mecanica-para-ingenieria-dinamica-anthony-bedford-5-edicion-solucionario

Problem 18.116 Model the excavator’s arm ABC as a y
C
single rigid body. Its mass is 1200 kg, and the moment
of inertia about its center of mass is I = 3600 kg-m2. If B 3.0 m
point A is stationary, the angular velocity of the arm is 2.4 m
zero, and the angular acceleration is 1.0 rad/s2 counter-

clockwise, what force does the vertical hydraulic cylin-

der exert on the arm at B?

Ax

Solution: The distance from A to the center of mass is 1.7 m 1.7 m

d = (3.4)2 + (3)2 = 4.53 m. Ax B mg
The moment of inertia about A is Ay
IA = I + d2m = 28,270 kg-m2.
From the equation of angular motion: 1.7B − 3.4mg = IAα.
Substitute α = 1.0 rad/s2, to obtain B = 40,170 N.

Problem 18.117 Model the excavator’s arm ABC as a
single rigid body. Its mass is 1200 kg, and the moment of
inertia about its center of mass is I = 3600 kg-m2. The
angular velocity of the arm is 2 rad/s counterclockwise
and its angular acceleration is 1 rad/s2 counterclockwise.
What are the components of the force exerted on the arm
at A?

Solution: The acceleration of the center of mass is


i jk

aG = α × rG/A − ω2rG/B =  0 0 α  − ω2(3.4i + 3j)
3.4 3 0

= −16.6i − 8.6j m/s2.

From Newton’s second law:

Ax = maGx = −19,900 N, Ay + B − mg = maGy .

From the solution to Problem 18.132, B = 40,170 N, from which
Ay = −38,720 N

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Problem 18.118 To decrease the angle of elevation of y
the stationary 200-kg ladder, the gears that raised it are Ox
disengaged, and a fraction of a second later a second
set of gears that lower it are engaged. At the instant
the gears that raised the ladder are disengaged, what is
the ladder’s angular acceleration and what are the com-
ponents of force exerted on the ladder by its support
at O? The moment of inertia of the ladder about O is
I0 = 14,000 kg-m2, and the coordinates of its center of
mass at the instant the gears are disengaged are x = 3 m,
y = 4 m.

Solution: The moment about O, −mgx = Ioα, from which y

α = − (200)(9.81)(3) = −0.420 rad/s2. (x , y)
14,000
mg
The acceleration of the center of mass is Fy

i jk x
Fx
aG = α × rG/O − ω2rG/O =  0 0 α  = −4αi + 3αj
340

aG = 1.68i − 1.26j (m/s2).

From Newton’s second law: Fx = maGx = 336 N, Fy − mg = maGy ,
from which Fy = 1710 N

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Problem 18.119 The slender bars each weigh 17.8 N 45°
and are 254 mm. long. The homogenous plate weighs 203.2 mm
44.5 N. If the system is released from rest in the posi-
tion shown, what is the angular acceleration of the bars 1016 mm
at that instant?
Ay Ax By Bx
Solution: From geometry, the system is a parallelogram, so that 17.8 N 17.8 N
FAx
the plate translates without rotating, so that the acceleration of every FBx
point on the plate is the same. FAy FBy
Newton’s second law and the equation of angular motion applied to FAy
the plate: −FAx − FBx = mpaP Gx , FAy + FBy − Wp = mpaP Gy . The FAx FBy FBx
motion about the center of mass:

− FAy (0.508) + FAx (0.102) + FBx (0.102)

+ FBy (0.508) = Ipα = 0.

Newton’s second law for the bars: −FAy + Ay − WB = mB aBGy , Substitute to obtain the nine equations in nine unknowns:
FAx + Ax = mB aBGx . −FBy + By − WB = mB aBGy . FBx + Bx = (1) −FAx − FBx = 0.254 mp sin θα,
mB aBGx . The angular acceleration about the center of mass:
(2) FAy + FBy − Wp = − 0.254 mp cos θα,
FAx (0.127) cos θ + FAy (0.127) sin θ − Ax (0.127) (3) −0.508FAy + 0 . 1 0 2 FAx + 0.508FBy + 0 . 1 0 2 FBx = 0,
cos θ + Ay (0.127) sin θ = IB α, (4) −FAy + Ay − WB = − ( 0 . 1 2 7 ) mB cos θα,

FBx (0.127) cos θ + FBy (0.127) sin θ − Bx (0.127)

cos θ + By (0.127) sin θ = IB α.

