag, once
the projectile is in motion, it is acted upon only by
the downward gravitational acceleration (i.e., its own
weight). Thus, projectile motion is a special case of
motion under constant acceleration.
Consider a general projectile set into motion at an an-
gle of
B
from the horizontal plane, and initial velocity
Vo ,
as shown in Fig. 14.4. In the absence of air drag,
the following rules apply to the case of travel over a
horizontal plane.
The trajectory is parabolic.
The impact velocity is equal to initial velocity, Vo
The range is maximum when
B =
45 .
_The time for the projectile to travel from the
launch point to the apex is equal to the time to
travel from apex to impact point.
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ P r o f e s s i o n a l P u b li c a t i o n s , I n c .
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1 4 - 4 F E R e v i e w M a n u a l
1 -
The time for the projectile to travel from the apex
of its flight path to impact is the same time an ini-
tially stationary object would take to fall straight
down from that height.
F ig u r e 14 .4 P r o i e c t i l e M o t i o n
y
v{t)
path of projectile
~----
x
The following solutions to most common projectile
problems are derived from the laws of uniform accel-
eration and conservation of energy.
ax =0 14.26
a
y
=g 1 4 . 2 7
Vx= Vxo= Vocosf 14 .28
Vy=VyO-
gt
=Vosine -
gt 1 4 . 2 9
x
= vxot = votcose
1 4 . 3 0
1
2
1
2
Y
= Vyot - 29t = vot sin e - 29t
1 4 . 3 1
s A M p L E P R O B L E M s
Problems 1-3 refer to a particle whose curvilinear mo-
tion is represented by the equation s
= 2 0t + 4t2 - 3t
3
.
1. What is the particle's initial velocity?
(A)
2 0
mjs
(B) 2 5 mjs
(C) 30 mfs
(D) 32
mfs
76394
S o l u t io n :
ds
2
v =d =
20
+ 8t - 9t
t .
At
t=
,
v
=
20
+
(8) (0) - (9) (0)2
=
20
mfs
Answer is A.
2 .
What is the acceleration of the particle at tiuie
t = O ?
(A ) 2 mfs2
(B) 3 mf
s
2
(0) 5
mfs2
(D) 8
mf
s2
S o lu t i o n :
At t
=
0,
a
=8
mffp
Answer
is D.
#17 394 .
3. What is the maximum speed reached by the par-
ticle?
(A)
21.8
mfs
(B)
27.9
mjs
(0) 34.6 mfs
(D) 48.0
mfs
#78394
S o l u t io n :
The maximum of the velocity function is found
b y
equating the derivative of the velocity function to zero
and solving for t.
v=
20 +
8t - 9t
2
d v
- = 18t =0
dt
8
t=-
s
=
0.444 s
18
Vmax
=
0
+
(8)(0.444 s) - (9)(0.444 S)2
= 21.8 mfs
Answer is A.
4. Choose the equation that best represents a rigid
body or particle under constant acceleration.
(A) a =
9.81
mjs2 + voft
(B) v =o +
aot
0)
v
= vo +
l o t a(t)dt
D
a
=
vVr
.
P ro f es s io n al P u b li ca tio n s , I n c . - - -
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14 6 FE R ev ie w M an u al _
S o l u t i o n :
gt
2
Y =
ot sin e 2
J .. .
t
2
- votsine
+
y
=
0
2
y = -1500 m since it is below the launch plane.
( 9
m )
~12
m
. 2 s
t
2
- (1000 -; )
t
sin 30 - 1500m=
m
2
m
4.905 -
t -
500 - t - 1500
m
=0
S2 S
-b ...jb
2 -
4ac
t
=. [quadratic formula]
2a
500
V
-500)2 - (4)(4.905)(-1500)
(2 ) (4.905)
= +104.85 s, -2.9166 s
x
=
votcos8
=
1000 :) (104.85 s) cos 30
=90803 m (90800 m)
Answer is D.
F E - S T Y L E E X A M P R O B L E M S
Problems 1 and 2 refer to a particle for which the posi-
tion is defined by
set )
=
2 sin
ti
+
4 cas tj [t in radians]
1. What is the magnitude of the particle's velocity at
t =4 rad?
(A) 2.61
(B) 2.75
(C) 3.30
(D) 4.12
8 2689
2. What is the magnitude of the particle's acceleration
at
t
= 7r?
(A) 2.00
(B) 2.56
(C) 3.14
(D ) 4.00
8 3689
3. For the reciprocating pump shown, the radius of
the crank 'is ' T ' = 0.3 m,and the rotational speed is
n
=
350
rpm. What is the tangential velocity of point
A
on the crank corresponding to an angle of
e
=35 from
the horizontal?
(A) 0 m/s
(B ) 1.1 mjs
(C) 10
m/s
(D) 11 m/s
4. A golfer on level ground attempts to drive a golf
ball across a 50 m wide pond, hitting the ball so that it
travels initially at 25
isi].
The ball travels at an initial
angle of 45 to the horizontal plane. How far will the
golf ball travel, and does it clear the pond?
(A) 32
m;
the ball does not clear the pond
(B) 45 m; the ball does not clear the pond
(C) 58 m; the ball clears the pond
(D) 64 m; the ball clears the pond
5. Rigid link AB is 12 m long. It rotates counterclock-
wise about point A at 12 rev/min. A thin disk with ra-
dius 1.75m is pinned at its center to the link at point B.
The disk rotates counterclockwise at 60 rev /min with
respect to point
B.
What is the maximum tangential
velocity seen by any point on the disk?
(A) 6 m/s
(B) 26 m/s
(C )
33
m/s
(D) 45 mjs
#84394
#85994
P r o f e s s i o n a l P u b l i c a t i o n s , I n c .
IIIIiIII .;.. __ .;.. -- -.~
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K in em a ti c s 1 4 - 7
-
For the following pr-ob lerns use the NCEES Hand-
book as your only reference.
6. A particle has a tangential acceleration of at (rep-
resented by the equation given) when it moves around
a point in a curve with instantaneous radius of 1 m.
What is the instantaneous angular velocity (in rad/s)
of the particle?
at =2t - sin t + 3 cot t [inm/s2]
A t
2
+ cost + 3 In [csc r]
(B) t
2
-cost+3lnlcsctl
C t2 - cost
+
3 In [ s in t]
(D) t
2
+cost+3Inlsintl
3555794
7. A stone is dropped down a well. 2.47 s after the
stone is released, a splash is heard. If the velocity of
sound in air is 342
ti s] ,
find the distance to the surface
of the water in the well.
A 2.4
m
B.
7.2 m
C 28 m
(D) 30 m
2878687
Problems 8 and 9 refer to the following situation.
A motorist is traveling at 70 km/h when he sees a traffic
light in an intersection 250 m ahead turn red. The
light's red cycle is 15 s. The motorist wants to enter
the intersection without stopping his vehicle, just as
the light turns green.
8. What. uniform deceleration of the vehicle will just
put the motorist in the intersection when the light turns
green? .
(A) 0.18 m/s
2
(B) 0.25 m/s2
(C) 0.37m/s
2
(D )
1.3 m/s2
28 84687
9. If the vehicle decelerates at a constant rate of
0.5 m/s2, what will be its speed when the light turns
green?
(A) 43 km/h
(B) 52 km/h
(C)
59 km/h
(D) 63 km/h
2885687
Problems 10 and 11 refer to the following information.
The position (in radians) of a car traveling around a
curve is described by the following function of time (in
seconds).
8 (t)
= t
3
- 2 e - 4t + 10
10. What is the angular velocity at
t
= s?
(A) -16 rad/s
(B) -4 rad/s
(C) 11 rad/s
(D)
15
rad/s
2 886687
11. What is the angular acceleration at t = 5 s? .
(A )
4 rad/s
2
(B)
6 rad/s
2
(C) 26 rad/s
2
(D) 30 rad/s
2
. 2886 687
12. The rotor of a steam turbine is rotating at
7200 rev /min when the steam supply is suddenly cut
off. The rotor decelerates at a constant rate and comes
to rest after 5 min. What was the angular deceleration
of the rotor?
(A) 0.40 rad/s?
(B)
2.5 rad/s
2
(C ) 5.8 rad/s
2
(D) 16 rad/s
2
2887687
13. A flywheel rotates at 7200 rev /min when the power
is suddenly cut off. The flywheel decelerates at a con-
stant rate of 2.1
rad/s
and comes to rest
6
min later.
How many revolutions does the flywheel make before
coming to rest?
(A )
18000 rev
(B) 22000 rev
(C)
72000 rev
(D) 390000 rev
2888687
Problems 14-i6 refer to the following situation.
A projectile has an initial velocity of 110 m/s and a
launch angle of 20 from the horizontal. The surround-
in IT terrain is level and air friction is to be disregarded.
b ,
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ P r o f e s s io n a l P u b lic a ti o n s ; I n c .
