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•21–1. Show that the sum of the moments of inertia of a
body, Ixx + Iyy + Izz, is independent of the orientation of
the x, y, z axes and thus depends only on the location of its
origin.
Ixx + Iyy + Izz =
Lm
= 2
(y2 + z2)dm +
Lm
(x2 + z2)dm +
(x2 + y2)dm
Lm
Lm
(x2 + y2 + z2)dm
However, x2 + y2 + z2 = r2, where r is the distance from the origin O to dm. Since
ƒ r ƒ is constant, it does not depend on the orientation of the x, y, z axis. Consequently,
Ixx + Iyy + Izz is also indepenent of the orientation of the x, y, z axis.
Q.E.D.
925
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21–2. Determine the moment of inertia of the cone with
respect to a vertical y axis that passes through the cone’s
center of mass.What is the moment of inertia about a parallel
axis y¿ that passes through the diameter of the base of the
cone? The cone has a mass m.
–y
y
y¿
a
x
The mass of the differential element is dm = rdV = r(py2) dx =
dIv =
=
Iv =
L
h2
h
x2dx.
1
dmy2 + dmx2
4
r pa2
a 2
1 rpa2 2
B 2 x dx R a xb + ¢ 2 x2 ≤ x2 dx
4
h
h
h
r pa2
=
rpa2
(4h2 + a2) x4 dx
4h4
dIv =
rpa2
4
4h
h
(4h2 + a2)
x4dx =
r pa2h
(4h2 + a2)
20
x2 dx =
r pa2h
3
L0
However,
m =
Lm
dm =
h
r pa2
2
h
L0
Hence,
Iy =
3m
(4h2 + a2)
20
Using the parallel axis theorem:
Iv = Iv + md2
3h 2
3m
(4h2 + a2) = Iv + ma b
20
4
Iv =
3m 2
(h + 4a2)
80
Ans.
Iv = Iy + md2
=
h 2
3m 2
(h + 4a2) + ma b
80
4
=
m
(2h2 + 3a2)
20
Ans.
926
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z
21–3. Determine the moments of inertia Ix and Iy of the
paraboloid of revolution. The mass of the paraboloid is m.
a
m = r
L0
a
pz2 dy = rp
L0
a
r2
r2
bydy = rra b a
a
2
a
Iv =
2
r
z2 ⫽ — y
a
r
a
1
1
1
r4
r4
z4dy = rp a 2 b
v2dy = rp a ba
dm z2 = rp
2
2
6
a L0
Lm 2
L0
y
x
a
Thus,
Ix =
1
mr2
3
a
Ix =
=
Ans.
a
1
1
a dm z2 + dm y2 b = rp
z4 dy + r
pz2 y2 dy
4
L0
Lm 4
L0
a
a
rpr 2a 3
rpr 4a
1
r4
r2
1
1
r pa 2 b
y2dy + rpa b
y3dy =
+
= mr 2 + ma 2
a L0
4
12
4
6
2
a L0
Ix =
m 2
(r + 3a2)
6
Ans.
z
*21–4. Determine by direct integration the product of
inertia Iyz for the homogeneous prism. The density of the
material is r. Express the result in terms of the total mass m
of the prism.
a
a
The mass of the differential element is dm = rdV = rhxdy = rh(a - y)dy.
h
a
ra2h
m =
dm = rh
(a - y)dy =
2
Lm
L0
x
Using the parallel axis theorem:
dIyz = (dIy¿z¿)G + dmyGzG
h
= 0 + (rhxdy) (y) a b
2
Iyz =
=
rh2
xydy
2
=
rh2
(ay - y2) dy
2
a
rh2
ra3h2
1 ra2h
m
= a
b(ah) =
ah
(ay - y2) dy =
2 L0
12
6
2
6
927
Ans.
y
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z
•21–5. Determine by direct integration the product of
inertia Ixy for the homogeneous prism. The density of the
material is r. Express the result in terms of the total mass m
of the prism.
a
a
The mass of the differential element is dm = rdV = rhxdy = rh(a - y)dy.
h
a
y
ra2h
m =
dm = rh
(a - y)dy =
2
Lm
L0
x
Using the parallel axis theorem:
dIxy = (dIx¿y¿)G + dmxGyG
x
= 0 + (rhxdy)a b(y)
2
=
rh2 2
x ydy
2
=
rh2 3
(y - 2ay2 + a 2 y) dy
2
Ixy =
a
rh
(y3 - 2ay2 + a 2 y) dy
2 L0
=
ra 4h
1 ra 2h 2
m 2
=
a
ba =
a
24
12
2
12
Ans.
z
21–6. Determine the product of inertia Ixy for the
homogeneous tetrahedron. The density of the material is r.
Express the result in terms of the total mass m of the solid.
Suggestion: Use a triangular element of thickness dz and
then express dIxy in terms of the size and mass of the
element using the result of Prob. 21–5.
a
y
r
1
dm = r dV = rc (a - z)(a - z) ddz = (a - z)2 dz
2
2
m =
a
a
r
ra3
(a2 - 2az + z2)dz =
2 L0
6
a
x
From Prob. 21–5 the product of inertia of a triangular prism with respect to the xz
ra 4h
rdz
(a - z)4. Hence,
and yz planes is Ixy =
. For the element above, dIxy =
24
24
Ixy =
a
r
(a4 - 4a3z + 6z 2a 2 - 4az 3 + z 4)dz
24 L0
Ixy =
ra 5
120
Ixy =
ma 2
20
or,
Ans.
928
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21–7. Determine the moments of inertia for the
homogeneous cylinder of mass m about the x¿ , y¿ , z¿ axes.
r
z¿
x¿
y
r
z
x
Due to symmetry
Ixy = Iyz = Izx = 0
Iy = Ix =
1
r 2
7mr 2
m(3r 2 + r 2) + ma b =
12
2
12
Iz =
1 2
mr
2
For x¿ ,
ux = cos 135° = -
1
22
uy = cos 90° = 0,
,
1
uz = cos 135° = -
22
Ix = Ix u2x + Iy u2y + Iz u2z - 2Ixy ux uy - 2Iyz uy uz - 2Izx uz ux
=
7mr 2
1 2
1
1 2
b + 0 + mr 2 a b - 0-0-0
¢12
2
22
22
=
13
mr 2
24
Ans.
For y¿ ,
Iy¿ = Iy =
7mr 2
12
Ans.
For z¿ ,
ux = cos 135° = -
1
22
,
uy = cos 90° = 0,
uz = cos 45° =
1
22
Iz¿ = Ix u2x + Iy u2y + Iz u2z - 2Ixy ux uy - 2Iyz uy uz - 2Izx uz ux
=
7mr 2
1
1 2
1 2
b + 0 + mr 2 a b - 0-0-0
¢12
2
22
22
=
13
mr 2
24
Ans.
929
y¿
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z
*21–8. Determine the product of inertia Ixy of the
homogeneous triangular block. The material has a
density of r . Express the result in terms of the total mass
m of the block.
b
h
The mass of the differential rectangular volume element shown in Fig. a is
x
dm = rdV = rbzdy. Using the parallel - plane theorem,
dIxy = dI x¿y¿ + dmxGyG
b
= 0 + [rbzdy]a by
2
=
However, z =
rb2
zydy
2
h
(a - y). Then
a
dIxy =
rb2 h
rb2h
c (a - y)y ddy =
A ay - y2 B dy
2 a
2a
Thus,
Ixy =
L
dIxy =
a
rb 2h
A ay - y2 B dy
2a L0
=
y3 a
rb2h ay2
b2
¢
2a
2
3 0
=
1
ra 2b 2 h
12
1
1
However, m = rV = a ahb b = rabh. Then
2
2
Ixy =
1
m
1
ra 2b2h§
¥ = mab
12
1
6
rabh
2
Ans.
930
y
a
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z
•21–9. The slender rod has a mass per unit length of
6 kg>m. Determine its moments and products of inertia
with respect to the x, y, z axes.
2m
x
The mass of segments (1), (2), and (3) shown in Fig. a is
m1 = m2 = m3 = 6(2) = 12 kg. The mass moments of inertia of the bent rod about
the x, y, and z axes are
1
1
(12) A 22 B + 12 A 12 + 02 B d + c (12) A 22 B + 12 C 22 +
12
12
= 80 kg # m2
A -1 B 2 D d
Ans.
Iy = ©Iy¿ + m A xG 2 + zG 2 B
= c
1
1
(12) A 22 B + 12 A 12 + 02 B d + c0 + 12 A 22 + 02 B d + c (12) A 22 B + 12 C 22 +
12
12
= 128 kg # m2
A -1 B 2 D d
Ans.
Iz = ©Iz¿ + m A xG 2 + yG 2 B
= c
1
1
(12) A 22 B + 12 A 12 + 02 B d + c (12) A 22 B + 12 A 22 + 12 B d + c0 + 12 A 22 + 22 B d
12
2
= 176 kg # m2
Ans.
Due to symmetry, the products of inertia of segments (1), (2), and (3) with respect to
their centroidal planes are equal to zero. Thus,
Ixy = ©Ix¿y¿ + mxGyG
= C 0 + 12(1)(0) D + C 0 + 12(2)(1) D + C 0 + 12(2)(2) D
= 72 kg # m2
Ans.
Iyz = ©Iy¿z¿ + myGzG
= C 0 + 12(0)(0) D + C 0 + 12(1)(0) D + C 0 + 12(2)(-1) D
= -24 kg # m2
Ans.
Ixz = ©Ix¿z¿ + mxGzG
= C 0 + 12(1)(0) D + C 0 + 12(2)(0) D + C 0 + 12(2)(-1) D
= -24 kg # m2
Ans.
931
y
2m
2m
Ix = ©Ix¿ + m A yG 2 + zG 2 B
= A0 + 0B + c
O
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
z
21–10. Determine the products of inertia Ixy, Iyz, and Ixz
of the homogeneous solid. The material has a density of
7.85 Mg>m3.
200 mm
200 mm
100 mm
x
The masses of segments (1) and (2) shown in Fig. a are m1 = r V1
= 7850(0.4)(0.4)(0.1) = 125.6 kg and m2 = r V2 = 7850(0.2)(0.2)(0.1) = 31.4 kg.
Due
to
symmetry
for
segment
(1)
and
Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0
Ix–y– = Iy–z– = Ix–z– = 0 for segment (2). Since segment (2) is a hole, it should be
considered as a negative segment. Thus
Ixy = ©Ix¿y¿ + mxGyG
= C 0 + 125.6(0.2)(0.2) D - C 0 + 31.4(0.3)(0.1) D
= 4.08 kg # m2
Ans.
Iyz = ©Iy¿z¿ + myGzG
= C 0 + 125.6(0.2)(0.05) D - C 0 + 31.4(0.1)(0.05) D
= 1.10 kg # m2
Ans.
Ixz = ©Ix¿z¿ + mxGzG
= C 0 + 125.6(0.2)(0.05) D - C 0 + 31.4(0.3)(0.05) D
= 0.785 kg # m2
Ans.
932
200 mm
200 mm
y
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z
21–11. The assembly consists of two thin plates A and B
which have a mass of 3 kg each and a thin plate C which has
a mass of 4.5 kg. Determine the moments of inertia Ix , Iy
and Iz .
B
0.4 m
60⬚
A
C
0.4 m
0.3 m
x
Ix¿ = Iz¿ =
Iy¿ =
1
(3)(0.4)2 = 0.04 kg # m2
12
1
(3)[(0.4)2 + (0.4)2] = 0.08 kg # m2
12
Ix¿y¿ = Iz¿y¿ = Iz¿x¿ = 0
For zG,
ux¿ = 0
uy¿ = cos 60° = 0.50
uz¿ = cos 30° = 0.8660
IzG = 0 + 0.08(0.5)2 + 0.04(0.866)2 - 0 - 0 - 0
= 0.05 kg # m2
IxG = Ix¿ = 0.04 kg # m2
For yG,
ux¿ = 0
uy¿ = cos 30° = 0.866
uz¿ = cos 120° = - 0.50
IyG = 0 + 0.08(0.866)2 + 0.04(-0.5)2 - 0 - 0 - 0
= 0.07 kg # m2
Ix =
1
(4.5)(0.6)2 + 2[0.04 + 3{(0.3 + 0.1)2 + (0.1732)2}]
12
Ix = 1.36 kg # m2
Iy =
Ans.
1
(4.5)(0.4)2 + 2[0.07 + 3(0.1732)2]
12
Iy = 0.380 kg # m2
Iz =
Ans.
1
(4.5)[(0.6)2 + (0.4)2] + 2[0.05 + 3(0.3 + 0.1)2]
12
Iz = 1.26 kg # m2
Ans.
933
60⬚
0.3 m
0.2 m
0.2 m
y
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
z
*21–12. Determine the products of inertia Ixy, Iyz, and Ixz,
of the thin plate. The material has a density per unit area of
50 kg>m2.
The masses of segments (1) and (2) shown in Fig. a are m1 = 50(0.4)(0.4) = 8 kg
and m2 = 50(0.4)(0.2) = 4 kg. Due to symmetry Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0 for
segment (1) and Ix–y– = Iy–z– = Ix–z– = 0 for segment (2).