From kinematics: the acceleration of the center of mass of the bars in (5) FAx + Ax = ( 0 . 1 2 7 ) mB sin θα,
terms of the acceleration at point A is
(6) FAx sin θ + FAy cos θ − Ax sin θ + Ay cos θ = ( 7 . 8 7 ) IB α,
 i j 
0 k (7) FBx + Bx = ( 0 . 1 2 7 ) mB sin θα,
aBG = α × rG/A − ω2rG/A =  0 α 
−0.127sin θ 0 (8) −FBy + By − WB = − ( 0 . 1 2 7 ) mB cos θα,
− 0.127 cos θ
(9) FBx cos θ + FBy sin θ − Bx cos θ + By sin θ = ( 7 . 8 7 ) IB α. The
= 0.127 sin θαi − 0.127 cos αj (m/s2). number of equations and number of unknowns can be reduced by
combining equations, but here the choice is to solve the system
From which by iteration using TK Solver Plus. The results: FAx = −9.83 N,
FAy = 7.47 N, FBx = −14.77 N, Ax = 14.77 N, Ay = 20.37 N,
aBGx = (0.127) sin θα, aBGy = − (0.127)cos θα, Bx = 19.7 N, By = 25.3 N. α = 30.17 rad/s2.

since ω = 0 upon release.
The acceleration of the plate:

 i j 
0 k
aP = α × rP /A − ω2rP /A =  0 α 
−0.254 sin θ 0
− 0.2 5 4 cos θ

= 0.2 5 4 sin θαi − 0.2 5 4 cos θαj (m/s2).

From which aP x = (0.254) sin θα, aPy = − (0.254) cos θα.

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Problem 18.120 A slender bar of mass m is released
from rest in the position shown. The static and kinetic
friction coefficients of friction at the floor and the wall
have the same value µ. If the bar slips, what is its angu-
lar acceleration at the instant of release?

Solution: Choose a coordinate system with the origin at the inter- l
θ
section of wall and floor, with the x axis parallel to the floor. Denote
the points of contact at wall and floor by P and N respectively, and
the center of mass of the bar by G. The vector locations are

= iL sin θ, rP = jL cos θ, rG = L + j cos θ).
rN (i sin θ
2

From Newton’s second law:

P − µN = maGx , N + µP − mg = maGy ,

where aGx , aGy are the accelerations of the center of mass. The
moment about the center of mass is

MG = rP /G × (P i + µP j) + rN/G × (Nj − µ Ni) : Substitute to obtain the three equations in three unknowns,

   (1) P − µN = mL cos θ α,
i j k i j k
PL  − sin θ cos θ 0  NL  sin − cos θ 0. 2
2 (2) µP + N = − mL sin θ α + mg.
MG = 1 µ + θ 1 0
0 2 −µ
PL 2 NL
2 2
(3) − (cos θ + µ sin θ) + (sin θ − µ cos θ) = IB α.

MG = − PL (cos θ + µ sin θ)k + NL (sin θ − µ cos θ)k
2 2
Solve the first two equations for P and N:

From the equation of angular motion, mL µmg
P = 2(1 + µ2) (cos θ − µ sin θ)α + (1 + µ2) .
PL NL
− 2 (cos θ + µ sin θ) + 2 (sin θ − µ cos θ) = IB α

mL mg
N = − µ2) (sin θ + µ cos θ )α + µ2) .
2(1 + (1 +
From kinematics: Assume that at the instant of slip the angular veloc-
ity ω = 0. The acceleration of the center of mass in terms of the Substitute the first two equations into the third, and reduce to obtain
acceleration at point N is