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1 4 - 8 F E R e v i e w M a n u a l
IIIIII I ___
14. What is the flight time of the projectile?
(A) 3.8 s
(B) 7.7 s
(C )
8.9
s
(D) 12
s
O OOOMM 298
15. What is the horizontal distance traveled by the
projectile?
(A) 80
m
(B) 400 m
(C ) 800 m
(D) 1200
m
OOO OMM 298
16. What is the maximum elevation achieved by the
projectile?
(A) 72 m
(B) 140 m
(C)
350
m
(D ) 620 m
OOOOMM 298
S O L U T I O N S T O F E - S T Y L E E X A M P R B L E M S
S o l u t i o n
1 :
()
. ds( t)
2. .
v t
=
--;It
= cos
zi -- 4sintJ
At
t=
4
rad,
v(4) =2cos (4 rad) - 4sin (4 rad)j
=
1.3li - (-3.03)j
Iv 4)1
=
V(-1.31)2
+
(3.03)2
=.30
Answer is C.
S o lu ti o n 2 :
From Prob.
1,
v(t) = 2cos ti - 4 sin tj
( )
dv(t) . .
at = -- = -2Sintl- 4costJ
dt
a(r r)
=
-2sin1l'i-4cos1l'j
= O i
+ 4.0j
\a(1l) \=V(O)2
+
(4.0)2
=.0
Answer is D.
S o l u t i o n 3 :
Use the relationship between the tangential and rota-
tional variables.
Vt =
w
w =
angular velocity of the crank
=350 :i:) (21l' ::~) (60 ~)
mm
=36.65
rad/s
(
rad)
Vt
=
0.3
m) 36.65 -s-
=11.0 m/s
This value is the same for any point on the crank at
r =
0.3 m.
Answer is D.
S o l u t i o n 4 :
The elevation of the ball above the ground is
gt
2
. gt
2
Y =Vyot - 2 =vat sin e - 2
When the ball hits the ground, y =0, and
gt
2
votsine
=
2
Solving for t, the time to impact is
2va sin ()
t= - ---
9
Substitute the time of impact into the expression for
x
to obtain an expression for the range.
(
2vosine)
x
=
vatcose =vc 9 cose
2v
2
=
_0sin e cos ()
9
(2 ) (25
m r
---'----;m~s~ sin 45 cos45
9.812'
s
=63.7 m
Answer is D.
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S o l u t i o n 5 : S o lu t i o n 7 :
The elapsed time is the sum of the time for the stone to
fall and the time for the sound to return to the listener.
The distance, x, traveled by the stone under the influ-
ence of a constant gravitational acceleration is
The maximum tangential velocity of point B with re-
spect to point A is
Vt;BIA
=
rw
= r(27rf
(
rad) ( rev)
(12 m) 27r - 12 -.
rev nun
1 2
X =o +
v o t
+ 2 ,gt
60~
min
Xo and Vo are both zero.
=
15.08
m/s
1
x =_gt
2
2
The maximum tangential velocity of the periphery of
the
disk with respect to point B is
Solving for
t,
the time for the stone to drop is
Vt,disklB = rw =r(27r f
(1.75 m) (27r rad)
rev
(
60 re.v)
min
The time for the sound (traveling at velocity c) to return
to the listener is
60 _s_
min
=11.00 s s s ]
The total time taken is
he velocities combine when the two velocity vectors
coincide in direction. The maximum velocity of the
periphery of the disk with respect to point
A
is the sum
of the magnitudes of the two velocities.
h + t2
=
2.47 s
f
x x
-- + -
= 2.47 s
c
Vt,disklB
=
Vt,BIA + Vt,disklB
m m
=15.08 -
+
11.00 -
s s
Substitute values for 9 and c.
R
x
m + ------m =.47 s
9.81 - 342-
S2 s
=6.08 m/s
Answer is B.
By trial and error with the answer choices given (or by
solving the quadratic equation),
x
=28 m
S o lu t i o n 6 :
Answer is C.
dVt
at=: dt
Vt
= at dt
=J(2t~sint+3 cott) dt
= t
2
+ cos t + 3 In [sin tl [inm/s)
(70 ~) (1000 ~)
v o
=
3600 ~
=19.44
mls
S o lu t i o n 8 :
The initial speed of the vehicle is
Vt t2 + cost + 3 In [sin r]
=- =- - - - - - - - - - - - ~- - ~
rim
t
2
+ cas t + 3 In
[ s in
tl [in rad/s)
The distance traveled under a constant deceleration is
Answer is D.
1 2
x =Xo + vot - 2 , at
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ P r o f e s s i o n a l P u b l i c a t i o n s , I n c .
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1 4 - 1 0 F E R e v i e w M a n u a l
-
Letting
Xo =
0, the acceleration required to travel a
distance x in time
t
starting with velocity
v o
is
(2)(vot -
x)
a =~------ .,,-----
t
2
(2) [( 19.44 ~) (15
s ) -
250
m ]
(15 S )2
=
0.37 m/s2
Answer is C.
S o l u t i o n 9 :
v
=
va
+
at
krn ( -0.5 ~) (158) (3600 ~)
=70 _ + .:....:._
, -8
------=--_~-
h 1000 k~
=
43 km/h
Answer is A.
S o l u t i o n 1 0 :
d B 2
W ( t) = - = 3t - 4t - 4
. dt ..
w (3 ) =
(3)(3)2 - (4)(3) - 4
=
11 rad/s
Answer is C.
S o l u t i o n 1 1 :
a( t)
=
dw (t)
dt
6t - 4
a( 5) =
6)(5) - 4
=6 rad/s
2
Answer is C.
S o lu t i o n 1 2 :
W = W o - at
(
1200
_r~v) (271 _rad)
mm rev (s )
o = ---6-0-~s~------- - a( 5 min) 60 ~
min
e x
=
2.51 rad/s
2
Answer is B.
S o l u t i o n 1 3 :
1 2
B
=
B o
+
wo t - 2 at
=0 + (1200 : : i : )
(2 71) (6
ruin) - ~ (2.1 r : 2 d )
x
[ ( 6
min) (60 rr:n)]
2
=135.4
X
10
3
rad
e
=
135.4 x 10
3
rad
. 271
=1.5 X 10
3
rev
Alternative solution: The average rotational speed dur- .
ing deceleration is
rev .
7200 -. -0
mm
P r o f e s s i o n a l P u b l i c a t i o n s , I n c . =
2
=
600 rev /min
(
rev)
e =t =
3600 min (6 min)
=1,600 rev
Answer is B.
S o l u t i o n 1 4 :
The vertical component of velocity is zero at the apex.
Vy =
a sine -
gt
0= n o
7 )
sin 20 - (9.81 ~) t
t
=.84 s
The projectile takes an equal amount of time to return
to the ground from the apex. The total flight time is
T
= (2)(3.84 s) = 7.68 s (7.7 s)
Answer is B.
8/9/2019 FE/EIT Dynamics sample problems
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K in e m a t i c s 1 4 - 1 1
s o l u t i o n 1 5 :
Calculate the range from the horizontal component of
velocity.
S o l u t i o n 1 6 :
The elevation at time
t
is
x
=
vxt
=Vo
coset
= (110 7 ) cos200(7.68 s)
=794m
y =
vot sin e ~ ~
gt
2
=
(110 7 ) (3.84 s) sin
20
- (~) (9.81 ~) (3.848)2
=72 m
Answer is C.
Answer is A.
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ P r o fe s s io n a l P u b li ca t io n s , In c
8/9/2019 FE/EIT Dynamics sample problems
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inetics
S u b s c r ip t s
0
initial
f
friction
k
dynamic
n
normal
r
radial
R resultant
s
static
t tangential
e
transverse
I N T R O D U C T I O N T O K I N E T I C S
Kinetics is the study of motion and the forces that cause
N o m e n d a t u r e
motion. Kinetics includes an analysis of the relationship
acceleration
ft/sec
2
m/s2
between the force and mass for translational motion and
a
between torque and moment of inertia for rotational
f
linear frequency
Hz
Hz
motion. Newtons laws form the basis of the governing
F force
lbf
N
9
gravitational
theory in the subject of kinetics.
acceleration
ft/sec
2
m/s2
.............................................
9c
gravitational
M O M E N T U M
constant (32.2)
lbm- ft /lbf-sec
2
k
spring constant
lbf/ft
N/m
The vector linear momentum (momentum) is defined
m mass lbm kg by Eq. 15.1. It has the same direction as the velocity
N
normal force
lbf
N
vector. Momentum has. units of force x time (e.g., lbf-
p
linear momentum
lbf-sec
N-s
sec or Ns).
r
position
ft
m
r
radius
ft
m
p=mv
[ 8 1 ]
1 5 . 1 0
R
resultant force
lbf
N
mv
[U .S .]