400 mm
200 mm
Ixy = ©Ix¿y¿ + mxGyG
= C 0 + 8(0.2)(0.2) D + C 0 + 4(0)(0.2) D
x
= 0.32 kg # m2
y
400 mm
Ans.
Iyz = ©Iy¿z¿ + myGzG
= C 0 + 8(0.2)(0) D + C 0 + 4(0.2)(0.1) D
= 0.08 kg # m2
Ans.
Ixz = ©Ix¿z¿ + mxGzG
= C 0 + 8(0.2)(0) D + C 0 + 4(0)(0.1) D
= 0
Ans.
z¿
•21–13. The bent rod has a weight of 1.5 lb>ft. Locate the
center of gravity G(x, y) and determine the principal
moments of inertia Ix¿ , Iy¿ , and Iz¿ of the rod with respect
to the x¿, y¿, z¿ axes.
z
1 ft
Due to symmetry
1 ft
A
_
y
x¿
x
y = 0.5 ft
Ans.
x =
Ans.
(-1)(1.5)(1) + 2 C ( -0.5)(1.5)(1) D
©x W
=
= -0.667 ft
©w
3 C 1.5(1) D
Ix¿ = 2c a
1.5
1
1.5
b(0.5)2 d +
a
b(1)2
32.2
12 32.2
= 0.0272 slug # ft2
Iy¿ = 2 c
Ans.
1 1.5
1.5
1.5
a
b (1)2 + a
b (0.667 - 0.5)2 d + a
b(1 - 0.667)2
12 32.2
32.2
32.2
= 0.0155 slug # ft2
Iz¿ = 2 c
+
Ans.
1 1.5
1.5
a
b (1)2 + a
b(0.52 + 0.16672) d
12 32.2
32.2
1
1.5
1.5
a
b(1)2 + a
b(0.3333)2
12 32.2
32.2
= 0.0427 slug # ft2
Ans.
934
G
y¿
_
x
y
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z
21–14. The assembly consists of a 10-lb slender rod and a
30-lb thin circular disk. Determine its moment of inertia
about the y¿ axis.
2 ft
The mass of moment inertia of the assembly about the x, y, and z axes are
1 ft
1 30
30
1 10
10
b A 12 B +
b A 22 B +
Ix = Iz = c a
A 22 B d + c a
A 12 B d
4 32.2
32.2
12 32.2
32.2
A
= 4.3737 slug # ft
Iy =
y
B
2
x
1 30
a
b A 12 B + 0 = 0.4658 slug # ft2
2 32.2
y¿
Due to symmetry, Ixy = Iyz = Ixz = 0. From the geometry shown in Fig. a,
1
u = tan-1 a b = 26.57°. Thus, the direction of the y¿ axis is defined by the unit
2
vector
u = cos 26.57°j - sin 26.57°k = 0.8944j - 0.4472k
Thus,
ux = 0
uy = 0.8944
uz = -0.4472
Then
Iy¿ = Ix ux 2 + Iy uy 2 + Iz uz 2 - 2Ixy ux uy - 2Iyz uy uz - 2Ixzux uz
= 4.3737(0) + 0.4658(0.8944)2 + 4.3737(-0.4472)2 - 0 - 0 - 0
= 1.25 slug # ft2
Ans.
z
21–15. The top consists of a cone having a mass of 0.7 kg
and a hemisphere of mass 0.2 kg. Determine the moment of
inertia Iz when the top is in the position shown.
Ix¿ = Iy¿ =
30 mm
2
3
3
(0.7) C (4)(0.3)2 + (0.1)2 D + (0.7)c (0.1) d
80
4
+ a
100 mm
30 mm
2
83
3
b (0.2)(0.03)2 + (0.2)c (0.03) + (0.1) d = 6.816 A 10-3 B kg # m2
320
8
y
3
2
Iz¿ = a b(0.7)(0.03)2 + a b (0.2)(0.03)2
10
5
45⬚
Iz = 0.261 A 10-3 B kg # m2
ux = cos 90° = 0,
Iz =
Ix¿ u2x¿
+
Iy¿ u2y¿
uy¿ = cos 45° = 0.7071,
+
Iz¿ u2z¿
uz¿ = cos 45° = 0.7071
x
- 2Ix¿y¿ux¿ uy¿ - 2Iy¿z¿ uy¿ uz¿ - 2Ix¿z¿ ux¿ uz¿
= 0 + 6.816 A 10-3 B (0.7071)2 + (0.261) A 10-3 B (0.7071)2 - 0 - 0 - 0
Iz = 3.54 A 10-3 B kg # m2
Ans.
935
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z
*21–16. Determine the products of inertia Ixy, Iyz, and
Ixz of the thin plate. The material has a mass per unit area
of 50 kg>m2.
200 mm
200 mm
200 mm
The masses of segments (1), (2), and (3) shown in Fig. a are m1 = m2
= 50(0.4)(0.4) = 8 kg and m3 = 50cp(0.1)2 d = 0.5p kg.
200 mm
Due to symmetry Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0 for segment (1), Ix–y– = Iy–z– = Ix–z– = 0
for segment (2), and Ix‡y‡ = Iy‡z‡ = Ix‡z‡ = 0 for segment (3). Since segment (3) is
a hole, it should be considered as a negative segment. Thus
Ixy = ©Ix¿y¿ + mxGyG
x
= C 0 + 8(0.2)(0.2) D + C 0 + 8(0)(0.2) D - C 0 + 0.5p(0)(0.2) D
= 0.32 kg # m2
100 mm
400 mm
y
400 mm
Ans.
Iyz = ©Iy¿z¿ + myGzG
= C 0 + 8(0.2)(0) D + C 0 + 8(0.2)(0.2) D - C 0 + 0.5p(0.2)(0.2) D
= 0.257 kg # m2
Ans.
Ixz = ©Ix¿z¿ + mxGzG
= C 0 + 8(0.2)(0) D + C 0 + 8(0)(0.2) D - C 0 + 0.5p(0)(0.2) D
= 0 kg # m2
Ans.
z
•21–17. Determine the product of inertia Ixy for the bent
rod. The rod has a mass per unit length of 2 kg>m.
Product of Inertia: Applying Eq. 21–4. we have
Ixy = © A Ix¿y¿ B G + mxG yG
400 mm
= [0 + 0.4 (2) (0) (0.5)] + [0 + 0.6 (2) (0.3) (0.5)] + [0 + 0.5 (2) (0.6) (0.25)]
= 0.330 kg # m2
y
Ans.
600 mm
500 mm
x
936
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z
21–18. Determine the moments of inertia Ixx, Iyy, Izz for
the bent rod. The rod has a mass per unit length of 2 kg>m.
400 mm
y
600 mm
Moments of Inertia: Applying Eq. 21–3, we have
500 mm
Ixx = ©(Ix¿x¿)G + m(y2G + z2G)
= c
x
1
(0.4) (2) A 0.42 B + 0.4 (2) A 0.52 + 0.22 B d
12
+ C 0 + 0.6 (2) A 0.52 + 02 B D
+ c
1
(0.5) (2) A 0.52 B + 0.5 (2) A 0.252 + 02 B d
12
= 0.626 kg # m2
Ans.
Ixy = ©(Iy¿y¿)G + m(x2G + z2G)
= c
1
(0.4) (2) A 0.42 B + 0.4 (2) A 02 + 0.22 B d
12
+ c
1
(0.6) (2) A 0.62 B + 0.6 (2) A 0.32 + 02 B d
12
+ C 0 + 0.5(2) A 0.62 + 02 B D
= 0.547 kg # m2
Ans.
Izz = ©(Iz¿z¿)G + m(x2G + y2G)
= C 0 + 0.4 (2) A 02 + 0.52 B D
+ c
1
(0.6) (2) A 0.62 B + 0.6 (2) A 0.32 + 0.52 B d
12
+ c
1
(0.5) (2) A 0.52 B + 0.5 (2) A 0.62 + 0.252 B d
12
= 1.09 kg # m2
Ans.
937
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z
21–19. Determine the moment of inertia of the rod-andthin-ring assembly about the z axis. The rods and ring have
a mass per unit length of 2 kg>m.
A
O
500 mm
400 mm
For the rod,
D
ux¿ = 0.6,
Ix = Iy =
uy¿ = 0,
uz¿ = 0.8
B
1
[(0.5)(2)](0.5)2 = 0.08333 kg # m2
3
120⬚
120⬚
x
Ix¿ = 0
Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0
From Eq. 21–5,
Iz = 0.08333(0.6)2 + 0 + 0 - 0 - 0 - 0
Iz = 0.03 kg # m2
For the ring,
The radius is r = 0.3 m
Thus,
Iz = mR2 = [2 (2p)(0.3)](0.3)2 = 0.3393 kg # m2
Thus the moment of inertia of the assembly is
Iz = 3(0.03) + 0.339 = 0.429 kg # m2
Ans.
938
y
120⬚
C
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Page 939
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z
*21–20. If a body contains no planes of symmetry, the
principal moments of inertia can be determined
mathematically. To show how this is done, consider the rigid
body which is spinning with an angular velocity V , directed
along one of its principal axes of inertia. If the principal
moment of inertia about this axis is I, the angular momentum
can be expressed as H = IV = Ivx i + Ivy j + Ivz k. The
components of H may also be expressed by Eqs. 21–10,
where the inertia tensor is assumed to be known. Equate the
i, j, and k components of both expressions for H and consider
vx, vy, and vz to be unknown. The solution of these three
equations is obtained provided the determinant of the
coefficients is zero. Show that this determinant, when
expanded, yields the cubic equation
V
O
x
I 3 - (Ixx + Iyy + Izz)I 2
+ (IxxIyy + IyyIzz + IzzIxx - I2xy - I2yz - I2zx)I
- (IxxIyyIzz - 2IxyIyzIzx - IxxI 2yz
- IyyI 2zx - IzzI 2xy) = 0
The three positive roots of I, obtained from the solution of
this equation, represent the principal moments of inertia
Ix, Iy, and Iz.
H = Iv = Ivx i + Ivy j + Ivz k
Equating the i, j, k components to the scalar equations (Eq. 21–10) yields
(Ixx - I) vx - Ixy vy - Ixz vz = 0
-Ixx vx + (Ixy - I) vy - Iyz vz = 0
-Izx vz - Izy vy + (Izz - I) vz = 0
Solution for vx, vy, and vz requires
3
(Ixx - I)
-Iyx
-Izx
-Ixy
(Iyy - I)
-Izy
-Ixz
-Iyz 3 = 0
(Izz - I)
Expanding
I3 - (Ixx + Iyy + Izz)I2 + A Ixx Iyy + Iyy Izz + Izz Ixx - I2xy - I2yz - I2zx B I
- A Ixx Iyy Izz - 2Ixy Iyz Izx - Ixx I2yz - IyyI2zx - Izz I2xy B = 0 Q.E.D.
939
y
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•21–21. Show that if the angular momentum of a body is
determined with respect to an arbitrary point A, then H A
can be expressed by Eq. 21–9. This requires substituting
R A = R G + R G>A into Eq. 21–6 and expanding, noting
that 1 R G dm = 0 by definition of the mass center and
vG = vA + V : R G>A.
Z
z
RG/A
G
RG
RA
P
y
A
Y
x
HA = a
Lm
rA dmb * vA +
= a
Lm
(rG + rG>A) dmb * vA +
= a
Lm
rG dmb * vA + (rG>A * vA)
+ a
Since
Lm
Lm
Lm
X
rA * (v * rA)dm
(rG + rG>A) * C v * rG + rG>A) D dm
Lm
dm +
rG * (v * rG) dm
Lm
Lm
rGdmb * (v * rG>A) + rG>A * a v *
rG dm = 0 and from Eq. 21–8 HG =
Lm
Lm
rG dm b + rG>A * (v * rG>A)
rG * (v * rG)dm
HA = (rG>A * vA)m + HG + rG>A * (v * rG>A)m
= rG>A * (vA + (v * rG>A))m + HG
= (rG>A * mvG) + HG
Q.E.D.
940
Lm
dm
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Page 941
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z
21–22. The 4-lb rod AB is attached to the disk and collar
using ball-and-socket joints. If the disk has a constant
angular velocity of 2 rad>s, determine the kinetic energy of
the rod when it is in the position shown. Assume the angular
velocity of the rod is directed perpendicular to the axis of
the rod.