aG = aN + α × rG/N − ω2rG/N α IB + mL2 1 − µ2 = mgL 1 − µ2
4 1 + µ2 2 1 + µ2

 i j 
0 k
= aN i +  0 θ L cos θ α  µ
L sin 0 sin θ − mgL 1 + µ2 cos θ.
2
− 2

Substitute IB = 1 mL2, reduce, and solve:
12

aG = aN − αL cos θ i+ − αL sin θ j, (3(1 − µ2) sin θ − 6µ cos θ)g
2 2 (2 − µ2)L
α =

from which aGy = −L sin θ α.
2

The acceleration of the center of mass in terms of the acceleration at µP
point P is aG = aP + α × rG/P . P

aG = aP + α × rG/P − ω2rG/P

 i j  θ
0 k mg
 0 − L cos θ α  , µN
= aP j + sin 2 0
L θ

2 N

aG = αL cos θ i+ aP + αL sin θ j,
2 2

from which aGx = L cos θ
α.

2

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Problem 18.121 Each of the go-cart’s front wheels 381 mm 101.6 mm
B
weighs 22.2 N and has a moment of inertia of 0.014 152.4 mm
kg-m2. The two rear wheels and rear axle form a single A
rigid body weighing 177.9 N and having a moment of
inertia of 0.136 kg-m2. The total weight of the go-cart 406.4 mm

and driver is 1067 N. (The location of the center of

mass of the go- cart and driver, not including the front
wheels or the rear wheels and rear axle, is shown.) If

the engine exerts a torque of 16.3 N-m on the rear axle,
what is the go-cart’s acceleration?

1524 mm

Solution: Let a be the cart’s acceleration and αA and αB the 16.3 N-m Bx
Ax
wheels’ angular accelerations. Note that
Ay 1067 – 222.4 N By
a = (0.152 )αA, (1)
16.3 N-m Ay By
a = ( 0.102) αB . (2) Ax 44.5 NBx
Front wheel:
177.9 N fB
Fx = Bx + fB = (44.5/9.81)a, (3) fA NB

NA

Fy = By + NB − 10 = 0, (4)

M = −fB (0.102) = (0.028) αB . (5)
Rear Wheel: (6)

Fx = Ax + fA = (177.9/9.81) a,

Fy = Ay + NA − 1 7 7 . 9 = 0, (7)

M = 16.3 − fA(0.152) = (0.136) αA. (8)

Cart:

Fx = −Ax − Bx = (844.6/9.81)a, (9)

Fy = −Ay − By − 844.6 = 0, (10)

M = Bx [(0.381 − 0.102) ] + By [(1.524 − 0.406)]
+ Ax [(0.381 − 0.152 )] − Ay (0.406) − 16.3 = 0. (11)

Solving Eqs. (1)–(11), we obtain
a = 0.91 m/s2.

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Problem 18.122 Bar AB rotates with a constant angu- y
lar velocity of 10 rad/s in the counterclockwise direc-
tion. The masses of the slender bars BC and CDE are B
2 kg and 3.6 kg, respectively. The y axis points upward.
Determine the components of the forces exerted on bar 400 mm C D E
BC by the pins at B and C at the instant shown. A x

10 rad/s

700 mm 400 700 mm
mm

Solution: The velocity of point B is By
Bx

ij k Cy
Cx
vB = ωAB × rB =  0 0 10 
0 0.4 0 Cy

= −0.4(10)i = −4i (m/s). WBC
Cx

The velocity of point C is

  WCE
i k
j ωBC  The acceleration of the center of mass of BC is
vC = vB + ωBC × rC/B = −4i +  0 0
−0.4 0
0.7

= −4i + 0.4ωBC i + 0.7ωBC j (m/s).  j 
i 0 k
From the constraint on the motion at point C, vC = vC j. Equate −0.2 αBC  − ωB2 C (0.35i − 0.2j),
components: 0 = −4 + 0.4ωBC , vC = 0.7ωBC , from which ωBC = aG = −40j +  0
10 rad/s, vC = 7 m/s. The velocity at C in terms of the angular velocity 0
ωCDE , 0.35

vC = vD + ωCDE × rC/D from which aG = 61.25i + 148.44j (m/s2)