1 5 .1 b
distance
ft
m
p=-
t
time
sec
gc
s
T
period
sec
s
Momentum is conserved when no external forces act on
v
velocity
ft/sec
m/s
W
weight
lbf
N
a particle. If no forces act on the particle, the velocity
and direction of the particle are unchanged. The
law o
conservation of momentum states that the linear mo-
S y m b o l s
mentum is unchanged if no unbalanced forces act on the
particle. This does not prohibit the mass and velocity
Q
angular
from changing, however. Only the product of mass and
acceleration rad/sec?
rad/s
2
velocity is constant.
S
deflection
ft
m
)
angular position
rad
rad
j . L
coefficient of
N E W r O N / S F i R S T A N D S E C O N D i A W S O F
M O T I O ~ . . . . . . .
friction
.r:
Newtons first law of motion
states that a particle will
ngle deg
deg
w
natural frequency
rad/sec
rad/s
remain in a state of rest or will continue to move with
P r of e ss io n a l p u b li ca t i o n s , I n c
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15 2 FER ev iew M an u a l _
constant velocity unless an unbalanced external force
acts on it.
This law can also be stated in terms of conservation of
momentum: If the resultant external force acting on a
particle is zero, then the linear momentum of the par-
ticle is constant.
Newtons second law of motion
states that the acceler-
ation of a particle is directly proportional to the force
acting on it and is inversely proportional to the particle
mass. The direction of acceleration is the same as the
direction of force.
This law can be stated in terms of the force vector re-
quired to cause a change in momentum: The resultant
force is equal to the rate of change of linear momentum.
F= dp
dt
1 5 . 2
For a fixed mass,
F _ dp _ d(mv)
- dt
;It
dv
=m-
dt
=ma [S I]
[US.]
1 5 . 3 a
F= ma
gc
1 5 . 3 b
s
W E I G H T
The
weight,
W, of an object is the force the object exerts
due to its position in a gravitational field, g.
W=mg
W=
mg
gc
1 5 . 4 a
S I]
[US.]
1 5 . 4 b
gc
is the gravitational constant, approximately 32.2 lbm-
ft/lbf-sec
2
.
Friction is a force that always resists motion or impend-
ing motion. It always acts parallel to the contacting
surfaces. If the body is moving, the friction is known
as
dynamic friction.
If the body is stationary, friction
is known as
static friction.
The magnitude of the frictional force depends on the
normal force,
N,
and the
coefficient of friction, IL,
be-
tween the body and the contacting surface.
1 5 . 5
The static coefficient of friction is usually denoted with
the subscript
s,
while the dynamic coefficient of friction
is denoted with the subscript
k. ILk
is often assumed
to be 75 percent of the value oflLs. These coefficients
are complex functions of surface properties. Experi-
mentally determined values for various contacting con-
ditions can be found in handbooks.
For a body resting on a horizontal surface, the normal
force
is the weight of the body. If the body rests on an
inclined surface, the normal force is calculated as the
component of weight normal to that surface, as illus-
trated in Fig. 15.1.
N=mgcos
N
=
mgcos
9 c
1 5 . 6 a
1 5 . 6 b
1 5 .1
[S I]
[US.]
F i g u r e 1 5 . / F r i c t i o n a l a n d N o r m a l F o r c e s
impending motion
P r o fe s s io n a l P u b li c a ti o n s, I n c. - - - -
The frictional force acts only in response to a disturbing
force, and it increases as the disturbing force increases.
The motion of a stationary body is impending when the
disturbing force reaches the maximum frictional force
ILsN.
Figure 15.1 shows the condition of impending
motion for a block on a plane. Just before motion starts,
the resultant,
R,
of the frictional force and normal force
equals the weight of the block. The angle at which
motion is just impending can be calculated from the
coefficient of static friction.
=an
-lJ.Ls
Once motion begins, the coefficient of friction droPS
slightly, and a lower frictional force opposes movement.
This is illustrated in Fig. 15.2.
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K in e t i c s 1 5 - 3
F i g u r e 1 5 . 2 F r i c t i o n a l F o r c e ~ e r s u s D i s t u r b i n g F o r c e . . . . .
. I impending motion
no rnotron ~ I
~e~~~b~~~~~~e~ I
~motion
4~
~sN
disturbing force
K i N E T I C S O F A P A R T i C L E
Newton's second law can be applied separately to any
direction in which forces are resolved into components.
The law can be expressed in rectangular coordinate
form (i.e., in terms of z- and y-component forces), in
polar coordinate form (i.e., in tangential and normal
components), or in radial and transverse component
form.
R e c ta n g u la r C o o r d in a t es
Equation 15.8 is Newton's second law in rectangular
coordinate form and refers to motion in the x-direction.
Similar equations can be written for the y-direction or
any other coordinate direction.
[ 8 1 ]
1 5 . 8
In general,
Fx
may be a function of time, displacement,
and/ or velocity. If
Fx
is a function of time only, then
the motion equations are
J
(F x(t))
vx(t) =xo + - - - : ; : n - dt
x( t) = o + vxot + J vx( t)dt
1 5 . 1 0
[ 8 1 ]
1 5 .9 '
If Fx is constant (i.e., is independent of time, displace-
ment, or velocity), then the motion equations become
[ 8 1 ]
1 5 . 1 1
vx (t)
=
v-o
+ (~) t
1 5 . 1 2
F. t
2
x(t) =Xo + vxot
+
2~
a
x
t
2
=Xo + vxo t + --
2
1 5 .1 3
T a n g e n t i a l a n d N o rm a l C o m p o n e n t s
For a particle moving along a circular path, the tangen-
tial and normal components of force, acceleration, and
velocity are related.
LFn = man=
m ( v I )
[ 8 1 ]
1 5 . 1 4
LF
t
=
m a t
=m
d ~ t )
[ 8 1 ]
1 5 . 1 5
R a d i a l a n d T r a n sv er se C o m p o n e n t s
For a particle moving along a circular path, the radial
and transverse components of force are
[ 8 1 ] 1 5 . 1 6
LFe =mae
[ 8 1 ]
1 5 . 1 7
F R E E V I B R A T I O N
.
: : ,
.
Vibration is an oscillatory motion about an equilibrium
point. If the motion is the result of a disturbing force
that is applied once and then removed, the motion is
known as
natural
(or free )
vibration.
If a continuous
force or single impulse is applied repeatedly to a system,
the motion is known as
forced vibration.
A simple application of free vibration is a mass sus-
pended from a vertical spring, as shown in Fig. 15.3.
After the mass is displaced and released, it will oscillate
up and down. If there is no friction (i.e., the vibration
is undamped), the oscillations will continue forever.
F i g u r e
1 5 . 3
S i m p l e M a s s -S p r i n g S y s t e m
The system shown in Fig. 15.3 is initially at rest. The
mass is hanging on the spring, and the equilibrium po-
sition is the static deflection,
8
st
.
This is the deflection
due to the gravitational force alone.
mg =k 8
st
[ 8 1 ]
1 5 . 1 8 0
[u.s.] IS . 1 8 b
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ P r o f e s s i o n a l P u b l i c a t i o n s I n c
8/9/2019 FE/EIT Dynamics sample problems
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1 5 - 4 F E R e v i e w M a n u a l _
The system is then disturbed by a downward force (i.e.,
the mass is pulled downward from its static deflection
and released). After the initial disturbing force is re-
moved, the mass will be acted upon by the restoring
force (-
kx )
and the inertial force (
-mg) . .Both
of these
forces are proportional to the displacement from the .
equilibrium point, and they are opposite in sign from
the displacement. The equation of motion is
F=ma
[ 8 1 J
1 5 .1 9
mg - k(x + Ost ) = m/i:
[ 8 1 ]
1 5 . 2 0
kO st - k(x + Ost ) = mii:
[ 8 1 J
1 5 .2 1
mx+kx = 0
[ 8 1 ]
1 5 .2 2
The solution to this second-order differential equation is
x(t)
=
C
1
coswt
+
C
2sinwt
1 5 .2 3
C
1
and C
2
are constants of integration that depend on
the initial displacement and velocity of the mass. w is
known as the natural frequency of vibration or angular
frequency. It has units of radians per second. It is not
the same as the linear frequency,
i
which has units of
hertz. The period of oscillation, T, is the reciprocal of
the linear frequency.
w = {
[ 8 1 ]
1 5 . 2 4 0
k
gc
[U .8 .]
1 5 . 2 4 b
= -
m
f= ~ = ~
1 5 .2 5
2 1 1
T
T
= ~ = 211
1 5 .2 6
f
w
For the general case where the initial displacement is
Xo
and the initial velocity is va, the solution of the equation
of motion is
x(t)
=
ocoswt + ~) sinwt 1 5 . 2 7
For the special case where the initial displacement is
Xo
and the initial velocity is zero, the solution of the
equation of motion is
x t) =xocos w t
1 5 .2 8
s A M p f p R O B I E M S .
....................
P ro fe s si on a l P u b li ca t i o n s , In c . - - - .