B
1 ft
A
vA = vB + v * rA>B
i
vAi = -(1)(2)j + 3 vx
3
j
vy
-1
k
vz 3
-1
x
Expand and equate components:
vA = -vy + vz
(1)
2 = vx + 3 vz
(2)
0 = -vx - 3 vy
(3)
Also:
v # rA>B = 0
(4)
3vx - vy - vz = 0
Solving Eqs. (1)–(4):
vx = 0.1818 rad>s
vy = -0.06061 rad>s
vz = 0.6061 rad>s
vA = 0.667 ft>s
v is perpendicular to the rod.
vz = 0.1818 rad>s,
vy = -0.06061 rad>s,
vz = 0.6061 rad>s
vB = {-2j} ft>s
rA>B = {3i - 1j - 1k} ft
vG = vB + v *
vG
i
13
= -2j +
0.1818
2
3
rA>B
2
j
-0.06061
-1
2 rad/s
k
0.6061 3
-1
vG = {0.333i - 1j} ft>s
vG = 2(0.333)2 + (-1)2 = 1.054 ft>s
v = 2(0.1818)2 + (-0.06061)2 + (0.6061)2 = 0.6356 rad>s
1
4
1
4
1
b (1.054)2 + a b c
a
b(3.3166)2 d(0.6356)2
T = a ba
2 32.2
2 12 32.2
T = 0.0920 ft # lb
Ans.
941
3 ft
y
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Page 942
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z
21–23. Determine the angular momentum of rod AB in
Prob. 21–22 about its mass center at the instant shown.
Assume the angular velocity of the rod is directed
perpendicular to the axis of the rod.
B
2 rad/s
1 ft
A
x
vA = vB + v * rA>B
i
vAi = -(1)(2)j + 3 vx
3
j
vy
-1
k
vz 3
-1
Expand and equate components:
vA = -vy + vz
(1)
2 = vx + 3 vz
(2)
0 = -vx - 3 vy
(3)
Also:
v # rA>B = 0
(4)
3vx - vy - vz = 0
Solving Eqs. (1)–(4):
vx = 0.1818 rad>s
vy = -0.06061 rad>s
vz = 0.6061 rad>s
vA = 0.667 ft>s
v is perpendicular to the rod.
rA>B = 2(3)2 + (-1)2 + (-1)2 = 3.3166 ft
IG = a
1
4
ba
b (3.3166)2 = 0.1139 slug # ft2
12 32.2
HG = IG v = 0.1139 (0.1818i - 0.06061j + 0.6061k)
HG = {0.0207i - 0.00690j + 0.0690k) slug # ft2>s
Ans.
942
3 ft
y
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Page 943
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z
*21–24. The uniform thin plate has a mass of 15 kg. Just
before its corner A strikes the hook, it is falling with a
velocity of vG = 5-5k6 m>s with no rotational motion.
Determine its angular velocity immediately after corner A
strikes the hook without rebounding.
vG
A
G
x 200 mm
Referring to Fig. a, the mass moments of inertia of the plate about the x, y, and z
axes are
1
(15) A 0.42 B + 15 A 0.22 + 02 B = 0.8 kg # m2
Ix = Ix¿ + m A yG 2 + zG 2 B =
12
Iy = Iy¿ + m A xG 2 + zG 2 B =
1
(15) A 0.62 B + 15c(-0.3)2 + 02 d = 1.8 kg # m2
12
Iz = Iz¿ + m A xG 2 + yG 2 B =
Due to symmetry, Ix¿y¿
1
(15) A 0.42 + 0.62 B + 15c( -0.3)2 + 0.22 d = 2.6 kg # m2
12
= Iy¿z¿ = Ix¿z¿ = 0. Thus,
Ixy = Ix¿y¿ + mxGyG = 0 + 15(-0.3)(0.2) = -0.9 kg # m2
Iyz = Iy¿z¿ + myGzG = 0 + 15(0.2)(0) = 0
Ixz = Ix¿z¿ + mxGzG = 0 + 15( -0.3)(0) = 0
Since the plate falls without rotational motion just before the impact, its angular
momentum about point A is
(HA)1 = rG>A * mvG = (-0.3i + 0.2j) * 15(-5k)
= [-15i - 22.5j] kg # m2>s
Since the plate rotates about point A just after impact, the components of its
angular momentum at this instant can be determined from
C (HA)2 D x = Ixvx - Ixyvy - Ixz vz
= 0.8vx - (-0.9)vy - 0(vz)
= 0.8vx + 0.9vy
C (HA)2 D y = -Ixyvx + Iyvy - Iyz vz
= -(-0.9)vx + 1.8vy - 0(vz)
= 0.9vx + 1.8vy
C (HA)2 D z = -Ixz vx + Iyzvy - Iz vz
= 0(vx) - 0(vy) + 2.6vz
= 2.6vz
Thus,
(HA)2 = (0.8vx + 0.9vy)i + (0.9vx + 1.8vy)j + 2.6vz k
Referring to the free-body diagram of the plate shown in Fig. b, the weight W is a
nonimpulsive force and the impulsive force FA acts through point A. Therefore,
angular momentum of the plate is conserved about point A. Thus,
(HA)1 = (HA)2
-15i - 22.5j = (0.8vx + 0.9vy)i + (0.9vx + 1.8vy)j + 2.6vz k
Equating the i, j, and k components,
-15 = 0.8vx + 0.9vy
(1)
-22.5 = 0.9vx + 1.8vy
(2)
0 = 2.6vz
(3)
Solving Eqs. (1) through (3),
vx = -10.71 rad>s
vy = -7.143 rad>s
vz = 0
Thus,
v = [-10.7i - 7.14j] rad>s
Ans.
943
300 mm
300 mm
200 mm
y
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Page 944
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z
•21–25. The 5-kg disk is connected to the 3-kg slender
rod. If the assembly is attached to a ball-and-socket joint at
A and the 5-N # m couple moment is applied, determine the
angular velocity of the rod about the z axis after the
assembly has made two revolutions about the z axis starting
from rest. The disk rolls without slipping.
M⫽5N⭈m
A
x
Ix = Iz =
Iv =
1
1
(5)(0.2)2 + 5(1.5)2 + (3)(1.5)2 = 13.55
4
3
1
(5)(0.2)2 = 0.100
2
v = -vy j¿ + vz k = -vy¿ j¿ + vz sin 7.595°j¿ + vz cos 7.595°k¿
= (0.13216vz - vy¿)j¿ + 0.99123 vz k¿
Since points A and C have zero velocity,
vC = vA + v * rC>A
0 = 0 + C (0.13216 vz - vy¿)j¿ + 0.99123vz k¿ D * (1.5j¿ - 0.2k¿)
0 = -1.48684vz - 0.026433 vz + 0.2 vy¿
vy¿ = 7.5664 vz
Thus,
v = -7.4342 vz j¿ + 0.99123 vz k¿
T1 + ©U1 - 2 = T2
0 + 5(2p) (2) = 0 +
1
1
(0.100)(-7.4342 vz)2 + (13.55)(0.99123 vz)2
2
2
vz = 2 58 rad>s
Ans.
944
1.5 m
B
0.2 m
y
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Page 945
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z
21–26. The 5-kg disk is connected to the 3-kg slender rod.
If the assembly is attached to a ball-and-socket joint at A
and the 5-N # m couple moment gives it an angular velocity
about the z axis of vz = 2 rad>s, determine the magnitude
of the angular momentum of the assembly about A.
M⫽5N⭈m
A
x
Ix = Iz =
Ix =
1
1
(5)(0.2)2 + 5(1.5)2 + (3)(1.5)2 = 13.55
4
3
1
(5)(0.2)2 = 0.100
2
v = -vy j¿ + vz k = -vy¿ j¿ + vz sin 7.595°j¿ + vz cos 7.595°k¿
= (0.13216vz - vy )j¿ + 0.99123 vz k¿
Since points A and C have zero velocity,
vC = vA + v * rC>A
0 = 0 + C (0.13216 vz - vy ) j¿ + 0.99123 vz k¿ D * (1.5j¿ - 0.2k¿)
0 = -1 48684vz - 0.26433vz + 0.2vy
vy¿ = -7.5664 vz
Thus,
v = -7.4342 vz j¿ + 0.99123 vz k¿
Since vz = 2 rad>s
v = -14.868j¿ + 1 9825k¿
So that,
HA = Ix¿ vx i¿ + Iy¿ vy¿ j + Iz¿ vz¿ k¿ = 0 + 0.100(-14.868)j¿ + 13.55(1.9825) k¿
= -1.4868j¿ + 26.862k¿
HA = 2(-1.4868)2 + (26.862)2 = 26.9 kg # m2>s
Ans.
945
1.5 m
B
0.2 m
y
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Page 946
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21–27. The space capsule has a mass of 5 Mg and the
radii of gyration are kx = kz = 1.30 m and ky = 0.45 m .
If it travels with a velocity vG = 5400j + 200k6 m>s ,
compute its angular velocity just after it is struck by a
meteoroid having a mass of 0.80 kg and a velocity
vm = 5-300i + 200j - 150k6 m>s . Assume that the
meteoroid embeds itself into the capsule at point A and
that the capsule initially has no angular velocity.
vG
G
x
Conservation of Angular Momentum: The angular momentum is conserved about
the center of mass of the space capsule G. Neglect the mass of the meteroid after the
impact.
(HG)1 = (HG)2
rGA * mm vm = IG v
(0.8i + 3.2j + 0.9k) * 0.8(-300i + 200j - 150k)
= 5000 A 1.302 B vx i + 5000 A 0.452 B vy j + 5000 A 1.302 B vz k
-528i - 120j + 896k = 8450vx i + 1012.5vy j + 8450 vz k
Equating i, j and k components, we have
-528 = 8450vx
-120 = 1012.5vy
896 = 8450vz
z
vm
vx = -0.06249 rad>s
vy = -0.11852 rad>s
vz = 0.1060 rad>s
Thus,
v = {-0.0625i - 0.119j + 0.106k} rad>s
Ans.
946
A (0.8 m, 3.2 m, 0.9 m)
y
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z
*21–28. Each of the two disks has a weight of 10 lb. The
axle AB weighs 3 lb. If the assembly rotates about the z
axis at vz = 6 rad>s, determine its angular momentum
about the z axis and its kinetic energy. The disks roll
without slipping.
2 ft
2 ft
B
1 ft
A
x
1
6
=
vx
2
vx = 12 rad>s
vA = {-12i} rad>s
Hz = c
vB = {12i} rad>s
1 10
1 10
a
b (1)2 d(12i) + c a
b (1)2 d( -12i)
2 32.2
2 32.2
+ 0 + b 2c
3
1 10
10
1
a
b (1)2 +
(2)2 d(6) +
a
b(4)2(6) r k
4 32.2
32.2
12 32.2
Hz = {16.6k} slug # ft2>s
T =
=
Ans.
1
1
1
I v2 + Iy v2y + Iz v2z
2 x x
2
2
1
1 10
c2 a a
b (1)2 b d(12)2 + 0
2
2 32.2
+
1 10
10
1
3
1
b (1)2 +
(2)2 d +
a
b(4)2 r (6)2
b 2c a
2
4 32.2
32.2
12 32.2
= 72.1 lb # ft
Ans.
947
1 ft
vz ⫽ 6 rad/s
y
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Page 948
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z
•21–29. The 10-kg circular disk spins about its axle with a
constant angular velocity of v1 = 15 rad>s. Simultaneously,
arm OB and shaft OA rotate about their axes with constant
angular velocities of v2 = 0 and v3 = 6 rad>s, respectively.
Determine the angular momentum of the disk about point O,
and its kinetic energy.
O
The mass moments of inertia of the disk about the centroidal x¿ , y¿ , and z¿ axes,
Fig. a, are
x
Ix¿ = Iy¿ =
1
1
mr2 = (10) A 0.152 B = 0.05625 kg # m2
4
4
Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0
Here, the angular velocity of the disk can be determined from the vector addition of
v1 and v3. Thus,
v = v1 + v2 = [6i + 15k] rad>s
The angular momentum of the disk about its mass center G can be obtained by
applying
Hx = Ix¿vx = 0.05625(6) = 0.3375 kg # m2>s
Hy = Iy¿vy = 0.05625(0) = 0
Hz = Iz¿vz = 0.1125(15) = 1.6875 kg # m2>s
Thus,
HG = [0.3375i + 1.6875k] kg # m2>s
Since the mass center G rotates about the x axis with a constant angular velocity of
v3 = [6i] rad>s, its velocity is
vG = v3 * rC>O = (6i) * (0.6j) = [3.6k] m>s
Since the disk does not rotate about a fixed point O, its angular momentum must be
determined from
HO = rC>O * mvC + HG
= (0.6j) * 10(3.6k) + (0.3375i + 1.6875k)
= [21.9375i + 1.6875k] kg # m2>s
Ans.
The kinetic energy of the disk is therefore
T =
=
1 #
v HO
2
1
(6i + 15k) # (21.9375i + 1.6875k)
2
= 78.5 J
Ans.
948
V2
B
150 mm
y
Due to symmetry, the products of inertia of the disk with respect to its centroidal
planes are equal to zero.
= [21.9i + 1.69k] kg # m2>s
V3
V1
1
1
mr2 = (10) A 0.152 B = 0.1125 kg # m2
2
2
Iz¿ =
A
600 mm
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z
21–30. The 10-kg circular disk spins about its axle with a
constant angular velocity of v1 = 15 rad>s. Simultaneously,
arm OB and shaft OA rotate about their axes with constant
angular velocities of v2 = 10 rad>s and v3 = 6 rad>s,
respectively. Determine the angular momentum of the disk
about point O, and its kinetic energy.