 The equations of motion: Bx + Cx = mBC aGx , By + Cy − mBC g =
ij k mBC aGy , where the accelerations aGx , aGy are known. The moment
equation, 0.35Cy + 0.2Cx − 0.2Bx − 0.35By = IBC αBC , where αBC ,
= 0 +  0 0 ωCDE  = −0.4ωCDE j, is known, and
−0.4 0 0
IBC = 1 mBC L2BC = 0.1083 kg-m2, 0.4Cy − 0.15mCE g = IDαCE ,
12

= 7
from which ωCDE − = −17.5 rad/s. 1
0.4 where ID = 12 mCE L2CE + (0.15)2mCE = 0.444 kg-m2,

The acceleration of point B is

aB = −ωA2 B rB = −(102)(0.4)j = −40j (m/s2). is the moment of inertia about the pivot point D, and 0.15 m is the
distance between the point D and the center of mass of bar CDE.
Solve these four equations in four unknowns by iteration:

The acceleration at point C is aC = aB + αBC × rC/B − ωB2 C rC/B . Bx = −1959 N,

 j  By = 1238 N,
i 0 k
−0.4 αBC  − ωB2 C (0.7i − 0.4j) (m/s2).
aC = −40j +  0 0

0.7

Cx = 2081 N,

aC = +(0.4αBC − 0.7ωB2 C )i + (−40 + 0.7αBC + 0.4ωB2 C )j (m/s2). Cy = −922 N.

The acceleration in terms of the acceleration at D is


ijk

aC =  0 0 αCDE  − ωC2 DE (−0.4i)
−0.4 0 0

= −0.4αCDE j + 0.4ωC2 DE i.
Equate components and solve:

αBC = 481.25 rad/s2, αCDE = −842.19 rad/s2.

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Problem 18.123 At the instant shown, the arms of the y

robotic manipulator have the constant counterclockwise 300 mm 250 mm
angular velocities ωAB = −0.5 rad/s, ωBC = 2 rad/s,
and ωCD = 4 rad/s. The mass of arm CD is 10 kg, and 30° B 20°
the center of mass is at its midpoint. At this instant, what
D
force and couple are exerted on arm CD at C? x

C

A 250 mm

Solution: The relative vector locations of B, C, and D are Cy Cx D
C 125
rB/A = 0.3(i cos 30◦ + j sin 30◦) mm
= 0.2598i + 0.150j (m), mg

rC/B = 0.25(i cos 20◦ − j sin 20◦)
= 0.2349i − 0.08551j (m),

rD/C = 0.25i (m).
The acceleration of point B is
aB = −ωA2 B rB/A = −(0.52)(0.3 cos 30◦i + 0.3 sin 30◦j),
aB = −0.065i − 0.0375j (m/s2).
The acceleration at point C is
aC = aB − ωB2 C rC/B = aB − ωB2 C (0.2349i − 0.08551j).
aC = −1.005i + 0.3045j (m/s2).
The acceleration of the center of mass of CD is
aG = aC − ωC2 D(0.125i) (m/s2),
from which
aG = −3.005i + 0.3045j (m/s2).
For the arm CD the three equations of motion in three unknowns are
Cy − mCD g = mCDaGy , Cx = mCD aGx , M − 0.125Cy = 0,
which have the direct solution:
Cy = 101.15 N,
Cx = −30.05 N.
M = 12.64 N-m,
where the negative sign means a direction opposite to that shown in
the free body diagram.

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Problem 18.124 Each bar is 1 m in length and has A
a mass of 4 kg. The inclined surface is smooth. If the
system is released from rest in the position shown, what
are the angular accelerations of the bars at that instant?

45° B
O 30°

Solution: For convenience, denote θ = 45◦, β = 30◦, and L = Ay
Ax Ax
1 m. The acceleration of point A is
Ay
 j  mg mg
i 0 k
L sin θ αOA  .
aA = αOA × rA/O =  0
0
L cos θ

aA = αOA(−iL sin θ + jL cos θ) (m/s2). B
The acceleration of A is also given by 30°

The equations of motion for the bars: for the pin supported left bar:

aA = aB + αAB × rA/B . L
(5) Ay L cos θ − Ax L sin θ − mg 2 cos θ = IOAαOA,
 
i j k where IOA = mL2 = 4 kg-m2.
0 αAB  . 3 3
aA = aB +  0 L sin θ
0
−L cos θ

aA = aB − iαAB L sin θ − jαAB L cos θ (m/s2). The equations of motion for the right bar:

From the constraint on the motion at B, Equate the expressions for (6) − Ax − B sin β = maGABx ,
the acceleration of A to obtain the two equations:

(1) − αOAL sin θ = aB cos β − αAB L sin θ, (7) − Ay − mg + B cos β = maGABy ,

and (2) αOAL cos θ = aB sin β − αAB L cos θ. L LL
The acceleration of the center of mass of AB is (8) Ay 2 cos θ + Ax 2 sin θ + B 2 sin θ cos β

aGAB = aA + αAB × rGAB/A L
− B 2 cos θ sin β = ICAB αAB ,

 i j  where IGAB = 1 mL2 = 1 kg-m2.
0 k 12 3
 0 − L sin θ αAB  ,
= aA + L cos θ 2 0

2 These eight equations in eight unknowns are solved by iteration: Ax =
−19.27 N, Ay = 1.15 N, αOA = 0.425 rad/s2, αAB = −1.59 rad/s2,
LαAB αAB L B = 45.43 N, aGABx = −0.8610 m/s2, aGABy = −0.2601 m/s2
2 2
aGAB = aA + sin θi + cos θj (m/s2),

from which

(3) aGABx = −αOAL sin θ + LαAB sin θ (m/s2),
2

(4) aGABy = αOAL cos θ + LαAB cos θ.
2

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Problem 18.125 Each bar is 1 m in length and has
a mass of 4 kg. The inclined surface is smooth. If the
system is released from rest in the position shown, what
is the magnitude of the force exerted on bar OA by the
support at O at that instant?

Solution: The acceleration of the center of mass of the bar OA is Ay
Ax
 i j 
0 k Fy
aGOA = αOA × rG/A = aA +  L 0 θ L sin θ αOA  , mg
cos 0
2 Fx
2
Use the solution to Problem 18.140: θ = 45◦, αGA = 0.425 rad/s2,
aGOA = −L sin θ αO A i + L cos θ αOAj (m/s2). Ax = −19.27 N, m = 4 kg, from which Fx = 18.67 N, Fy =
2 2 38.69 N, from which |F| = Fx2 + Fy2 = 42.96 N

The equations of motion:

Fx + Ax = maGOAx , Fy + Ay − mg = maGOAy .

Problem 18.126 The fixed ring gear lies in the Planet gear 140 340
Hub gear mm mm
horizontal plane. The hub and planet gears are bonded 240 mm
Connecting
together. The mass and moment of inertia of the rod 720 mm
combined hub and planet gears are mHP = 130 kg and
IHP = 130 kg-m2. The moment of inertia of the sun gear Sun gear
is Is = 60 kg-m2. The mass of the connecting rod is Ring gear
5 kg, and it can be modeled as a slender bar. If a 1 kN-
Hub & Planet Gears
m counterclockwise couple is applied to the sun gear, Q

what is the resulting angular acceleration of the bonded FR
F
hub and planet gears?
M
Solution: The moment equation for the sun gear is Sun Gear

(1) M − 0.24F = Is αs . R
For the hub and planet gears: Connecting Rod

(2) (0.48)αHP = −0.24αs ,

(3) F − Q − R = mHP (0.14)(−αHP ),

(4) (0.14)Q + 0.34F − IHP (−αHP ).
For the connecting rod:

(5) (0.58)R = ICRαCR,

where ICR = 1 mGR(0.582) = 0.561 kg-m2.
3

(6) (0.58)αCR = −(0.14)αHP .
These six equations in six unknowns are solved by iteration:

F = 1482.7 N, αs = 10.74 rad/s2,
αHP = −5.37 rad/s2, Q = 1383.7 N,

R = 1.25 N, αCR = 1.296 rad/s2.

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Problem 18.127 The system is stationary at the instant 50 mm B 125 mm
A 40° C
shown. The net force exerted on the piston by the
M
exploding fuel-air mixture and friction is 5 kN to the
left. A clockwise couple M = 200 N-m acts on the crank
AB. The moment of inertia of the crank about A is
0.0003 kg-m2. The mass of the connecting rod BC is

0.36 kg, and its center of mass is 40 mm from B on

the line from B to C. The connecting rod’s moment
of inertia about its center of mass is 0.0004 kg-m2.