1.
For which of the following situations is the net force
acting on a particle necessarily equal to zero?
(A) The particle is traveling a~ constant velocity
around a circle. .
(B)
The particle has constant linear momentum.
C The particle has constant kinetic energy.
(D) The particle has constant angular momentum.
#86691
S o l u t i o n :
This is a restatement of Newton's first law of motion
which says that if the resultant external force
acting
on a particle is zero, then the linear momentum of the
particle is constant.
Answer is B.
2.
One newton is the force required to
(A) give a 1
g
mass an acceleration of 1m/s2.
(B)
accelerate a
10
kg mass at a rate of
0.10
mjs2.
(C) accelerate a 1 kg mass at a rate of 1.00 em/s2.
(D) accelerate a 1 kg mass at a rate of 9.81 m/s2.
#87689
S o l u t i o n :
Newton's second law can be expressed in the form of
F =ma. The unit of force in S1 units is the newton,
which has fundamental units of kgm/s
2
. A newton is
the force required to accelerate a 1kg mass at a rate of
1
m/s2 or a
10
kg mass at a rate of
0.10
m/s2.
Answer is B.
3. A 550 kg mass initially at rest is acted upon by a
force of 50e
t
N. What are the acceleration, speed, and
displacement of the mass at
t =4
s?
(A) 4.96 m/s2, 4.87 mis, 19.5 m
(B )
4.96 m/s2, 4.96 mis, 19.8 m
(C) 4.96
m/s2,
135.5
mis,
1466
m
D
4.96
m/s
2
,
271
us],
3900
in
#88691
S o l u t i o n :
F 50e
4
N
a = - = - - -
550 kg
=4.96
rn/s
2
8/9/2019 FE/EIT Dynamics sample problems
15/37
K i n e t i c s 1 5 - 5
r
4
50e
t
N e
t
4 1
v = )0 550 kg dt = 1110 = 4.96 - 11
= 4.87 m/s
i
t i t 50e
t
N
i t
(t) = --k-
de
=
(4.87 m
dt
a a 550 gas
s =
~87tl~
=
(4.87
7 )
(4 s ) - 0
=19.48 rn
Answer is A.
Problems 4 and 5 refer to the following situation.
A 5 kg block begins from rest and slides down an
inclined plane .
After 4 s, the block has a velocity of 6 m/s.
4. If the angle of inclination is 45, how far has the
block traveled after 4 s?
(A ) 1.5 m
(B) 3
m
(C) 6
m
(D ) 12 m
89689
S o l u t i o n :
v(t) =a
+
aot
m
v(t) -va 6 -;-0
ao= =
_o? _
t 4
s
=.5 m/s2
1 2
s C t =o + vat + 2 ao t
=0+0+ (~) (1.5 ~) (4S)2
=
12
m
Answer is D.
5. What is the coefficient of friction between the plane
- and the block?
(A ) 0.15
(B) 0.22
(C)
0.78
(D ) 0.85
90 689
S o l u t i o n :
Choose a coordinate system so that the x-direction is
. parallel to the inclined plane.
LFx
= ma
x
= mg
x - Ff
ma
x
=mgsin45 - p,mgcos45
mgsin45 -:-ma
x
p,=
mgcos45
gsin45 - ax
9 cos 45
(9.81 ~) s in 45 -1.5 ~
=
(9.81 : Z cos 45
=
0.78
Answer is C.
6. A constant force of 750 N is applied through a pulley
system to lift a mass of 50 kg as shown. Neglecting
the mass and friction of the pulley system, what is the
acceleration of the 50 kg mass?
F=
750 N
(A ) 5.20 m/s2
(B )
8.72 m/s2
(C) 16.2 m/s2
(D) 20.2 m/s2
91 694
S o l u t i o n :
Apply Newtons second law to the mass and to the
two frictionless, massless pulleys. Refer to the following
free-body diagrams.
T,
1 .
F
T,
mass A
pulley B
pulley C
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1 5 6 F E R e v i e w M a n u a l - - - - - - - - - - - - - - - - - - - - - - - _
mass A: Tl - mg =ma
pulley
B: 2T2 - T , =0
pulley C: T2 = F = 750 N
Tl -
mg
2T2 -
mg
2F -
mg
a=
m m
m
(2) (750 N) - (50
kg)
(9.81 ~)
50
kg
=
20.2
m/s2
Answer is
7. A mass of 10 kg is suspended from a vertical spring
with a spring constant of 10 N/m. What is the period
of vibration?
A 0.30 s
(B) 0.60 s
(C )
0.90
s
(D) 6.3
s
9 2394
S o l u t i o n :
m ~
T =
2 7 rV
k =
27 r ~
10 ~
=6.3 s
Answer is
F E -S T Y L E E X A M P R O B L E M S
1.
If the sum of the forces on a particle is not equal to
zero, the particle is
(A)
moving with constant velocity in the direction
of the resultant force.
(B)
accelerating
in
a direction opposite to the re-
sultant force.
(C)
accelerating in the same direction as the resul-
tant force.
(D)
moving with a constant velocity opposite to
the direction of the resultant force.
93689
2.
A varying force acts on a 40 kg weight as shown in
the following force versus time diagram. What is the
object's velocity at
t =
4 s if the object starts from
rest?
F N
3
2
2
t (5)
(A) 0 m/s
B
0.075
m/s
(C ) 0.15 m/s
(D) 0.30 ta]
9 4691
Problems 3 and 4 refer to the following situation.
The 52 kg block shown starts from rest at posi-
tion A and slides down the inclined plane to posi-
tion
B.
When the block reaches position B, a 383 N hori-
zontal force is applied.
The block comes to a complete stop at position
C.
The coefficient of friction between the block and
the plane is f .- L =0.15.
c
3.
Find the velocity at position B.
(A) 2.41 m/s
(B) 4.12
m/s
(C) 6.95
m/s
(D)
9.83
m/s
. 9568 1
4. Find the distance between positions Band C.
(A) 3.23 m
(B) 4.78 rn
(C )
7.78 m
(D) 10.1
m
966 81
P r o f e s s i o na l P u b l ic a t io n s , I n c . .- - -
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problems 5 and 6 refer to the following pulley system.
In standard gravity, block
A
exerts a force of 10000 N
and block
B
exerts a force of 7500 N. Both blocks are
initially held stationary. There is no friction and the
pulleys have no mass. ,~
10000 N
5.
Find the acceleration of block
A
after the blocks are
released.
(A) 0 m/s2
(B ) 1.4 m/s2
(C ) 2.5 m/s2
(D) 5.6
m/s2
9799 4
6 .
Find the velocity of block
A
2.5 s after the blocks
are released.
\
(A )
B
(C )
(D)
o m/s
3.5 m/s
4.4 m/s
4.9 m/s
9 8994
For the following problems use the NCEES Hand-
book as your only reference.
7. What is the period of a pendulum that passes the
center point 20 times a minute?
(A) 0.2 s
(B) 0.3 s
(C ) 3 s
(D) 6 s
2142689
8. A variable force of (40 N)cos
e
is attached to the end
of a spring whose spring constant is 50 N
[ux.
There is
no deflection when e =90. At what angle, e , will the
spring deflect 20 em from its equilibrium position?
(A) -14
(B) 25
(C ) 64
(D) 76
39331294
K ineti s 15 7
9. A spring has a constant of 50 N/m. The spring is
hung vertically, and a mass is attached to its end. The
spring end displaces 30 em from its equilibrium position.
The same mass is removed from the first spring and
attached to the end of a second (different) spring, and
the displacement is 25 em. What is the spring constant
of the second spring?
(A )
(B )
C
(D)
46
N/m
56 N/m
60 N/m
6 3 N /m
39351294
10. A cannonball of mass 10 kg is fired from a cannon
of mass 250 kg. The initial velocity of the cannonball
is 1000
km/h.
All of the cannon's recoil is absorbed by
a spring with a spring constant of 520 NIcm. What is
the maximum recoil distance of the cannon?
A 0.35 m
B
0.59
m
(C ) 0.77 m
(D )
0.92 m
3 942295
11. A child keeps a 1 kg toy airplane flying horizontally
in a circle by holding onto a 1.5 m long string attached
to its wing tip. The string is always in the plane of the
circular flight path.
If
the plane flies at 10
us],
what
is the tension in the string?
(A) 7 N
(B ) 15 N
(C )
28
N
(D ) 67 N
3 471 193
12. A car with a mass of 1530 kg tows a trailer (mass
of 200 kg) at 100 km/h. What is the total momentum
of the car-trailer combination?
(A) 4600 N-s
(B) 22000 N-s
(C) 37000Ns
(D) 48 000 Ns
1584 689
13. A car is pulling a trailer at 100 km/h. A 5 kg cat
riding on the roof of the car jumps from the car to the
trailer. What is the change in the eat's momentum?