O
The mass moments of inertia of the disk about the centroidal x¿ , y¿ , and z¿ axes.
Fig. a, are
x
Ix¿ = Iy¿ =
Iz¿ =
1
1
mr2 = (10) A 0.152 B = 0.1125 kg # m2
2
2
Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0
Here, the angular velocity of the disk can be determined from the vector addition of
v1, v2, and v3. Thus,
v = v1 + v2 + v3 = [6i + 10j + 15k] rad>s
The angular momentum of the disk about its mass center G can be obtained by
applying
Hx = Ix¿vx = 0.05625(6) = 0.3375 kg # m2
Hy = Iy¿vy = 0.05625(10) = 0.5625 kg # m2
Hz = Iz¿vz = 0.1125(15) = 1.6875 kg # m2
Thus,
HG = [0.3375i + 0.5625j + 1.6875k] kg # m2
Since the mass center G rotates about the fixed point O with an angular velocity of
Æ = v2 + v3 = [6i + 10j], its velocity is
vG = Æ * rG>O = (6i + 10j) * (0.6j) = [3.6k] m>s
Since the disk does not rotate about a fixed point O, its angular momentum must be
determined from
HO = rC>O * mvG + HG
= (0.6j) * 10(3.6k) + (0.3375i + 0.5625j + 1.6875k)
= [21.9375i + 0.5625j + 1.6875k] kg # m2>s
Ans.
The kinetic energy of the disk is therefore
=
1 #
v HO
2
1
(6i + 10j + 15k) # (21.9375i + 0.5625j + 1.6875k)
2
= 81.3J
Ans.
949
V2
B
150 mm
y
Due to symmetry, the products of inertia of the disk with respect to its centroidal
planes are equal to zero.
T =
V3
V1
1
1
mr2 = (10) A 0.152 B = 0.05625 kg # m2
4
4
= [21.9i + 0.5625j + 1.69k] kg # m2>s
A
600 mm
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z, z¿
Vz¿ ⫽ 1250 rad/s
21–31. The 200-kg satellite has its center of mass at point G.
Its radii of gyration about the z¿ , x¿ , y¿ axes are
kz¿ = 300 mm, kx¿ = ky¿ = 500 mm, respectively. At the
instant shown, the satellite rotates about the x¿, y¿ , and z¿
axes with the angular velocity shown, and its center of mass
G has a velocity of vG = 5—250i + 200j + 120k6 m>s.
Determine the angular momentum of the satellite about
point A at this instant.
G
Vx¿ ⫽ 600 rad/s
The mass moments of inertia of the satellite about the x¿ , y¿ , and z¿ axes are
x¿
Ix¿ = Iy¿ = 200 A 0.52 B = 50 kg # m2
x
Iz¿ = 200 A 0.32 B = 18 kg # m2
Due to symmetry, the products of inertia of the satellite with respect to the x¿ , y¿ ,
and z¿ coordinate system are equal to zero.
Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0
The angular velocity of the satellite is
v = [600i + 300j + 1250k] rad>s
Thus,
vx¿ = 600 rad>s
vy¿ = -300 rad>s
800 mm
vz¿ = 1250 rad>s
Then, the components of the angular momentum of the satellite about its mass
center G are
(HG)x¿ = Ix¿vx¿ = 50(600) = 30 000 kg # m2>s
(HG)y¿ = Iy¿vy¿ = 50(-300) = 15 000 kg # m2>s
(HG)z¿ = Iz¿vz¿ = 18(1250) = 22 500 kg # m2>s
Thus,
HG = [30 000i + 15 000j + 22 500k] kg # m2>s
The angular momentum of the satellite about point A can be determined from
HA = rG>A * mvG + HG
= (0.8k) * 200(-250i + 200j + 120k) + (30 000i + 15 000j + 22 500k)
= [-2000i - 25 000j + 22 500k] kg # m2>s
Ans.
950
A
vG
Vy ¿ ⫽ 300 rad/s
y¿
y
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z, z¿
Vz¿ ⫽ 1250 rad/s
*21–32. The 200-kg satellite has its center of mass at point G.
Its radii of gyration about the z¿ , x¿ , y¿ axes are kz¿ = 300 mm,
kx¿ = ky¿ = 500 mm, respectively. At the instant shown, the
satellite rotates about the x¿ , y¿ , and z¿ axes with the angular
velocity shown, and its center of mass G has a velocity of
vG = 5—250i + 200j + 120k6 m>s. Determine the kinetic
energy of the satellite at this instant.
G
Vx¿ ⫽ 600 rad/s
The mass moments of inertia of the satellite about the x¿ , y¿ , and z¿ axes are
x¿
Ix¿ = Iy¿ = 200 A 0.52 B = 50 kg # m2
x
Iz¿ = 200 A 0.32 B = 18 kg # m2
Due to symmetry, the products of inertia of the satellite with respect to the x¿ , y¿ ,
and z¿ coordinate system are equal to zero.
Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0
The angular velocity of the satellite is
v = [600i - 300j + 1250k] rad>s
Thus,
vx¿ = 600 rad>s
vy¿ = -300 rad>s
vz¿ = 1250 rad>s
Since vG 2 = (-250)2 + 2002 + 1202 = 116 900 m2>s2, the kinetic energy of the
satellite can be determined from
T =
=
800 mm
1
1
1
1
mvG 2 + Ix¿vx¿ 2 + Iy¿vy¿ 2 + Iz¿vz¿ 2
2
2
2
2
1
1
1
1
(200)(116 900) + (50) A 6002 B + (50)(-300)2 + (18) A 12502 B
2
2
2
2
= 37.0025 A 106 B J = 37.0MJ
Ans.
951
A
vG
Vy ¿ ⫽ 300 rad/s
y¿
y
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z
•21–33. The 25-lb thin plate is suspended from a ball-andsocket joint at O. A 0.2-lb projectile is fired with a velocity
of v = 5 -300i - 250j + 300k6 ft>s into the plate and
becomes embedded in the plate at point A. Determine the
angular velocity of the plate just after impact and the axis
about which it begins to rotate. Neglect the mass of the
projectile after it embeds into the plate.
0.5 ft
O
0.5 ft
y
x
0.75 ft
0.25 ft
Angular momentum about point O is conserved.
A
(HO)2 = (HO)1 = rOA * mp vp
(HO)1 = (0.25j - 0.75k) * a
0.25 ft
0.2
b( -300i - 250j + 300k) = {-0.6988i + 1.3975j + 0.4658k} lb # ft # s
32.2
Ix = a
1
25
25
ba
b C (1)2 + (1)2 D + a
b(0.5)2 = 0.3235 slug # ft2
12 32.2
32.2
Iy = a
25
1
25
ba
b (1)2 + a
b (0.5)2 = 0.2588 slug # ft2
12 32.2
32.2
Iz = a
25
1
ba
b(1)2 = 0.06470 slug # ft2
12 32.2
(HO)1 = (HO)2
-0.6988i + 1.3975j + 0.4658k = 0.3235vx i + 0.2588vy j + 0.06470vz k
vx =
-0.6988
= -2.160 rad>s
0.3235
vy =
1.3975
= 5.400 rad>s
0.2588
vz =
0.4658
= 7.200 rad>s
0.06470
v = {-2.16i + 5.40j + 7.20k} rad>s
Ans.
Axis of rotation line is along v:
uO =
-2.160i + 5.400j + 7.200k
2(-2.160)2 + (5.400)2 + (7.200)2
= -0.233i + 0.583j + 0.778k
Ans.
952
v
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z
21–34. Solve Prob. 21–33 if the projectile emerges from
the plate with a velocity of 275 ft>s in the same direction.
0.5 ft
O
0.5 ft
y
x
0.75 ft
0.25 ft
uv = a
250
300
-300
bi - a
bj + a
b k = -0.6092i - 0.5077j + 0.6092k
492.4
492.4
492.4
Ix = a
25
1
25
ba
b C (1)2 + (1)2 D + a
b(0.5)2 = 0.32350 slug # ft2
12 32.2
32.2
A
0.25 ft
v
1
25
25
Iy = a b a
b (1)2 + a
b (0.5)2 = 0.25880 slug # ft2
12 32.2
32.2
Iz = a
1
25
ba
b(1)2 = 0.06470 slug # ft2
12 32.2
H1 + ©
L
MO dt = H2
(0.25j - 0.75k) * a
0.2
b( -300i - 250j + 300k) + 0 = 0.32350vx i + 0.25880vy j + 0.06470vz k
32.2
+(0.25j - 0.75k) * a
0.2
b (275)(-0.6092i - 0.5077j + 0.6092k)
32.2
Expanding, the i, j, k, components are:
-0.6988 = 0.32350vx - 0.390215
1.3975 = 0.25880vy + 0.78043
0.4658 = 0.06470vz + 0.26014
vx = -0.9538,
vy = 2.3844,
vz = 3.179
v = {-0.954i + 2.38j + 3.18k} rad>s
Ans.
Axis of rotation is along v:
uA =
-0.954i + 2.38j + 3.18k
2(-0.954)2 + (2.38)2 + (3.18)2
uA = -0.233i + 0.583j + 0.778k
Ans.
953
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z
21–35. A thin plate, having a mass of 4 kg, is suspended
from one of its corners by a ball-and-socket joint O. If a
stone strikes the plate perpendicular to its surface at an
adjacent corner A with an impulse of Is = 5-60i6 N # s,
determine the instantaneous axis of rotation for the plate
and the impulse created at O.
O
200 mm
x
y
A
(HO)1 + ©
L
MO dt = (HO)2
Is ⫽ {⫺60i} N⭈s
0 + rA>O * IS = (HO)2
200 mm
0 + ( -0.2(0.7071)j - 0.2(0.7071)k) * (-60i) = (IO)x vx i + (IO)y vy j + (IO)z vz k
Expand and equate components:
Iy¿z¿ = 0,
Ix¿y¿ = 0,
Iy¿ = a
0 = (IO)x vx
(1)
8.4853 = (IO)y vy
(2)
-8.4853 = (IO)z vz
(3)
Ix¿z¿ = 0
1
b(4)(0.2)2 = 0.01333,
12
ux¿ = cos 90° = 0,
Iz¿ = a
1
b(4)(0.2)2 = 0.01333
12
uy¿ = cos 135° = -0.7071,
uz¿ = cos 45° = 0.7071
(IG)z = Ix¿ u2x¿ + Iy¿ u2y¿ + Iz¿ u2z¿ - 2Ix¿y¿ux¿uy¿ - 2Iy¿z¿uy¿uz¿ - 2Iz¿x¿uz¿ux¿
= 0 + (0.01333)(-0.7071)2 + (0.01333)(0.7071)2 - 0 - 0 - 0
(IG)z = (IO)z = 0.01333
For (IO)y, use the parallel axis theorem.
(IO)y = 0.01333 + 4 C 0.7071(0.2) D 2,
(IO)y = 0.09333
Hence, from Eqs. (1) and (2):
vx = 0,
vy = 90.914,
vz = -636.340
The instantaneous axis of rotation is thus,
ulA =
90.914j - 636.340k
2(90.914)2 + (-636.340)2
= 0.141j - 0.990k
Ans.
The velocity of G just after the plate is hit is
vG = v * rG>O
vG = (90.914j - 636.340k) * (-0.2(0.7071)k) = -12.857i
m(vG)1 + ©
0 - 60i +
L
L
L
F dt = m(vG)2
FO dt = -4(12.857)i
FO dt = {8.57i} N
#
s
Ans.
954
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z
*21–36. The 15-lb plate is subjected to a force F = 8 lb
which is always directed perpendicular to the face of the
plate. If the plate is originally at rest, determine its angular
velocity after it has rotated one revolution (360°). The plate
is supported by ball-and-socket joints at A and B.
A
F ⫽ 8 lb
Due to symmetry
1.2 ft
Ix¿y¿ = Iy¿z¿ = Iz¿x¿ = 0
Ix¿ =
Iy¿ =
1
15
a
b (1.2)2 = 0.05590 slug # ft2
12 32.2
0.4 ft
15
1
a
b(1.22 + 0.42) = 0.06211 slug # ft2
12 32.2
B
x
y
15
1
Iz¿ =
a
b (0.4)2 = 0.006211 slug # ft2
12 32.2
For z axis
uy¿ = cos 90° = 0
ux¿ = cos 71.57° = 0.3162
uz¿ = cos 18.43° = 0.9487
Iz = Ix¿ u2x¿ + Iy¿ u2y¿ + Iz¿ u2z¿ - 2Ix¿y¿ux¿uy¿ - 2Iy¿z¿uy¿uz¿ - 2Iz¿x¿uz¿ux¿
= 0.05590(0.3162)2 + 0 + 0.006211(0.9487)2 - 0 - 0 - 0
= 0.01118 slug # ft2
Principle of work and energy:
T1 + ©U1 - 2 = T2
0 + 8(1.2 sin 18.43°)(2p) =
1
(0.01118)v2
2
v = 58.4 rad>s
Ans.
z
•21–37. The plate has a mass of 10 kg and is suspended
from parallel cords. If the plate has an angular velocity of
1.5 rad>s about the z axis at the instant shown, determine
how high the center of the plate rises at the instant the plate
momentarily stops swinging.