The mass of the piston is 4.6 kg. What is the piston’s

acceleration? (Neglect the gravitational forces on the

crank and connecting rod.)

Solution: From the law of sines: from which
(3) aGCRx = aC − 0.085αBC sin β (m/s2),
sin β = sin 40◦ (4) aGCRy = −0.085αBC cos β (m/s2).
, The equations of motion for the crank:
0.05 0.125 (5) By (0.05 cos 40◦) − Bx (0.05 sin 40◦) − M = IAαAB
For the connecting rod:
from which β = 14.9◦. The vectors (6) − Bx + Cx = mCR aGCRx
(7) − By + Cy = mCR aGCRy
rB/A = 0.05(i cos 40◦ + j sin 40◦)rB/A (8) Cy (0.085 cos β) + Cx (0.085 sin β) + Bx (0.04 sin β)

= 0.0383i + 0.0321j (m). + By (0.04 cos β) = IGCRαBC
For the piston:
rB/C = 0.125(−i cos β + j sin β) (m). (9) − Cx − 5000 = mP aC .
These nine equations in nine unknowns are solved by iteration:
rB/C = −0.121i + 0.0321, (m).
The acceleration of point B is αAB = 1255.7 rad/s2, αBC = −398.2 rad/s2,
aGCRx = −44.45 m/s2, aGCRy = 32.71 m/s2,
aB = αAB × rB/A − ωA2 B rB/A,
By = 1254.6 N, Bx = −4739.5 N,
 j  Cx = −4755.5 N, Cy = 1266.3 N,
i 0 k aC = −53.15 m/s2.
0.0321 αAB 
aB =  0
0
0.0383

− ωA2 B (0.0383i + 0.0321j) (m/s2).

The acceleration of point B in terms of the acceleration of point C is

 j 
i 0 k
0.0321 αBC 
aB = aC + αBC × rB/C = aC i +  0
0
−0.121

− ωB2 C (−0.121i + 0.0321j) (m/s2). B 0.125 Bx Cy
Equate the two expressions for the acceleration of point B, note ωAB = 0.05 β G Cx
ωBC = 0, and separate components:
40 C By
(1) − 0.05αAB sin 40◦ = aC − 0.125αBC sin β, A

(2) 0.05αAB cos 40◦ = −0.125αBC cos β.

The acceleration of the center of mass of the connecting rod is

aGCR = aC + αBC × rGCR/C − ωB2 C rGCR/C ,

 j  Cx 5000 N By
i 0 k Bx
0.085 sin β αBC 
aGCR = aC i +  0 Ay Ax
0
−0.085 cos β M

− ωB2 C (−0.085 cos βi + 0.085 sin βj) (m/s2), Cy N

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Problem 18.128 If the crank AB in Problem 18.127
has a counterclockwise angular velocity of 2000 rpm at
the instant shown, what is the piston’s acceleration?

Solution: The angular velocity of AB is

2π
ωAB = 2000 60 = 209.44 rad/s.

The angular velocity of the connecting rod BC is obtained from the
expressions for the velocity at point B and the known value of ωAB :

 j 
i k
0 ωAB  .
vB = ωAB × rB/A =  0 0.05 sin 40◦
0
0.05 cos 40◦

vB = −0.05 sin 40◦ωAB i + 0.05 cos 40◦ωAB j (m/s).

 j 
i 0 k
0.125 sin β ωBC  ,
vB = vC i +  0
0
−0.125 cos β

vB = vC i − 0.125 sin βωBC i − 0.125 cos βωBC j (m/s).
From the j component, 0.05 cos 40◦ωAB = −0.125 cos βωBC , from
which ωBC = −66.4 rad/s. The nine equations in nine unknowns
obtained in the solution to Problem 18.127 are

(1) − 0.05αAB sin 40◦ − 0.05ωA2 B cos 40◦

= aC − 0.125αBC sin β + 0.125ωB2 C cos β,
(2) 0.05αAB cos 40◦ − 0.05ωA2 B sin 40◦