(A) -25 N-s(loss)
(B) a N-s
(C) 25 Ns (gain)
{D) 1300 N-s (gain)
#4.169 795
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1 5 - 8 F E R e v i e w M a n u a l 1 _
-
14. A 3500 kg car accelerates from rest. The constant
forward tractive force of the car is 1000 N, and the con-:
stant drag force is 150 N. What distance will the car
travel in 3 s?
A 0.19
m
(B )
1.1
m
C
1.3
m
(D )
15 m
2108 689
S O L U T I O N S T O F E - S T Y L E E X A M P R O B L E M S
S o lu ti o n 1 :
Newton's second law, F =ma, can be applied sepa-
rately to any direction in which forces are resolved into
components, including the resultant direction.
Since force and acceleration are both vectors, and mass
is a scalar, the direction of acceleration is the same as
the resultant force.
Answer is C.
S o l u t i o n 2 :
Use the impulse-momentum principle. The impulse is
the area under the F-t curve. There are two right tri-
angles.
F= dp
d t
FD.. t = m :::.v
FD.. t
(2) [(~) (3N)(2S)]
D ..v
= -- =
--=-~--:,------=-
tri
40 kg
=
0.15
m/s
Answer is C.
S o l u t i o n 3 :
Choose a coordinate system parallel and perpendicular
to the plane, as shown.
L F x =max
Wx - flN =ma
x
mg sin e - flmg cas e = ma;
ax =
9
sin e -
fl9
cas e
=9.81 ~) [1
5
3 - (0.15) G ~ ]
=
2.415
m/s2
v
2
=v~
+
2ao( s - so )
v a
=
o =
v
2
= 2 aos
= (2) (2.415 ~) (20
m)
=
96.6
m2/s2
v
= J
96.6 ~2
=
9.83 iis]
Answer is D.
S o l u t i o n 4 :
L F x =ma
mg
sin
e - pcose - fl mgcose + Psine)
=
ma
a = (~) [mgsine - pcose - fl mgcose + P sin e)]
P
g sin e - flg cos e - - (cose
+
fl sin e)
m
=9.81 ~) [ 1
5
3 - (0.15TG~)]
_ (383
N ) [12 + (0.15) (~)]
52 kg 13 13
=
4.809
m/s2
P r o f e s s io na l P ub li c a t i o n s , I n c . - - -
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K inetin 1 5 9
2 2
v =vo+2aO(S-30)
vo =
9.83 m/s
[from Prob. 3]
v = 30 = 0
2 _ (9.83 m ) 2
-vo S
3 =
2ao
= (2 ) (-4.809 ;)
=
10.05
m :
Answer is D.
S o lu t io n 5 :
Refer to the following free-body diagrams.
T
T
Apply Newton's second law to the free body of mass A.
\
But
m=Wig,
so
1 0 0 0 0 : )
(aA)
=
10000
N -
TA
9.812
s
Apply Newton's second law to the free body of
mass B.
Combine the equations and solve for
aA,
setting
TA =
TB, and aA =-aB'
W
A
- mAaA = WB - mBaB = WE + mBaA
WA - WB g(WA - WB)
aA= =
mB+mA WB+WA
m (10000 N -7500 N)
= 9.81 s2 7500 N + 10000 N _
= 1 .4 m /s2
A l t e r n a t e S o lu ti o n :
But F
=
ma, so
Since the tension in the rope is the same everywhere,
10000 N - T T - 7500 N
10000 N 7500 N
T
=
8571 N
From block A,
=
g
g(WA -
T)
A
(9.81 ~) (10000 N - 8571 N)
10000 N
=1.4
m/s2
Answer is B.
S o lu ti on 6 :
VA
=
Vo
+ aAt
=
+
1 . ~ ; (2.5 s)
=3.5 ui]
Answer is B.
S o lu t io n 7 :
A pendulum will pass the center point two times during
each complete cycle. Therefore,
10
cycles are completed
in
60 s.
elapsed time
T = - ------,--
no. of cycles
60
s
- 10
= 6
s
Answer is D.
___________________________________ . ~h~oo~~~~m~ ~
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15 10 FE R ev iew M anua l _
S o lu t io n 8 :
From Hookes law, the relationship between force, F,
and deflection,
x,
or a linear spring is
F=kx
(50 :) (20 cm)
(40 N )c o se
=
cm
100 -
m
c as e =
0.25
e
=
cos-1(0.25)
-;; 75.5 (76)
Answer is D.
S o lu ti on 9 :
The gravitational force on the mass is the same for both
springs. From Hooke ~ law, .
F =
k1Xl
=
kZX2
kz =k1Xl
X2
(50 ~) (30 cm)
25 cm
=0 N/m
Answer is C.
Solution 1 0 :
Use the conservation of momentum equation to deter-
mine the velocity of the cannon after the ball is fired.
Initially, the cannon and cannonball are both at rest.
Since the cannon recoils, its velocity direction will be
opposite (i.e., negative) to the direction of the cannon-
ball.
(10 kg) (0)+(250 kg) (0)
=
10 kg) (1000 ~) + (250 kg) (v.)
Vc
=
40 km/h
Use the conservation of energy principle to determine
the compression of the spring. The kinetic energy of
the cannon will be equal to the elastic potential energy
stored in the spring.
KE=PE
(6.5)(250 kg) [(40~) (l~OO ~) r
3600
h J
= (0.5) (520 c:) (100 :)
x2
:r;
=
0.77 m
Answer is C.
So lu tio n 1 1:
The normal acceleration (perpendicular to the path of
the airplane) is
v
Z
an
=
.l.
T
(10
~ 2
1.5 m
=66.7 m/s2
The tension in the string is equal to the centripetal
force.
=
1 kg) (66.7 ~)
= 66.7 N
Answer is D.
Solution 12 :
(100 k:) (1000 :)
v = ------- ---.,. ----
3600 ~
= 27.78
sii]
P=mv
= (1530 kg + 200 kg) (27.78 7 )
=
8060
N-s
Answer is D.
S o lu ti on 1 3 :
The law of conservation of momentum states that the
linear momentum is unchanged if no unbalanced forces
act on an object. This does not prohibit the mass and
velocity from changing; only the product .of mass and
P ro f e s s io n a l P u b li ca l io n s, In c .
8/9/2019 FE/EIT Dynamics sample problems
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locity is constant. In this case, both the total mass
d the velocity are constant. Thus, there is no change.
nswer is
B
l u ti o n 1 4 :
F=1000 N - 150 N
=
850 N
F
a
m
850 N
3500 kg
=
0.243
m/s2
1
2
S
= v o t
+ zat
=+ ~
(0.243 ~) (3 s)2
=
1.09
m
nswer is B
K i n e t i c s 1 5 - 1 1
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Kinetics of
Rotational Motion
S u b s c r i p t s
o initial
c centroidal
f friction
n normal or natural
o origin or center
S static
t tangential or torsional
N o m e n c l a tu r e
a
acceleration
ft/sec
2
m/s2
A
area
ft2 m2
M A S S M O M E NT O F I N E R T I A
d
distance
ft
m
F
force lbf N
The
mass moment of inertia
measures a solid object's
9
gravitational
acceleration
ft/sec
2
m/s2
resistance to changes in rotational speed about a specific
axis.
Ix , Iy,
and
Iz
are the mass moments of inertia with
. g e
gravitational
respect to the
x-, y-,
and z-axes, respectively. They are
constant
(32.2) lbm-ft.Zlbf-sec
not components of a resultant value.
G
shear modulus
lbf/ft
2
Pa
h
angular momentum
ft-lbf-sec
Nms
I mass moment
i: = J
(y2 + z2)dm
1 6 . 1
of inertia
Ibm-ft2
kg.m
2
J
area polar moment
of inertia
ft4
m4
t;
=
J
(x
2
+ z2)dm
1 6 .2
k
t
torsional spring
constant
ft-lbf/rad
N-m/rad
t=
J (x
2
+
y2)dm
length
ft
rn
1 6 .3
m
mass
lbm kg
M
moment
ft-lbf
N-m
r
radius
ft
rn
The
centroidal mass moment of inertia, Ie ,
is obtained
r
radius of gyration
ft
m
when the origin of the axes coincides with the object's
R
moment arm
ft
m
center of gravity. Once the centroidal mass moment of
t
time
sec s
inertia is known, the
parallel axis theorem
is used to find
v
velocity
ft/sec
m/s
the mass moment of inertia about any parallel axis. In
W
weight
lbf N
Eq.
16.4,
d is the distance from the center of mass to
the parallel
ax is .
S y m b o l s
. 2
1 6 .4
Q
angular acceleration
rad/sec''
rad/s
2
Iny parallel axis = Ie +md
o
angular position
rad
rad
)
superelevation angle
deg deg
For a composite object, the parallel axis theorem must
f J ,
coefficient of friction
be applied for each of the constituent objects.
p
density
Ibm/ft
3
kg/m
3
w
angular velocity
rad/sec
radj's
to
natural frequency
rad/sec rad/s
-2 . 2
1 6 . 5
I = Ie ,l +m]:d
1
+ Ic ,2 +m2
d
2 + ...