Consevation Energy: Datum is set at the initial position of the plate. When the
plate is at its final position and its mass center is located h above the datum. Thus,
its gravitational potential energy at this position is 10(9.81)h = 98.1h. Since the
plate momentarily stops swinging, its final kinetic energy T2 = 0. Its initial kinetic
energy i
120⬚
250 mm
1
1 1
T1 = IG v2 = c (10) A 0.252 B d A 1.52 B = 0.3516 J
2
2 2
120⬚
y
120⬚
1.5 rad/s
T1 + V1 = T2 + V2
x
0.3516 + 0 = 0 + 98.1h
h = 0.00358 m = 3.58 mm
Ans.
955
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z
21–38. The satellite has a mass of 200 kg and radii of
gyration of kx = ky = 400 mm and kz = 250 mm. When it
is not rotating, the two small jets A and B are ignited
simultaneously, and each jet provides an impulse of
I = 1000 N # s on the satellite. Determine the satellite’s
angular velocity immediately after the ignition.
500 mm
500 mm
500 mm
The mass moments of inertia of the satellite about the x, y, and z axes are
Ix = Iy = 200 A 0.4 B = 32 kg # m2
Iz = 200 A 0.252 B = 12.5 kg # m2
x
2
I
Due to symmetry,
Ixy = Iyz = Ixz = 0
Thus, the angular momentum of the satellite about its mass center G is
Hx = Ixvx = 32vx
Hy = Iyvy = 32vy
Hz = Izvz = 12.5vz
Applying the principle of angular impulse and momentum about the x, y, and z axes,
t2
(Hx)1 + ©
Mx dt = (Hx)2
Lt1
0 + 0 = 32vx
vx = 0
A Hy B 1 + ©
My dt = A Hy B 2
t2
Lt1
0 - 1000(0.4) - 1000(0.5) = 32vy
vy = -28.125 rad>s
t2
(Hz)1 + ©
Lt1
My dt = (Hz)2
0 + 1000(0.5) + 1000(0.5) = 12.5vz
vz = 80 rad>s
Thus
v = {-28.1j + 80k} rad>s
Ans.
956
G
B
⫺I
A
400 mm
y
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z
21–39. The bent rod has a mass per unit length of 6 kg>m,
and its moments and products of inertia have been
calculated in Prob. 21–9. If shaft AB rotates with a constant
angular velocity of vz = 6 rad>s, determine the angular
momentum of the rod about point O, and the kinetic energy
of the rod.
Vz
2m
A
O
B
x
y
2m
Here, the angular velocity of the rod is
v = [6k] rad>s
2m
Thus,
vx = vy = 0
vz = 6 rad>s
The rod rotates about a fixed point O. Using the results of Prob. 20–91
Hx = Ixvx - Ixyvy - Ixzvz
= 80(0) - 72(0) - (-24)(6) = 144 kg # m2>s
Hy = -Ixyvx + Iyvy - Iyzvz
= -72(0) + 128(0) - (-24)(6) = 144 kg # m2>s
Hz = -Ixzvx - Iyzvy + Izvz
= -(-24)(0) - (-24)(0) + 176(6) = 1056 kg # m2>s
Thus,
HO = [144i + 144j + 1056k] kg # m2>s
Ans.
The kinetic energy of the rod can be determined from
T =
=
1 #
v HO
2
1
(6k) # (144i + 144j + 1056k)
2
= 3168 J = 3.17 kJ
Ans.
957
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*21–40. Derive the scalar form of the rotational equation
of motion about the x axis if æ Z V and the moments and
products of inertia of the body are not constant with respect
to time.
In general
d
(Hx i + Hy j + Hz k)
dt
#
#
#
= A Hx i + Hy j + Hz k B xyz + Æ * (Hx i + Hyj + Hz k)
M =
Substitute Æ = Æ x i + Æ y j + Æ z k and expanding the cross product yields
#
#
M = a A Hx B xyz - Æ z Hy + Æ yHz b i + a A Hy B xyz - Æ x Hz + Æ zHx bj
#
+ a A Hz B xyz - Æ y Hx + Æ x Hy bk
Subsitute Hx, Hy and Hz using Eq. 21–10. For the i component
©Mx =
d
(I v - Ixy vy - Ixz vz) - Æ z (Iy vy - Iyz vz - Iyxvx)
dt x x
+ Æ y (Iz vz - Izx vx - Izy vy)
Ans.
One can obtain y and z components in a similar manner.
•21–41. Derive the scalar form of the rotational
equation of motion about the x axis if æ Z V and the
moments and products of inertia of the body are constant
with respect to time.
In general
d
(Hx i + Hy j + Hz k)
dt
#
#
#
= A Hx i + Hyj + Hz k B xyz + Æ * (Hx i + Hyj + Hz k)
M =
Substitute Æ = Æ x i + Æ y j + Æ z k and expanding the cross product yields
#
#
M = a A Hx B xyz - Æ z Hy + Æ yHz b i + a A Hy B xyz - Æ x Hz + Æ zHx bj
#
+ a A Hz B xyz - Æ y Hx + Æ x Hy bk
Substitute Hx, Hy and Hz using Eq. 21–10. For the i component
©Mx =
d
(I v - Ixy vy - Ixz vz) - Æ z (Iy vy - Iyz vz - Iyxvx)
dt x x
+ Æ y (Iz vz - Izx vx - Izy vy)
For constant inertia, expanding the time derivative of the above equation yields
#
#
#
©Mx = (Ix vx - Ixy vy - Ixzvz) - Æ z (Iy vy - Iyz vz - Iyx vx)
+ Æ y (Iz vz - Izxvx - Izy vy)
One can obtain y and z components in a similar manner.
958
Ans.
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21–42. Derive the Euler equations of motion for æ Z V ,
i.e., Eqs. 21–26.
In general
d
(Hx i + Hy j + Hz k)
dt
#
#
#
= A Hx i + Hyj + Hz k B xyz + Æ * (Hx i + Hyj + Hz k)
M =
Substitute Æ = Æ x i + Æ y j + Æ z k and expanding the cross product yields
#
#
M = a A Hx B xyz - Æ z Hy + Æ yHz b i + a A Hy B xyz - Æ x Hz + Æ zHx bj
#
+ a A Hz B xyz - Æ y Hx + Æ x Hy bk
Substitute Hx, Hy and Hz using Eq. 21–10. For the i component
©Mx =
d
(I v - Ixy vy - Ixz vz) - Æ z (Iy vy - Iyz vz - Iyxvx)
dt x x
+ Æ y (Iz vz - Izx vx - Izy vy)
Set Ixy = Iyz = Izx = 0 and require Ix, Iy, Iz to be constant. This yields
#
©Mx = Ix vx - Iy Æ z vy + Iz Æ yvz
Ans.
One can obtain y and z components in a similar manner.
959
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21–43. The uniform rectangular plate has a mass of
m = 2 kg and is given a rotation of v = 4 rad>s about its
bearings at A and B. If a = 0.2 m and c = 0.3 m, determine
the vertical reactions at A and B at the instant the plate
is vertical as shown. Use the x, y, z axes shown and note that
mac c2 - a2
Izx = - a
ba 2
b.
12
c + a2
x
A
V
B
c
y
vx = 0,
vy = 0,
vz = -4
#
vx = 0,
#
vy = 0,
#
vz = 0
#
#
©My = Iyy vy - (Izz - Ixx)vz vx - Iyz avz - vx vy b - Izx A v2z - v2x B
#
- Ixy avx + vy vz b
1
1
a 2
c 2 2
a 2
c 2 2
Bx B a b + a b R - Ax B a b + a b R = -Izx (v)2
2
2
2
2
Bx - Ax = a
mac
c2 - a2
≥v2
b£
2
2 32
6
Ca + c D
©Fx = m(aG)x ;
Ax + Bx - mg = 0
Substitute the data,
Bx - Ax =
2(0.2)(0.3)
(0.3)2 - (0.2)2
C
S(-4)2 = 0.34135
2
2 32
6
(0.3)
+
(0.2)
C
D
Ax + Bx = 2(9.81)
Solving:
Ax = 9.64 N
Ans.
Bx = 9.98 N
Ans.
960
a
z
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*21–44. The disk, having a mass of 3 kg, is mounted
eccentrically on shaft AB. If the shaft is rotating at a constant
rate of 9 rad>s, determine the reactions at the journal bearing
supports when the disk is in the position shown.
1m
A
1.25 m
v ⫽ 9 rad/s
50 mm
vx = 0,
vy = -9,
vz = 0
#
©Mx = Ix vx - (Iy - Iz)vy vz
Bz (1.25) - Az (1) = 0 - 0
#
©Mz = Iz vz - (Ix - Iy)vx vy
Ax (1) - Bx (1.25) = 0 - 0
©Fx = max ;
Ax + Bx = 0
©Fz = maz ;
Az + Bz - 3(9.81) = 3(9)2(0.05)
Solving,
Ax = Bx = 0
Ans.
Az = 23.1 N
Ans.
Bz = 18.5 N
Ans.
961
75 mm
B
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•21–45. The slender rod AB has a mass m and it is
connected to the bracket by a smooth pin at A. The bracket
is rigidly attached to the shaft. Determine the required
constant angular velocity of V of the shaft, in order for the
rod to make an angle of u with the vertical.
L
3
V
The rotating xyz frame is set with its origin at the rod’s mass center, Fig. a. This
frame will be attached to the rod so that its angular velocity is Æ = v and the x, y, z
axes will always be the principal axes of inertia. Referring to Fig. b,
A
u
v = -v cos uj + v sin uk
L
Thus,
vx = 0
vy = -v cos u
vz = v sin u
B
#
Since both the direction and the magnitude is constant v = 0. Also, since Æ = v,
#
#
A vxyz B = v = 0. Thus,
#
#
#
vx = vy = vz = 0
The mass moments of inertia of the rod about the x, y, z axes are
Ix = Iz =
1
mL2
12
Iy = 0
Applying the equation of motion and referring to the free-body diagram of the rod,
Fig. a,
L
1
#
mL2 ≤ ( -v cos u)(v sin u)
©Mx = Ixvx - A Iy - Iz B vyvz; - Az a b = 0 - ¢ 0 2
12
Az =
mv2L
sin u cos u
6
(1)
The acceleration of the mass center of the rod can be determined from
L
L
v2L
(3 sin u + 2) and is directed as shown in
aG = v2 r = v2 a sin u + b =
2
3
6
Fig. c. Thus,
©Fz = m(aG)z ;
Az - mg sin u = -
mv2L
(3 sin u + 2) cos u
6
Az = mg sin u = -
mv2L
(3 sin u + 2) cos u (2)
6
Equating Eqs. (1) and (2),
mv2L
mv2L
sin u cos u = mg sin u (3 sin u + 2) cos u
6
6
v =
3g tan u
A L(2 sin u + 1)
Ans.
962
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21–46. The 5-kg rod AB is supported by a rotating arm.The
support at A is a journal bearing, which develops reactions
normal to the rod. The support at B is a thrust bearing, which
develops reactions both normal to the rod and along the axis
of the rod. Neglecting friction, determine the x, y, z
components of reaction at these supports when the frame
rotates with a constant angular velocity of v = 10 rad>s.
z
0.5 m
B
0.5 m
G
y
A
1
Iy = Iz =
(5)(1)2 = 0.4167 kg # m2
12
Ix = 0
ω = 10 rad/s
x
#
#
#
Applying Eq. 21–25 with vx = vy = 0 vz = 10 rad>s vx = vy = vz = 0
#
©Mx = Ix vx - (Iy - Iz)vy vz ;
0 = 0
#
©My = Iy vy - (Iz - Ix)vz vx ;
Bz (0.5) - Az(0.5) = 0
(1)
#
©Mz = Iz vz - (Ix - Iy)vx vy ;
Ay (0.5) - By(0.5) = 0
(2)
Also,
©Fx = m(aG)x ;
Bx = -5(10)2 (0.5)
©Fy = m(aG)y ;
Ay + By = 0
(3)
©Fz = m(aG)z ;
Az + Bz - 5(9.81) = 0
(4)
Bx = -250N
Ans.
Solving Eqs. (1) to (4) yields:
Ay = By = 0
Az = Bz = 24.5 N
Ans.
z
21–47. The car travels around the curved road of radius r
such that its mass center has a constant speed vG. Write the
equations of rotational motion with respect to the x, y, z
axes. Assume that the car’s six moments and products of
inertia with respect to these axes are known.
Applying Eq. 21–24 with vx = 0,
vy = 0,
vz =
y
G
vG
,
r
r
vx = vy = vz = 0
©Mx = -Iyz B 0 - a
©My = -Izx B a
Iyz
vG 2
b R = 2 v2G
r
r
Ans.
Izx
vG 2
b - 0 R = - 2 v2G
r
r
Ans.