= −0.125αBC cos β − 0.125ωB2 C sin β,

(3) aGCRx = aC − 0.085αBC sin β + 0.085ωB2 C cos β (m/s2),

(4) aGCRy = −0.085αBC cos β − 0.085ωB2 C sin β (m/s2),
(5) By (0.05 cos 40◦) − Bx (0.05 sin 40◦) − M = IAαAB ,

(6) − Bx + Cx = mCRaGCRx ,

(7) − By + Cy = mCRaGCRy ,

(8) Cy (0.085 cos β) + Cx (0.085 sin β)

+ Bx (0.04 sin β) + By (0.04 cos β) = IGCRαBC .

(9) − Cx − 5000 = mP aC .
These nine equations in nine unknowns are solved by iteration:

αAB = −39, 386.4 rad/s2αBC = 22,985.9 rad/s2,
aGCRx = −348.34 m/s2, aGCRy = −1984.5 m/s2,

By = 1626.7 N, Bx = −3916.7 N,

Cx = −4042.1 N, Cy = 912.25 N,

ac = −208.25 (m/s2)

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to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.

Problem 19.1 The moment of inertia of the rotor of
the medical centrifuge is I = 0.2 kg-m2. The rotor starts
from rest and the motor exerts a constant torque of
0.8 N-m on it.

(a) How much work has the motor done on the rotor
when the rotor has rotated through four revolu-
tions?

(b) What is the rotor’s angular velocity (in rpm) when
it has rotated through four revolutions?

Solution:

(a) W = (0.8 N-m)(4 rev) 2π rad = 20.1 N-m W = 20.1 N-m
1 rev

(b) 1 I ω2 = W, 1 (0.2 kg-m2)ω2 = 20.1 N-m ⇒ ω= 2(20.1 N-m)
2 2 0.2 kg-m2

ω = 14.2 rad/s

Problem 19.2 The 17.8 N slender bar is 0.61 m in length. y B
A x
It started from rest in an initial position relative to the
inertial reference frame. When it is in the position shown,
the velocity of the end A is 6.71i + 4.27j (m/s) and the
bar has a counterclockwise angular velocity of 12 rad/s.
How much work was done on the bar as it moved
from its initial position to its present position?

30Њ

Solution: Work = Change in kinetic energy. To calculate the

kinetic energy we will first need to find the velocity of the center
of mass

vG = vA + ω × rG/A

= (6.71i + 4.27j)(m/s) + (12 rad/s)k × (0.305 m)(cos 30◦i + sin 30◦j)

= (−4.27i + 8.5j) m/s
Now we can calculate the work, which is equal to the kinetic energy.
W = 1 mv2 + 1 I ω2

22

=1 17.8 N [(−4.27 m/s)2 + (8.5 m/s)2]
2 9.81 m/s2

+ 1 1 17.8 N (0 . 6 1 m)2 (12 rad/s)2
2 12 9.81 m/s2

= 86.1 N-m.
W = 86.1 N-m.

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to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.

Problem 19.3 The 20-kg disk is at rest when the 10 N-m
constant 10 N-m counterclockwise couple is applied.
Determine the disk’s angular velocity (in rpm) when 0.25 m
it has rotated through four revolutions (a) by applying
the equation of angular motion M = 1α, and (b) by
applying the principle of work and energy.

Solution:

(a) First we find the angular acceleration
M = Iα

(10 N-m) = 1 kg)(0.25 m)2α ⇒ α = 16 rad/s
(20
2

Next we will integrate the angular acceleration to find the angular
velocity.

dω = α ⇒ ω θ α dθ ⇒ ω2 = αθ
ω
ωdω =
dθ 0 0 2

√ rev 60 s = 271 rev
ω = 2αθ = 2(16 rad/s2)(4 rev) min min
2π rad

ω = 271 rpm.

(b) Applying the principle of work energy
W = 1 I ω2
2

(10 N-m)(4 rev) 2π rad = 1 1 (20 kg)(0.25 m)2 ω2
rev 2 2

ω = 28.4 rad/s rev 60 s = 271 rpm
2π rad min

ω = 271 rpm.

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to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.