P r o f e s s i o l o l P ub li c ati o n s, I n c
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1 6 - 2 F E R e v i e w M a n u a l _
The radius of gyration, r , of a solid object represents the
distance from the rotational axis at which the objects
entire mass could be located without changing the mass
moment of inertia.
For a rigid body rotating about an axis passing through
its center of gravity located at point 0, the scalar value
of angular momentum is given by Eq. 16.9.
ho =Iw
[S I]
] 6 . 9 0
r=ff
] 6 .6
Iw
[U .S .]
] 6 . 9 b
o=-
gc
1= r2m
] 6 .7
Table 16.1 (at the end of this chapter) lists the mass
moments of inertia and radii of gyration for some stan-
dard shapes.
P L A t ~ f M o t i o N O F
A
R I G ID B O D Y
General rigid body plane motion, such as rolling wheels,
gear sets, and linkages, can be represented in two dimen-
sions (i.e., the plane of motion). Plane motion can be
considered as the sum of a translational component and
a rotation about a fixed axis, as illustrated in Fig. 16.1.
F i g u r e ]
6.]
C o m p o n e n t s
of
P l a n e M o ti o n
p
plane motion
p
translation
rotation
R o t a t i o n A b o u t a F i x e d A x i s
Rotation about a fixed axis describes a motion in which
all particles within the body move in concentric circles
about the axis of rotation.
The angular momentum taken about a point is the
moment of the linear momentum vector. Angular mo-
mentum has units of distance x force x time (e.g.,
ft-lbf-sec or Ncn-s). It has the same direction as the
rotation vector and can be determined from the vectors
by use of the right-hand rule (cross product).
ho
=
r X mv
mv
ho=rx-
. 9c
]6 . 8 0
1 6 . 8 b
[S I]
[ U .S .]
Although Newton's laws do not specifically deal with
rotation, there is an analogous relationship between ap-
plied moment and change in angular momentum. For
a rotating body, the moment (torque),
M,
required to
change the angular momentum is
dho
M=-
dt
] 6 . ]0
The rotation of a rigid body will be about the center
of gravity unless the body is constrained otherwise. If
the moment of inertia is constant, the scalar form of
Eq.
16.10
is
[S I] ] 6 . 1 1 0
[U .S .] ] 6 . 1 1 b
Velocity and position in terms of rotational variables
can be determined by integrating the expression for ac-
celeration.
] 6 .1 2
1 6 .] 3
] 6 .1 4
I n s t a n t a n e o u s C e n t e r o f R o t a t io n
Analysis of the rotational component of a rigid body's
plane motion can sometimes be simplified if the loca-
tion of the body's
instantaneous center
is known. Using
the instantaneous center reduces many relative motion
problems to simple geometry. The instantaneous center
(also known as the
instant center
and
Ie)
is a point at
which the body could be fixed (pinned) without chang-
ing the instantaneous angular velocities of any point on
the body. Thus, for angular velocities, the body seems
to rotate about a fixed instantaneous center.
The instantaneous center is located by finding two
points for which the absolute velocity directions are
known. Lines drawn
per pen dicu lar to these two velo c -
ities will intersect at the instantaneous center. (This
P r o f e s s i o n a l P u b l ic a t i o n s , In c . --
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_ . _ K in et ic s o f R o t at io n al M o tion 163
crraphic procedure is slightly different if the two veloci-
o
ties are parallel, as Fig. 16.2 shows.) For a rolling wheel,
the instantaneous center is the point of contact with the
supporting surface.
.. ,... .
F i g u r e 1 6 . 2 . G r a p h i c M e th o d o f F in d i n g t h e I n s t a n ta n e o u s C e n t e r
IC
IC
The absolute velocity of any point, P, on a wheel rolling
(Fig. 16.3) with translational velocity, vo, can be found
by geometry. Assume that the. wheel is pinned at point
C and rotates with its actual angular velocity,
e
=
w =
voir. The direction of the point's velocity will be per-
pendicular to the line of length
I
between the instanta-
neous center and the point.
lvo
v=lw= -
r
1 6 . 1 5
F i g u r e
1 6 . 3
I n s t a n t a n e o u s C e nt e r o f a R o ll i n g W h e e l
C E N t R I F U G A C F O R C f
t
Newton's second law states that there is a force for every
acceleration that a body experiences. For a body mov-
ing around a curved path, the total acceleration can be
separated into tangential and normal
components,
By
Newton's second law, there are corresponding forces in
the tangential and normal directions. The force associ-
ated with the normal acceleration is known as the
cen-
tripetal force. The centripetal force is a real force on the
body toward the center of .rotation. The so-called cen-
trifugal force
is an apparent force on the body directed
away from the center of rotation. The centripetal and
centrifugal forces are equal in magnitude but opposite
in sign.
Equation 16.16 gives the centrifugal force on a body of
mass m with distance r from the center of rotation to
the center of mass.
mvi 2
F e = man = -- = mrw
r
2 2
F
_ man _ mv
t _
mrw
g e g e
r
ge
[S I] 1 6 . 1 6 0
[u.s.]
1 6 .1 6 b
B A N K IN G O F C U R V E S .
If a vehicle travels in a circular path on a fiat plane with
instantaneous radius r and tangential velocity
Vt,
it will
experience an apparent centrifugal force. The centrifu-
gal force is resisted by acombination of roadway bank-
ing (superelevation) and sideways friction. The vehicle
weight, W, corresponds to the normal force. For small
banking angles, the maximum frictional force is
Ff
=J.lsN
=
.lsW
1 6 . 1 7
For large banking angles, the centrifugal force contributes
to the normal force. If the roadway is banked so that
friction is not required to resist the centrifugal force, the
superelevation angle,
e ,
can be calculated from Eq, 16.18.
v
2
tan 8
= -.i.
gr
1 6 . 1 8
T O R S I O N A L F R E E V I B R A T IO N
The torsional pendulum in Fig. 16.4 can be analyzed
in a manner similar to the spring-mass combination.
Disregarding the mass and moment of inertia of the
shaft, the differential equation is
1 6 . 1 9
F i g u r e
1 6 . 4
T o r s i o n a l P e n d u l u m
L
G
)9
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,
1 6 - 4 F E R e v i e w M a n u a l _
For the torsional pendulum, the torsional spring con-
stant
k
t
can be written
1 6 . 2 0
The solution to Eq.
16.20
is directly analogous to the
solution for the spring-mass system.
B (t) =ocosw n t + (::) sinwnt
1 6 .2 1
S A M p L E P R O B L E M s
1.
Why does a spinning ice skater's angular velocity
increase as she brings her arms in toward her body?
(A) Her mass moment of inertia is reduced.
(B) Her angular momentum is constant.
(e) Her radius of gyration is reduced.
(D )
all of the above
99689
S o l u t i o n :
As the skater brings her arms in, her radius of gyra-
tion and mass moment of inertia decrease. However, in
the absence of friction, her angular momentum,
h,
is
constant. From Eq. 16.9,
h
w
I
Since angular velocity, w, is inversely proportional to
the mass moment of inertia, the angular velocity in-
creases when the mass moment of inertia decreases.
Answer is D.
2. Link AB of the linkage mechanism shown in the
illustration rotates with an instantaneous counterclock-
wise angular velocity of 10 rad/s, What is the instan-
taneous angular velocity of link Be when link AB is
horizontal and link CD is vertical?
,W A B
= 10 rad/s counterclockwise)
~/~5m
--5-m--I ~
I
4 C . -
5m
D._
~
(A) 2.25 rad/s (clockwise)
(B) 3.25 rad/s (counterclockwise)
(e) 5.50 rad/s (clockwise)
(D) 12.5 rad/s
(clockwise)
1 00694
S o l u t i o n :
Find the instantaneous center of rotation. The absolute
velocity directions at points Band e are known. The
instantaneous center is located
by
drawing perpendicu-
lars to these velocities as shown. The angular velocity
of any point on rigid body link Be is the same at this
instant.
VB = 50 m/s
t
B
I/IC
4m
II
- - - - - - - - - - - - 1 -
1
I
I
13m
I
I
I
4 I
vc-
C
VB =ABwAB
(
rad)
= (5 m) 10 -s- = 50m/s
50 m
VB
wBC = OB = 4 ~ = 12.5 rad/s
[clockwise]
Answer is D.
3. Two 2 kg blocks are linked as shown. Assuming
that the surfaces are frictionless, what is the velocity of
block B if block A is moving at a speed of 3
m/s?
P r of e ss io n a l P u b li ca l io n s , I n c. - -
8/9/2019 FE/EIT Dynamics sample problems
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16 6 FE R ev iew M an u a l _
1. A 2 kg mass swings in a vertical plane at the end
of a 2 m cord. When e = 30, the magnitude of the
tangential velocity of the mass is
1
ui]. What is the
tension in the cord at this position?