Ans.
©Mz = 0
Note: This result indicates the normal reactions of the tires on the ground are not all
necessarily equal. Instead, they depend upon the speed of the car, radius of
curvature, and the products of inertia Iyz and Izx. (See Example 13–6.)
963
x
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z
*21–48. The shaft is constructed from a rod which has a
mass per unit length of 2 kg>m. Determine the x, y, z
components of reaction at the bearings A and B if at the
instant shown the shaft spins freely and has an angular
velocity of v = 30 rad>s. What is the angular acceleration of
the shaft at this instant? Bearing A can support a component
of force in the y direction, whereas bearing B cannot.
v ⫽ 30 rad/s
A
O
0.6 m
x
0.1 m
B
0.2 m
y
0.2 m
0.6 m
©W = C 3(0.2) + 1.2 D (2)(9.81) = 35.316 N
©xW = 0 C 1.2(2)(9.81) D + 0.1 C 0.4(2)(9.81) D + 0.2 C 0.2(2)(9.81) D = 1.5696 N # m
x =
1.5696
© xW
=
= 0.04444 m
©W
35.316
Ix = 2 c
1
C 0.2(2) D (0.2)2 d + C 0.2(2) D (0.2)2 = 0.02667 kg # m2
3
#
#
Applying Eq. 21–25 with vx = vz = 0 vy = 30 rad>s vx = vz = 0
#
©Mx = Ix vx - (Iy - Iz)vy vz ;
Bz (0.7) - Az(0.7) = 0
#
©My = Iy vy - (Iz - Ix)vz vx;
#
35.316(0.04444) = 0.02667vy
(1)
#
vy = 58.9 rad>s2
#
©Mz = Iz vz - (Ix - Iy)vx vy ;
Ans.
Bx (0.7) - Ax(0.7) = 0
(2)
Also,
©Fx = m(aG)x ;
-Ax - Bx = -1.8(2)(0.04444)(30)2
©Fy = m(aG)y ;
Ay = 0
©Fz = m(aG)z ;
Az + Bz - 35.316 = -1.8(2)(0.04444)(58.9)
(3)
Ans.
(4)
Solving Eqs. (1) to (4) yields:
Ax = Bx = 72.0 N
Az = Bz = 12.9 N
Ans.
964
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•21–49. Four spheres are connected to shaft AB. If
mC = 1 kg and mE = 2 kg, determine the mass of spheres
D and F and the angles of the rods, uD and uF, so that the
shaft is dynamically balanced, that is, so that the bearings at
A and B exert only vertical reactions on the shaft as it
rotates. Neglect the mass of the rods.
z
D
C
uD
A
x
V
0.1 m
0.1 m
(0.1 cos 30°)(2) - (0.1 sin uF)mF - (0.2 sin uD)mD = 0
(1)
For z = 0; ©z1 m1 = 0
(0.1)(1) - (0.1 sin 30°)(2) + (0.2 cos uD)mD + (0.1 cos uF)mF = 0
(2)
For Ixz = 0; ©x1z1 m1 = 0
-(0.2)(0.2 sin uD)mD + (0.3)(0.1 cos 30°)(2) - (0.4)(0.1 sin uF)mF = 0
(3)
For Ixy = 0; ©x1y1 m1 = 0
(0.1)(0.1)(1) + (0.2)(0.2 cos uD)mD - (0.3)(0.1 sin 30°)(2)
+ (0.1 cos uF)(0.4)(mF) = 0
(4)
Solving,
uD = 139°
Ans.
mD = 0.661 kg
Ans.
uF = 40.9°
Ans.
mF = 1.32 kg
Ans.
965
F
uF
0.1 m
B
0.1 m
y
0.1 m
30⬚
For x = 0; ©x1 m1 = 0
0.2 m
0.1 m
E 0.1 m
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21–50. A man stands on a turntable that rotates about a
vertical axis with a constant angular velocity of
vp = 10 rad>s. If the wheel that he holds spins with a
constant angular speed of vs = 30 rad>s, determine the
magnitude of moment that he must exert on the wheel to
hold it in the position shown. Consider the wheel as a thin
circular hoop (ring) having a mass of 3 kg and a mean radius
of 300 mm.
500 mm
300 mm
vs ⫽ 30 rad/s
The rotating xyz frame will be set with an angular velocity of Æ = vP = [10k] rad>s.
Since the wheel is symmetric about its spinning axis, the x, y, and z axes will remain as
the principle axes of inertia. Thus,
Iy = Iz =
vp ⫽ 10 rad/s
1
1
mr2 = (3) A 0.32 B = 0.135 kg # m2
2
2
Ix = mr2 = 3 A 0.32 B = 0.27 kg # m2
The angular velocity of the wheel is v = vs + vP = [-30i + 10k] rad>s. Thus,
vx = -30 rad>s
vy = 0
vz = 10 rad>s
Since the directions of vs and vp do not change with respect to the xyz frame and
#
their magnitudes are constant, vxyz = 0. Thus,
#
#
#
vx = vy = vz = 0
Applying the equations of motion and referring to the free-body diagram shown in
Fig. a,
#
©Mx = Ixvx - Iy Æ zvy + Iz Æ yvz;
Mx = 0
#
©My = Iyvy - Iz Æ xvz + Ix Æ zvx;
My = 0 - 0 + 0.27(10)(-30) = -81.0 N # m
#
©Mz = Izvz - Ix Æ yvx + Iy Æ xvy;
Mz = 0
Thus,
M = 2Mx 2 + My 2 + Mz 2 = 202 + (-81.0)2 + 02 = 81.0 N # m
966
Ans.
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z
21–51. The 50-lb disk spins with a constant angular rate of
v1 = 50 rad>s about its axle. Simultaneously, the shaft
rotates with a constant angular rate of v2 = 10 rad>s.
Determine the x, y, z components of the moment developed
in the arm at A at the instant shown. Neglect the weight of
arm AB.
2 ft
A
x
The rotating xyz frame is established as shown in Fig. a. This frame will be set to
have an angular velocity of Æ = v2 = [10i] rad>s. Since the disk is symmetric about
its spinning axis, the x, y, and z axes will remain as the principle axes of inertia. Thus,
Ix = Iy =
Iz =
1 50
a
b A 0.752 B = 0.2184 slug # ft2
4 32.2
1 50
a
b A 0.752 B = 0.4367 slug # ft2
2 32.2
The angular velocity of the disk is v = vs + vp = [10i + 50k] rad>s. Thus,
vx = 10 rad>s
vy = 0
vz = 50 rad>s
Since the directions of v1 and v2 do not change with respect to the xyz frame and
#
their magnitudes are constant, vxyz = 0. Thus,
#
#
#
vx = vy = vz = 0
Applying the equations of motion and referring to the free-body diagram shown in
Fig. a,
#
©Mx = Ixvx - Iy Æ zvy + Iz Æ yvz;
MX - AZ(2) = 0
#
©My = Iyvy - Iz Æ xvz + Ix Æ zvx;
MY = 0 - 0.4367(10)(50) + 0
(1)
MY = -218.36 lb # ft = -218 lb # ft
#
©Mz = Izvz - Ix Æ yvzx + Iy Æ xvy;
Ans.
MZ = 0 - 0 + 0
MZ = 0
Ans.
Since the mass center of the disk rotates about the X axis with a constant angular
#
velocity of v1 = [10i] rad>s, its acceleration is aG = v2 * rG - v2rG
2
2
= 0 - 10 (2j) = [-200j] ft>s . Thus,
©FZ = m(aG)Z ;
AZ - 50 =
50
(0) AZ = 50 lb
32.2
Substituting this result into Eq. (1), we have
MX = 100 lb # ft
Ans.
967
v2 ⫽ 10 rad/s
0.75 ft
B
v1 ⫽ 50 rad/s
y
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u
*21–52. The man stands on a turntable that rotates about a
vertical axis with a constant angular velocity of v1 = 6 rad>s. If
he tilts his head forward at a constant angular velocity of
v2 = 1.5 rad>s about point O, determine the magnitude of the
moment that must be resisted by his neck at O at the instant
u = 30°. Assume that his head can be considered as a uniform
10-lb sphere, having a radius of 4.5 in. and center of gravity
located at G, and point O is on the surface of the sphere.
G
4.5 in.
O
The rotating xyz frame shown in Fig. a will be attached to the head so that it rotates
with an angular velocity of Æ = v, where v = v1 + v2. Referring to Fig. b,
v1 = [6 cos 30° j + 6 sin 30° k] rad>s = [5.196j + 3k] rad>s. Thus, v = [-1.5i
+ 5.196j + 3k] rad>s. Then
vx = -1.5 rad>s
vy = 5.196 rad>s
V1 ⫽ 6 rad/s
vz = 3 rad>s
#
The angular acceleration of the head v with respect to the XYZ frame can be
obtained by setting another x¿y¿z¿ frame having an angular velocity of
Æ¿ = v1 = [5.196j + 3k] rad>s. Thus
#
#
v = A vx¿y¿z¿ B + Æ¿ * v
#
#
= (v1)x¿y¿z¿ + (v2)x¿y¿z¿ + Æ¿ * v1 + Æ¿ * v2
= 0 + 0 + 0 + (5.196j + 3k) * ( -1.5i)
= [-4.5j + 7.794k] rad>s2
#
#
Since Æ = v, vx¿y¿z¿ = v = [-4.5j + 7.794k] rad>s2. Thus,
#
vx = 0
#
vy = -4.5 rad>s2
#
vz = 7.794 rad>s2
Also, the x, y, z axes will remain as principal axes of inertia. Thus,
Ix = Iz =
Iy =
2 10
10
b A 0.3752 B = 0.06114 slug # ft2
A 0.3752 B + a
5 32.2
32.2
2 10
a
b A 0.3752 B = 0.01747 slug # ft2
5 32.2
Applying the moment equations of motion and referring to the free-body diagram
shown in Fig. a,
#
©Mx = Ixvx - A Iy - Iz B vyvz ;
Mx - 10 sin 30°(0.375) = 0 - (0.01747 - 0.06114)(5.196)(3)
Mx = 2.556 lb # ft
#
©My = Iyvy - (Iz - Ix)vzvx ;
My = 0.01747(-4.5) - 0
#
©Mz = Izvz - (Ix - Iy)vxvy ;
Mz = 0.06114(7.794) - (0.06114 - 0.01747)(-1.5)(5.196)
My = -0.07861 lb # ft
= 0.8161 lb # ft
Thus,
MA = 2Mx 2 + My 2 + Mz 2 = 22.5562 + (-0.07861)2 + 0.81612 = 2.68 lb # ft
968
Ans.
V2 ⫽ 1.5 rad/s
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z
•21–53. The blades of a wind turbine spin about the shaft S
with a constant angular speed of vs, while the frame
precesses about the vertical axis with a constant angular
speed of vp. Determine the x, y, and z components of
moment that the shaft exerts on the blades as a function of u.
Consider each blade as a slender rod of mass m and length l.
Vp
u
x
u
S
Vs
The rotating xyz frame shown in Fig. a will be attached to the blade so that it rotates
with an angular velocity of Æ = v, where v = vs + vp. Referring to Fig. b
vp = vp sin ui + vp cos uk. Thus, v = vp sin ui + vs j + vp cos uk. Then
vx = vp sin u
vy = vs vz = vp cos u
#
The angular acceleration of the blade v with respect to the XYZ frame can be
obtained by setting another x¿y¿z¿ frame having an angular velocity of
Æ¿ = vp = vp sin ui + vp cos uk. Thus,
#
#
v = A vx¿y¿z¿ B + Æ¿ * v
#
#
= (v1)x¿y¿z¿ + (v2)x¿y¿z¿ + Æ¿ * vS + Æ¿ * vP
= 0 + 0 + A vp sin ui + vp cos uk B * (vsj) + 0
= -vsvp cos ui + vsvp sin uk
#
#
Since Æ = v, vx¿y¿z¿ = v. Thus,
#
vx = -vsvp cos u
#
vy = 0
#
vz = vsvp sin u
Also, the x, y, and z axes will remain as principle axes of inertia for the blade. Thus,
Ix = Iy =
1
2
(2m)(2l)2 = ml2
12
3
Iz = 0
Applying the moment equations of motion and referring to the free-body diagram
shown in Fig. a,
#
©Mx = Ixvx - A Iy - Iz B vyvz;
Mx =
2 2
2
ml (-vsvp cos u) - a ml2 - 0b (vs)(vp cos u)
3
3
4
= - ml2 vsvp cos u
3
#
©My = Iyvy - A Iz - Ix B vzvx;
My = 0 - a0 =
#
©Mz = Izvz - (Ix - Iy)vxvy ;
Ans.
2 2
ml b(vp cos u)(vp sin u)
3
1 2 2
ml vp sin 2u
3
Ans.
Mz = 0 - 0 = 0
Ans.
969
y
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21–54. Rod CD of mass m and length L is rotating with a
constant angular rate of v1 about axle AB, while shaft EF
rotates with a constant angular rate of v2. Determine the X,
Y, and Z components of reaction at thrust bearing E and
journal bearing F at the instant shown. Neglect the mass of
the other members.
z
a
a
B
F
#
The angular acceleration of the rod v with respect to the XYZ frame can be
obtained by using another rotating x¿y¿z¿ frame having an angular velocity of
Æ¿ = v2 = v2i. Fig. a. Thus,
x
#
#
v = A vx¿y¿z¿ B + Æ¿ * v
#
#
= (v1)x¿y¿z¿ + (v2)x¿y¿z¿ + Æ¿ * v1 + Æ¿ * v2
= 0 + 0 + A v2i B * (-v1j) + 0
= -v1v2k
#
#
Since Æ = v, vx¿y¿z¿ = v. Thus,
#
#
vx = vy = 0
#
vz = -v1v2
Also, the x, y, and z axes will remain as principal axes of inertia for the rod. Thus,
Ix = 0
Iy = Iz =
1
mL2
12
Applying the equations of motion and referring to the free-body diagram shown in
Fig. a,
#
©Mx = Ixvx - A Iy - Iz B vyvz; 0 = 0
#
©My = Iyvy - A Iz - Ix B vzvx; EZ (a) - FZ (a) = 0 - 0
EZ - FZ = 0
(1)
1
1
#
mL2 (-v1 v2) - a0 mL2 b(v2)(-v1)
©Mz = Izvz - (Ix - Iy)vxvy ; FY(a) - EY(a) =
12
12
FY - EY = -
mL2v1v2
6a
(2)
Since the mass center G does not move, aG = 0. Thus,
©FX = m(aG)X;
EX = 0
©FY = m(aG)Y;
FY + EY = 0
(3)
©FZ = m(aG)Z ;
FZ + EZ - mg = 0
(4)
Ans.
Solving Eqs. (1) through (4),
FY = -
mL2 v1v2
12a
EZ = FZ =
EY =
mL2v1v2
12a
Ans.
mg
2
Ans.
970
E
C
vz = 0
vy = -v1
L
2
D
The rotating xyz frame shown in Fig. a will be attached to the rod so that it rotates
with an angular velocity of Æ = v, where v = v1 + v2 = v2i - v1j. Thus,
vx = v2
L
2
V2
A
V1
y
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z
21–55. If shaft AB is driven by the motor with an angular
velocity of v1 = 50 rad>s and angular acceleration of
#
v1 = 20 rad>s2 at the instant shown, and the 10-kg wheel
rolls without slipping, determine the frictional force and the
normal reaction on the wheel, and the moment M that must
be supplied by the motor at this instant. Assume that the
wheel is a uniform circular disk.
0.3 m
0.1 m
B
0.1 m
A
x
y
The rotating xyz frame is established to coincide with the fixed XYZ frame at the
instant considered, Fig. a. This frame will be set to have an angular velocity of
Æ = v1 = [50k] rad>s. Since the wheel is symmetric about its spinning axis, the x, y,
z axes will remain as the principal axes of inertia. Thus,
Iz = Iy =
Ix =
M
1
(10) A 0.12 B + 10 A 0.32 B = 0.925 kg # m2
4
1
(10) A 0.12 B = 0.05 kg # m2
2
Since the wheel rolls without slipping, the instantaneous axis of zero velocity is
shown in Fig. b. Thus
v2
3
=
v1
1
v2 = 3v1 = 3(50) = 150 rad>s
The angular velocity of the wheel is v = v1 + v2 = [150i + 50k] rad>s. Then,
vx = 150 rad>s
vy = 0
vz = 50 rad>s
Since the directions of v1 and v2 do not change with respect to the xyz frame,
#
#
#
#
#
vxyz = (v1)xyz + (v2)xyz where (v2)xyz = 3(v1)xyz = 3(20) = 60 rad>s2. Thus,
#
vxyz = [60i + 20k] rad>s2, so that
#
vx = 60 rad>s2
#
vy = 0
#
vz = 20 rad>s2
Applying the equations of motion and referring to the free-body diagram shown in
Fig. a,
#
©Mx = Ixvx - Iy Æ zvy + Iz Æ yvz;
F(0.1) = 0.05(60) - 0 + 0
#
©My = Iyvy - Iz Æ xvz + Ix Æ zvx;
N(0.3) - 10(9.81)(0.3) = 0 - 0 + 0.05(50)(150)
F = 30N
N = 1348.1 N = 1.35 kN
#
©Mz = Izvz - Ix Æ yvx + Iy Æ xvy;
Ans.
Ans.
M - 30(0.3) = 0.925(20) - 0 + 0
M = 27.5 N # m
Ans.
971
V1 ⫽ 50 rad/s
V1 ⫽ 20 rad/s2
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*21–56. A stone crusher consists of a large thin disk which
is pin connected to a horizontal axle. If the axle rotates at a
constant rate of 8 rad>s, determine the normal force which
the disk exerts on the stones. Assume that the disk rolls
without slipping and has a mass of 25 kg. Neglect the mass
of the axle.
Ix = Iz =
8 rad/s
0.8 m
1
(25)(0.2)2 + 25(0.8)2 = 16.25 kg # m2
4
0.2 m
1
Iy = (25)(0.2)2 = 0.5 kg # m2
2
v = -vy j + vz k, where vz = 8 rad>s
v = 0.8vz = (0.8)(8) = 6.4 m>s
vy = -
6.4
= -32 rad>s
0.2
Thus,
v = -32j + 8k
#
#
v = vxyz + Æ * v = 0 + (8k) * (-32j + 8k) = 256i
vx = 256 rad>s2
©Mx = Ix vx - (Iy - Iz) vy vz
ND (0.8) - 25(9.81)(0.8) = (16.25)(256) - (0.5 - 16.25)(-32)(8)
ND = 405 N
Ans.
972
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•21–57. The 25-lb disk is fixed to rod BCD, which has
negligible mass. Determine the torque T which must be
applied to the vertical shaft so that the shaft has an angular
acceleration of a = 6 rad>s2. The shaft is free to turn in its
bearings.
Iz =
2 ft
1 ft
B
T
1 ft
D
1 25
25
a
b (1)2 + a
b (2)2 = 3.4938 slug # ft2
2 32.2
32.2
C
A
#
Applying the third of Eq. 21–25 with Ix = Iy, vx = vy = 0, vz = 6 rad>s2
#
©Mz = Iz vz - (Ix - Iy)vx vy ;
T = 3.4938(6) = 21.0 lb # ft
Ans.
21–58. Solve Prob. 21–57, assuming rod BCD has a weight
per unit length of 2 lb>ft.
2 ft
1 ft
B
T
1 ft
D
C
A
Iz =
1(2)
1 25
25
1 2(2)
a
b(1)2 + a
b(2)2 + a
b(2)2 + a
b(2)2 = 3.9079 slug # ft2
2 32.2
32.2
3 32.2
32.2
#
Applying the third of Eq. 21–25 with Ix = Iy, vx = vy = 0, vz = 6 rad>s2
#
©Mz = Iz vz - (Ix - Iy) vx vy ;
T = 3.9079(6) = 23.4 lb # ft
973
Ans.
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z
21–59. If shaft AB rotates with a constant angular velocity
of v = 50 rad>s, determine the X, Y, Z components of
reaction at journal bearing A and thrust bearing B at the
instant shown. The thin plate has a mass of 10 kg. Neglect
the mass of shaft AB.
450 mm
150 mm
150 mm
A
450 mm
x
V ⫽ 50 rad/s
150 mm
The rotating xyz frame is set with its origin at the plate’s mass center as shown on
the free-body diagram, Fig. a. This frame will be fixed to the plate so that its angular
velocity is Æ = v and the x, y, and z axes will always be the principal axes of inertia
of the plate. Referring to Fig. b,
v = [-50 sin 60° j + 50 cos 60° k] rad>s = [-43.30j + 25k] rad>s
Thus,
vx = 0
vy = -43.30 rad>s
vz = 25 rad>s
#
Since v is always directed towards the –Y axis and has a constant magnitude, v = 0.
#
#
Also, since Æ = v, vxyz = v = 0. Thus,
#
#
#
vx = vy = vz = 0
The mass moments of inertia of the plate about the x, y, and z axes are
Ix = Iz =
1
(10) A 0.32 B = 0.075 kg # m2
12
Iy =
1
(10) A 0.32 + 0.32 B = 0.15 kg # m2
12
Applying the equations of motion,
#
©Mx = Ixvx - A Iy - Iz B vyvz;
BZ(0.45) - AZ(0.45) = 0 - (0.15 - 0.075)(-43.30)(25)
BZ - AZ = 180.42
(1)
#
©My = Iyvy - (Iz - Ix)vzvx; -AX(0.45 cos 60°) - BX(0.45 cos 60°) = 0 - 0
AX = -BX
(2)
#
©Mz = Izvz - A Ix - Iy B vxvy; -AX(0.45 sin 60°) - BX(0.45 sin 60°) = 0 - 0
AX = -BX
©FX = m(aG)X;
BX - AX = 0
©FY = m(aG)Y;
BY = 0
©FZ = m(aG)Z ;
AZ + BZ - 10(9.81) = 0
(3)
Ans.
(4)
Solving Eqs. (1) through (4),
AZ = -41.16N = -41.6 N
BZ = 139.26 N = 139 N
AX = BX = 0
Ans.
Ans.
The negative sign indicates that AZ acts in the opposite sense to that shown on the
free-body diagram.
974
150 mm
B
60⬚
y
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*21–60. A thin uniform plate having a mass of 0.4 kg spins
with a constant angular velocity V about its diagonal AB. If
the person holding the corner of the plate at B releases his
finger, the plate will fall downward on its side AC.
Determine the necessary couple moment M which if
applied to the plate would prevent this from happening.
V
y
B
C
x
G
300 mm
Using the principal axis shown,
Ix =
1
(0.4)(0.3)2 = 3 A 10 - 3 B kg # m2
12
Iy =
1
(0.4)(0.15)2 = 0.75 A 10 - 3 B kg # m2
12
Iz =
1
(0.4) C (0.3)2 + (0.15)2 D = 3.75 A 10 - 3 B kg # m2
12
u = tan - 1 a
A
75
b = 26.57°
150
vx = v sin 26.57°,
#
vx = 0
vy = v cos 26.57°,
vy = 0
vz = 0,
150 mm
#
vz = 0
©Mx = Ix vx - (Iy - Iz)vy vz
Mx = 0
©My = Iy vy - (Iz - Ix)vz vz
My = 0
©Mz = Iz vz - (Ix - Iy)vx vy
Mz = 0 - C 3 A 10 - 3 B - 0.75 A 10 - 3 B D v2 sin 26.57°cos 26.57°
Mz = -0.9 A 10 - 3 B v2 N # m = -0.9v2 mN # m
Ans.
The couple acts outward, perpendicular to the face of the plate.
975
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21–61. Show that the angular velocity of a body, in
terms # of Euler angles
as
# f , u , and# c , can be expressed
#
v# = (f sin u# sin c + u cos c)i + (f sin u cos c - u sin c)j +
(f cos u + c)k, where i, j, and k are directed along the x, y,
z axes as shown in Fig. 21–15d.
#
#
From Fig. 21–15b. due to rotation f, the x, y, z components of f are simply f along
z axis.
#
#
#
From Fig 21–15c,# due to rotation u, the x, y, z #components of f and u are f sin u in
the y direction, f cos u in the z direction, and u in the x direction.
Lastly, rotation c. Fig. 21–15d, produces the final components which yields
#
#
#
#
#
#
v = A f sin u sin c + u cos c B i + A f sin u cos c - u sin c B j + A f cos u + c B k Q.E.D.
21–62. A thin rod is initially coincident with the Z axis
when it is given three rotations defined by the Euler angles
f = 30°, u = 45°, and c = 60°. If these rotations are given in
the order stated, determine the coordinate direction angles a,
b , g of the axis of the rod with respect to the X, Y, and Z
axes. Are these directions the same for any order of the
rotations? Why?
u = (1 sin 45°) sin 30° i - (1 sin 45°) cos 30°j + 1 cos 45° k
u = 0.3536i - 0.6124j + 0.7071k
a = cos - 1 0.3536 = 69.3°
Ans.
b = cos - 1(-0.6124) = 128°
Ans.
g = cos - 1(0.7071) = 45°
Ans.
976
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21–63. The 30-lb wheel rolls without slipping. If it has a
radius of gyration kAB = 1.2 ft about its axle AB, and the
vertical drive shaft is turning at 8 rad>s, determine the
normal reaction the wheel exerts on the ground at C.
Neglect the mass of the axle.
y
V
Æ y = v = 8 rad>s
B
3(8)
= -13.33 rad>s
vz = 1.8
©Mz = Iz Æ y vz;
A
30(3) - Nw(3) = c a
30
b(1.2)2 d(8)( -13.33)
32.2
Nw = 77.7 lb
z
1.8 ft
C
Ans.
*21–64. The 30-lb wheel rolls without slipping. If it has a
radius of gyration kAB = 1.2 ft about its axle AB,
determine its angular velocity V so that the normal reaction
at C becomes 60 lb. Neglect the mass of the axle.