(A ) 18.0 N
(B) 19.6 N
(C) 24.5 N
(D) 29.4 N
104 694
2. A 2 kg mass swings in the horizontal plane of a circle
of radius 1.5 m and is held by a taut cord. The tension
in the cord is 100 N. What is the angular momentum
of the mass?
(A) 5.77 Nms
(B) 26.0
N-m s
(C) 113Nun-s
(D)
150 N-ms
105691
3. A disk rolls along a fiat surface at a constant speed
of 10 m/s. Its diameter is 0.5 m. At a particular in-
stant, point P on the edge of the disk is 45 from the
horizontal. What is the velocity of point P at that in-
s tan t?
o
f------------j
10
m/s ~
(A) 10.0 m/s
(B ) 1 5 .0 m/s
(C) 16.2
m/s
(D) 18.5 m/s
106193
4. A car travels around an unbanked 50 m radius curve
without skidding. The coefficient of friction between
the tires and road is 0.3. What is the car's maximum
speed?
(A)
14
km/h
(B) 25 km/h
(C) 44 km/h
(D) 54 km/h
107394
5. Traffic travels at 100 km/h around a banked high-
way curve with a radius of
1000
m. What banking angle
is necessary such that friction will not be required to re-
sist the centrifugal force?
(A )
1.4
(B) 2.8
(C) 4.5
(D) 46
2107689
Forthe following problems use the NCEES Hand-
book as your only reference.
6. The center of gravity of a roller coaster car is 0.5 ill
above the rails. The rails are 1m apart. What is the
maximum speed that the car can travel around an un-
banked curve of radius 15 m without the inner wheel
losing contact with the top of the rail?
(A) 8.58
m/s
B
12.1
m/s
(C) 17.2
m/s
(D) 24.2 m/s
2133689
7. A 50 kg cylinder has a height of 3 m and a radius of
50 em. The cylinder sits on the z-axis and is oriented
with its major axis parallel to the y-ax is . W hat is the
mass moment of inertia about the z-axis?
P r o fe s s io n a l P u b l k c fi o n s, I n c . -- -
8/9/2019 FE/EIT Dynamics sample problems
27/37
A
4.1 kgm
2
(B )
16 kgm
2
(C ) 41 kgm
2
(D )
150 kgm
2
3553794
8. A
uniform thin disk has a radius of
30 em
and a mass
of
2
kg.
A
constant force of
10 N
is applied
tangentially
at a varying, but unknown, distance from the center
of the disk. The disk accelerates about its axis at 3t
rad/s
2
. What is the distance from the center of the
disk at which the force is applied at t
= 12
s?
Cl W
~
R
= 30 em lever
(A )
32.4 em
(B) 3.6.0 em
(C )
54.0 em
(D ) 108
em
3975395
9. A torsional pendulum consists of a 5 kg uniform disk
with a diameter of
50
em attached at its center to a rod
1.5
m in length. The torsional spring constant is
0.625
Nrri/rad. Disregarding the mass of the rod, what is the
natural frequency of the torsional pendulum?
(A) 1.0 rad/s
(B) 1.2 radys
(C ) 1.4 rad/s
(D) 2.0 rad/s
4170 795
10. A 3 kg disk with a diameter of
0.6
m is rigidly at-
tached at point B to a 1 kg rod 1 m in length. The
rod-disk combination rotates around point A. What is
the mass moment of inertia about point A for the com-
bination?
0.6 m
(A) 0.47 kgm
2
(B) 0.56 kgm
2
(C )
0.87
kgrrr
(D) 3.7 kgm
2
2095689
11. A 1kg uniform rod 1 m long is suspended from
the ceiling by a frictionless hinge. The rod is free to
pivot. What is the product of inertia of the rod about
the pivot point?
(A) 0
kgm
2
(B) 0.045 kg-rrr
(C)
0.13 kg.m
2
(D) 0.33 kgm
2
2096689
12.
A wheel with a radius of 0.75
m
starts from rest
and accelerates clockwise. The angular acceleration (in
rad/s'') of the wheel is defined by a=6t - 4. What is
the resultant linear acceleration of a point on the wheel
rim at
t
=
2
s?
(A ) 6
m/s2
(B ) 12 m/s2
(C ) 13
m/s2
(D ) 18m/s2
2 893 687
13. A uniform rod (AB) of length L and weight W
is pinned at point C and restrained by cable OA. The
cable is suddenly cut. The rod starts to rotate about
point C, with point A moving down and point B moving
up. What is the instantaneous linear acceleration of
point B?
~
a
I__
Cl ~
L
- .. ; . P ro f e s sio n a l P u b li ca t io n s, I n c
8/9/2019 FE/EIT Dynamics sample problems
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16 8 FE R ev iew M an ua l 1 _
(A)
g
16
B
g
4
(C )
g
7
D
g
4
2911 687
14. A
uniform rod
(AB)
of length
L
and weight
W
is pinned at point
C.
The rod is accelerating with an
instantaneous angular acceleration (in
rad/s'')
of
0:
=
12g
/7
L. What is the instantaneous reaction at point
C?
A
1 .
4
B
L
A)
W
-
4
(B)
W
-
3
(C )
4W
-
7
(D)
7W
12
2912687
15. A 1530 kg car is towing a 300 kg trailer. The coef-
ficient of friction between all tires and the road is
0.80.
How fast can the car and trailer travel around an un-
banked curve of radius
200
m without either the car or
trailer skidding?
(A) 40.0 km/h
(B) 75.2
km/h
(C )
108.1
km/h
(D) 143
km/h
1579689
16. A 1530 kg car is towing a 300 kg trailer. The coef-
ficient of friction between all tires and the road is
0.80.
The car and trailer are traveling at
100
km/h around
a banked curve of radius
200
m. What is the neces-
sary banking angle such that tire friction will not be
n eces s ary to p rev en t
skidding?
(A) 8
(B) 21
(C) 36
(D) 78
158 2689
17. A wheel with a
0.75
m radius has a mass of
200
kg.
The wheel is pinned at its center and has a radius of
gyration of 0.25 m. A rope is wrapped around the wheel
and supports a hanging
100
kg block. When the wheel
is released, the rope begins to unwind. What is the
angular acceleration of the wheel?
A
5.9
rad/s
2
(B) 6.5
rad/s
2
(C ) 11 rad/s
2
(D) 14 rad/s
2
2907687
S O L U T I O N S T O F E -S T Y L E E X A M P R O B L E M S
So lu ti on 1:
Use tangential and normal components.
P r of es s io n a l P u b li ca t io n s , In c . - - -
v2 1 7 ) 2
an=
. . . 1 .
=---- -.:....-
r
2 m
=.5
m/s2
Sum forces in the normal direction.
LPn
=
man
=
T - Wsin60
T=man
+
mg sin 60
=
(2
kg)
(0.5 ~)
+
(2
kg)
(9.81 ~)
(sin
60)
=8.0 N
Answer is A.
So lu tio n 2 :
tension
=
centripetal force
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- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - lK in e t i c s o f R o t a t i o n a l M o t io n 1 6 - 9
mv
2
T= __
t
r
r
mv
2
-- =f1mg
r
v = vf1gr
= V
(0.3) (9.81 ~) (50 m )
=
12.13
m / s
(12.13 ~) (3600 ~)
v=
1000
r:
=
43.67
km/h
Answer is C.
S o lu t i o n 5 :
Since there is no friction force, the superelevation angle,
B ,
can be determined directly.
v
2
tanB = ~
gr
e
= tan
( ; )
(
(100 ~) (1000 ~))
2
3600 ~
=tan-l~--~--~~------~~
(9.81 ~) (1000 m)
100 N
(2
kg)(1.5 m)
=5.77
rad/s
ho
=
rmv
=
r
2
mw =
(1.5 m)2 (2 kg) (5.77 r:d)
=26.0 Nms
Answer is B.
S o l u t io n 3 :
Use the instantaneous center of rotation to solve this
problem. Assume the wheel is pinned at point A.
Z 2
=
(2)(0.25
m)2 -
(2)(0.25
m)2(cos
135)
[lawof cosines]
= 0.2134 m 2
I =
VO.2134m 2
=
0.462 m
Zvo (0.462 m) (10 ~)
vp = ---;;:- 0.25 m
=
18.5
m/s
The velocity of point P is perpendicular to the line AP.
Answer
is D.
S o lu t i o n 4 :
The car uses friction to resist the centrifugal force.
=
4.50
Answer
is
C.
S o lu t i o n 6 :
The wheel will lose contact with the top of the rail when
the reaction on the rail is zero. Refer to the following
illustration. Wheel A is the inner wheel.
YCG
=0.5 m
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ P r o fe s s io n a l P u b l i c a ti o n s , I n c .
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1 6 - 1 0 F E R e v i e w M a n u a l
1 _
The forces acting on the car are the centrifugal force, F e ,
its weight, and the reaction at the outer wheel. (The
reaction at the inner wheel is zero.) Take moments
about rail B.
LMB = 0 =
WXCG - F eYCG
mv;YcG
=mgxCG-
_.. .::.. : :.... .:....:c .
r
S o l u t io n 9 :
The radius of the disk is
R= .~D
2
50 em
- - 2 (100 ::)
=0.25 m
v =g XCGr
YCG
(9.81 ~) (~) (15 m)
0.5 m
The mass moment of inertia of the disk is
I = ~MR2
2
=(~) (5 kg)(0.25 m)2
=0.15625 kg-rrr
Natural frequency can be determined from the equation
for the torsional spring constant.
k
t
=
w
2
I
0.15625 kgm
2
= rad/s
= 12.1 m/s
Answer is B.
S o l u t i o n 7 :
Find the formula for
Ix
in the Mass Moments of Inertia
table.
m 3R2 + 4h2)
Ix
=----------
12
(50 kg)[(3)(0.5 m)2 + (4)(3 m)2]
12
=153.1 kg-rrr (150 kgrn )
w = /
~----,--,--
0.625 Nm
rad
Answer is D.
Answer is D.
S o l u t io n 8 :
The centroidal moment of inertia is
I
=
~mR2
2
=0.5)(2 kg)(0.3 m)2
= 0.09 kgm
2
The acceleration at t=12 s is
a =
3t
= ( 3 r:d) (12 s)
=
6 rad/s
2
Mo=F r=Ia
Ia
r
F
(0.09 kgm
2
) (36 ~)
10 N
=.324 m (32.4 em)
So lution 1 0 :
The mass moment of inertia of the rod about its end is
ML2
Irod,A
=3
(1 kg)(1 m)2
3
=0.33 kg.m
2
The mass moment of inertia of the disk about its own
center is
MR2
Idisk,c
=
-2-
(3 kg) Tr
2
=
0.135 kgm
2
Answer is A.
The distance AC is
AC =JAB2
+
BC
2
1 m)2 + (O.~ ill
r
=
1.04
rn
.
- - - - - - - - _ _ - - - - - - - - - - - - - - - _ . .
_
.
ro f e s si on a l P u b li ca t i o n s , I n c .
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Using the parallel axis theorem, the mass moment of
inertia of the disk about point A is
Idisk,A =Idisk,e +mdisk
AC2
=0.135
kgm
2
+
(3
kg)
(1.04
m)2
=3.38 kgrrr
The total moment of inertia of the rod and disk is
IA =
Irod,A
+
Idisk,A
=0.33kgm
2
+
3.38
kgm
=
3.71
kg.m
2
Answer is D.
S o lu t i o n 1 1 :
The product of inertia for the rod is zero because the
pivot point lies on an axis of symmetry.
Answer is A.
S o lu t i o n 1 2 :
The angular acceleration at t
=
2 s is
a(2 s)
=
(6 )(2 s) - 4 =8 rad/s
2
The equation for the angular velocity is
w
=
a(t)dt
J(6t - 4)dt
3t
2
- 4t +wo
However, the wheel starts from rest, so W o =
O .
At t = 2 s, the angular velocity is
w (2
s ) = (3)(2 S)2 - (4)(2 s )
= rad/s
The tangential acceleration of the point is
(
rad)
=
(0.75
m ) 8 ~
=6 m/s2
The normal acceleration (directed toward the center of
the wheel) is
(
rad)2
=(0.75 m) 4S
=2 m/s2
The resultant acceleration is
a
=
V a ; + a ~
( 6 ~ )
2
+ ( 1 2 ~ )
2
=13.4
m/s2
Answer is
C.
S o lu t i o n 1 3 :
Point C is L/4 from the center of gravity of the rod.
The mass moment of inertia about point C is
Ie =eG + md
2
=
1
1
2 )
m L
2
+ m ( ~
r
=
m L 2
C ~
+
1
1
6 )
: 8 ) m L
2
The sum of moments on the rod is
W L m g L
4 4
The angular acceleration is
L M e
a
Ie
m g L
4
: 8 ) m L 2
12 g
7L
The tangential acceleration of point B is
= (~) ~1
g
7
Answer is C.
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1 6 - 1 2 F E R e v i e w M a n u a l - - - - - - - - - - - - - - - - - - - - - - - - - - - - - _
S o l u ti o n 1 4 :
The mass moment of inertia of the rod about its center
of gravity is
Take moments about the center of gravity of the rod.
All moments due to gravitational forces will cancel. The
only unbalanced force action on the rod will be the re-
action force, Rc, at point C.
LMCG =Rc (~)
LMCG =CGacG
Rc= 4W
7
Answer is C
S o lu ti o n 1 5 :
To keep the vehicle and trailer from skidding, the cen-
tripetal force must be less than or equal to the frictional
force. At the limit,
Fe =F
f
p,N
a
n
m
p,mg
= -- =p,g
m
=
0.8) (9.81 ~)
=
7.848
m/s2
The normal acceleration can be calculated from the tan-
gential velocity.
Vt =
Janr
=
J
(7.848 ~) (200
m )
=
39.6 m/s
(39.6 ~) (3600 ~)
1000 :
=143
km/h
Answer is D
S o lu tio n 1 6 :
The velocity is
(100 k:) (1000 :)
v - -- --------;;----
- 3600 ~
=
27.78
m/s
The necessary superelevation angle is
Answer is B
S o l u ti o n 1 7 :
The mass moment of inertia of the wheel is
The unbalanced torque (moment) on the wheel is
M=
F R
=
mg - ma)R = mRg - a)
= mblockRg - Ra)
The acceleration is given by
M=Ia
mblockRg - Ra)
= =
mwheelr2a
mblockRg
_ (100 kg)(0.75
m )
(9.81 ~)
- (200 kg)(0.25 m )2 + (100 kg)(0.75 m )2
=
0.7
rad/s
2
Answer is C
P r o f e s s i o n a l P u b l i c a ti o n s , I n c . - - - -
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nergy and Work
N o m e n c l a t u r e
a
acceleration
e
coefficient of
restitution
E
energy
F
force
g gravitational
acceleration
\
gravitational
gc
constant
(32.2)
h
height above
datum
I
mass moment
of inertia
Imp
impulse
k
spring constant
m
mass
p
linear momentum
r
distance
t
time
v
velocity
W
work
X
displacement
S y m b o l s
w
angular velocity
E N E R G Y A N D W O R I C
ft-lbf
lbf
J
N
Ibm-ftj'lbf-sec-'
ft
m
lbm-ft2
Ibf-sec
lbf/ft
lbm
lbm-ft/sec
ft
kgm
2
Ns
N/ m
kg
kg-m.'s
m
s
m/s
J
sec
ft/sec
ft-lbf
ft
m
rad/sec
rad/s
The
energy
of a mass represents the capacity of the mass
to do work. Such energy can be stored and released.
There are many forms that the stored energy can take,
including mechanical, thermal, electrical, and magnetic
energies. Energy is a positive, scalar quantity, although
the change in energy can be either positive or negative .
Wark, W,
is the act of.changing the energy of a mass.
Work is a signed, scalar quantity. Work is positive when
a fo rce acts in the d irec tio n o f m o tio n an d m ov es a m as s
from one location to another. Work is negative when
a force acts to oppose motion. (Friction, for example,
always opposes the direction of motion and can only do
negative work.) The net work done on a mass by more
than one force can be found by superposition.
The work performed by a force is calculated as a dot
product of the force acting through a displacement.
w=
l=
1 7 . 1
K i n e t i c E n e r g y
Kinetic energy
is a form of mechanical energy associ-
ated with a moving or rotating body. The
linear kinetic
energy of a body moving with instantaneous linear ve-
locity v is
1
KE =
Zmv
2
[S I]
1 7 . 2 0
2
KE=
mv
[U .S . ] 1 7 . 2 b
2g
c
The
rotational kinetic energy
of a body moving with
instantaneous angular velocity
w
is
KE = ~Iw2
[S 1]
1 7 . 3 0
KE = Iw2
[U .S .]
/ 7 . 3 b
2gc
For general plane motion in which there are transla-
tional and rotational components, the kinetic energy is
the sum of the translational and rotational forms.
The change in kinetic energy is calculated from thedif-
ference of squares ofvelocity, not from the square of the
velocity difference.
1 7 . 4 0
1 7 . 4 b
P r o f e s s i o n a l P u b l i c a t i o n s , l i n e .
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1 7 - 2 F E R e v i e w M a n u a l
1 _
P o te n t ia l E n e rg y
Potential energy (also known as gravitational potential
energy)
is a form of mechanical energy possessed by
a mass due to its relative position in a gravitational
field. Potential energy is lost when the elevation of a
mass decreases. The lost potential energy usually is