©Mx = Ix Æ yvz ;
30(3) - 60(3) = c
x
3 ft
y
V
30
(1.2)2 dv(-1.667v)
32.2
v = 6.34 rad>s
B
Ans.
A
z
3 ft
1.8 ft
C
977
x
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•21–65. The motor weighs 50 lb and has a radius of
gyration of 0.2 ft about the z axis. The shaft of the motor is
supported by bearings at A and B, and spins at a constant
rate of V s = 5100k6 rad>s, while the frame has an angular
velocity of V y = 52j6 rad>s. Determine the moment which
the bearing forces at A and B exert on the shaft due to this
motion.
y
0.5 ft
0.5 ft
B
Vs A
Applying
Eq. 21–30:
For the coordinate system shown u = 90° f = 90°
#
#
#
z
u = 0 f = 2 rad>s c = 100 rad>s.
#
#
#
#
©Mx = -If2 sin u cos u + Iz f sin u(f cos u + c) reduces to
# #
©Mx = Izfc;
Mx = c a
x
Vy
50
b(0.2)2 d(2)(100) = 12.4 lb # ft
32.2
Ans.
Since vx = 0
©My = 0;
My = 0
Ans.
©Mz = 0;
Mz = 0
Ans.
21–66. The car travels at a constant speed of
vC = 100 km>h around the horizontal curve having a radius
of 80 m. If each wheel has a mass of 16 kg, a radius of
gyration kG = 300 mm about its spinning axis, and a radius
of 400 mm, determine the difference between the normal
forces of the rear wheels, caused by the gyroscopic effect.
The distance between the wheels is 1.30 m.
1.30 m
vC ⫽ 100 km/h
I = 2[16(0.3)2] = 2.88 kg # m2
80 m
100(1000)
= 69.44 rad>s
vs =
3600(0.4)
vp =
100(1000)
= 0.347 rad>s
80(3600)
M = I vs vp
¢F(1.30) = 2.88(69.44)(0.347)
¢F = 53.4 N
Ans.
978
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21–67. The top has a mass of 90 g, a center of mass at G,
and a radius of gyration k = 18 mm about its axis of
symmetry. About any transverse axis acting through point O
the radius of gyration is kt = 35 mm. If the top is connected
to a ball-and-socket joint at O and the precession is
vp = 0.5 rad>s, determine the spin V s.
Vp
Vs
45⬚
vp = 0.5 rad>s
#
#
#
#
©Mx = -If2 sin u cos u + Izf sin u af cos u + c b
O
G
60 mm
0.090(9.81)(0.06) sin 45° = -0.090(0.035)2 (0.5)2 (0.7071)2
vs = c = 3.63 A 103 B rad>s
#
+ 0.090(0.018)2(0.5)(0.7071) C 0.5(0.7071) + c D
Ans.
Vs
*21–68. The top has a weight of 3 lb and can be considered
as a solid cone. If it is observed to precess about the vertical
axis at a constant rate of 5 rad>s, determine its spin.
I =
1.5 in.
3
3
1.5 2
6 2
3 4.5 2
a
b B4a
b + a b R +
a
b = 0.01419 slug # ft2
80 32.2
12
12
32.2 12
6 in.
3
3
1.5 2
a
ba
b = 0.43672 A 10 - 3 B slug # ft2
10 32.2
12
#
#
#
©Mx = -If2 sin u cos u + Iz f sin u af cos u + c b
Iz =
(3) a
5 rad/s
#
4.5
b (sin 30°) - (0.01419)(5)2 sin 30° cos 30° + 0.43672 A 10 - 3 B (5) sin 30° a5 cos 30° + c b
12
c = 652 rad>s
Ans.
979
30⬚
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•21–69. The empty aluminum beer keg has a mass of m,
center of mass at G, and radii of gyration about the x and
y axes of kx = ky = 54 r, and about the z axis of kz = 14 r,
respectively. If the keg rolls without slipping with a constant
angular velocity, determine its largest value without having
the rim A leave the floor.
y
Z
h
r
a
A
Since the beer keg rolls without slipping, the instantaneous axis of zero velocity is
indicated in Fig. a. Thus, vp = vs sin a.
#
#
Since c = vs, f = -vp = -vs sin a, and u = 90° - a are constant, the beer keg
undergoes steady precession.
5 2
25
1 2
1
I = Ix = Iy = m a rb =
mr2. Referring to the
mr2 and Iz = ma rb =
4
16
4
16
free-body diagram of the beer keg in Fig. b,
#
#
#
#
©Mx = -If2 sin u cos u + Izf sin u(f cos u + c);
FB cos a(r) - NB sin a(r) = -
+
25
mr2(-vs sin a)2 sin (90° - a) cos (90° - a)
16
1
mr2(-vs sin a) sin (90° -a)[( -vs sin a) cos (90° -a) + vs]
16
FB cos a - NB sin a =
1
mr v2s sin a cos a A 26 sin2 a - 1 B
16
(1)
Since the mass center G of the beer keg rotates about the Z axis, Fig. a, its
r cos2 a
aG = v2pR = (-vs sin a)2 ¢
acceleration
can
be
found
from
≤
sin a
= v2s r sin a cos2 a and it is directed towards the negative Y¿ axis. Fig. a. Since the
mass center does not move along the Z¿ , (aG)Z¿ = 0. Thus,
©FY¿ = m(aG)Y¿ ;
-FB = -m A v2s r sin a cos2 a B
FB = mv2s r sin a cos2 a
©FZ¿ = m(aG)Z¿ ;
NB - mg = 0
NB = mg
Substituting these results into Eq. (1),
vS =
z
G
16g
B r cos a A 16 cos2 a - 26 sin2a + 1 B
Ans.
980
B
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21–70. The 10-kg cone spins at a constant rate of
vs = 150 rad>s. Determine the constant rate vp at which it
precesses if f = 90°.
Vp
#
#
Here, u = 180° - 90° = 90°, c = vs = -150 rad>s, and f = vp are constants. Thus,
3
this is a special case of steady precession. Iz =
(10) A 0.12 B = 0.03 kg # m2,
10
#
#
Æ y = -f = -vp, and vz = c = -150 rad>s. Thus,
©Mx = Iz Æ yvz;
-10(9.81)(0.225) = 0.03 A -vp B ( -150)
A
300 mm f
Ans.
vp = -4.905 rad>s
Vs
100 mm
21–71. The 10-kg cone is spinning at a constant rate of
vs = 150 rad>s. Determine the constant rate vp at which it
precesses if f = 30°.
Vp
#
#
Since c = vs = -150 rad>s, f = vp, and u = 180° - 30° = 150° are constants, the
3
(10) A 0.22 B = 0.03 kg # m2 and I = Ix =
cone undergoes steady precession. Iz =
10
3
Iy =
(10)c4 A 0.12 B + 0.32 d + 10 A 0.2252 B = 0.555 kg # m2.
80
A
300 mm f
Thus,
#
#
#
#
©Mx = -If2 sin u cos u + Izf sin u A f cos u + c B
-10(9.81) sin 30°(0.225) = -0.555v2p sin 150° cos 150° + 0.03vp sin 150° C vp cos 150° + (-150) D
0.2273v2p - 2.25vp + 11.036 = 0
Solving,
Ans.
vp = 13.5 rad>s or 3.60 rad>s
981
Vs
100 mm
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
z
*21–72. The 1-lb top has a center of gravity at point G. If it
spins about its axis of symmetry and precesses about the
vertical axis at constant rates of vs = 60 rad>s and
vp = 10 rad>s, respectively, determine the steady state
angle u. The radius of gyration of the top about the z axis is
kz = 1 in., and about the x and y axes it is kx = ky = 4 in.
u
vp ⫽ 10 rad/s
vs ⫽ 60 rad/s
G
#
#
Since c = vs = 60 rad>s and f = vp = -10 rad>s and u are constant, the top
undergoes steady precession.
Iz = a
1
1
b a b = 215.67 A 10 - 6 B slug # ft2
32.2 12
2
= 3.4507 A 10
-3
and
B slug # ft2.
I = Ix = Iy = a
4
1
ba b
32.2 12
3 in.
O
2
x
y
Thus,
#
#
#
#
©Mx = -If2 sin u cos u + Izf sin u A f cos + c B
-1 sin u(0.25) = -3.4507 A 10 - 3 B ( -10)2 sin u cos u + 215.67 A 10 - 6 B (-10) sin u[(-10) cos u + 60]
u = 68.1°
Ans.
•21–73. At the moment of take off, the landing gear of an
airplane is retracted with a constant angular velocity of
vp = 2 rad>s, while the wheel continues to spin. If the plane
takes off with a speed of v = 320 km>h, determine the
torque at A due to the gyroscopic effect. The wheel has a
mass of 50 kg, and the radius of gyration about its spinning
axis is k = 300 mm.
When
the
plane
travels
with
a
speed
of
A
v = a320
Vp
km 1000 m
ba
b
h
1 km
a
1h
b = 88.89 m>s, its wheel spins with a constant angular velocity of
3600 s
v
88.89
Æ y = vp = 2 rad>s
vs = =
= 222.22 rad>s. Here,
u = 90°,
and
r
0.4
vz = vs = 222.22 rad>s are constants. This is a special case of steady precession.
Iz = 50 A 0.32 B = 4.5 kg # m2. Thus,
©Mx = Iz Æ yvz;
Mx = 4.5(2)(222.22) = 2000 N # m = 2kN # m
982
Ans.
Vs
B
0.4 m
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21–74. The projectile shown is subjected to torque-free
motion. The transverse and axial moments of inertia are I
and Iz, respectively. If u represents the angle between the
precessional axis Z and the axis of symmetry z, and b
is the angle between the angular velocity V and the z
axis, show that b and u are related by the equation
tan u = (I>Iz) tan b .
From Eq. 21–34 vy =
y Z
V
x
u
z
b
G
vy
Iz
HG sin u
HG cos u
=
tan u
and vz =
Hence
I
Iz
vz
I
However, vy = v sin b and vz = v cos b
vy
vz
= tan b =
tan u =
Iz
I
tan u
I
tan b
Iz
Q.E.D.
21–75. The space capsule has a mass of 3.2 Mg, and about
axes passing through the mass center G the axial and
transverse radii of gyration are kz = 0.90 m and kt = 1.85 m,
respectively. If it spins at vs = 0.8 rev>s, determine its
angular momentum. Precession occurs about the Z axis.
Vs
Z
6⬚
G
Gyroscopic
Motion:
Here,
the
spinning
angular
velocity
c = vs = 0.8(2p) = 1.6p rad>s. The moment of inertia of the satelite about the
z axis is Iz = 3200 A 0.92 B = 2592 kg # m2 and the moment of inertia of the satelite
about its transverse axis is I = 3200 A 1.852 B = 10 952 kg # m2. Applying the third of
Eq. 21–36 with u = 6°, we have
I - Iz
#
HG cos u
c =
IIz
1.6p B
z
10 952 - 2592
R HG cos 6°
10 952(2592)
HG = 17.16 A 103 B kg # m2>s = 17.2 Mg # m2>s
Ans.
983
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z
*21–76. The radius of gyration about an axis passing through
the axis of symmetry of the 2.5-Mg satellite is kz = 2.3 m,
and about any transverse axis passing through the center of
mass G, kt = 3.4 m. If the satellite has a steady-state
precession of two revolutions per hour about the Z axis,
determine the rate of spin about the z axis.
Z
10⬚
2 rev/h
Iz = 2500(2.3)2 = 13 225 kg # m2
G
I = 2500(3.4)2 = 28 900 kg # m2
Use the result of Prob. 21–74.
tan u = a
tan 10° = a
I
b tan b
Iz
28 900
btan b
13 225
b = 4.613°
From the law of sines,
sin 5.387°
sin 4.613°
=
c
2
c = 2.33 rev>h
Ans.
•21–77. The 4-kg disk is thrown with a spin vz = 6 rad>s.
If the angle u is measured as 160°, determine the precession
about the Z axis.
Z
125 mm
I =
1
4
(4)(0.125)2 = 0.015625 kg # m2
Iz =
1
2
(4)(0.125)2 = 0.03125 kg # m2
u
vz ⫽ 6 rad/s
#
Applying Eq. 21–36 with u = 160° and c = 6 rad>s
I - Iz
#
HO cos u
c =
IIz
6 =
z
0.015625 - 0.03125
HO cos 160°
0.015625(0.03125)
HG = 0.1995 kg # m2>s
f =
HG
0.1995
=
= 12.8 rad>s
I
0.015625
Ans.
Note that this is a case of retrograde precession since Iz 7 I.
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z
21–78. The projectile
precesses about the Z axis at a
#
constant rate of f = 15 rad>s when it leaves the barrel of a
#
gun. Determine its spin c and the magnitude of its angular
momentum H G. The projectile has a mass of 1.5 kg and radii
of gyration about its axis of symmetry (z axis) and about
its transverse axes (x and y axes) of kz = 65 mm and
kx = ky = 125 mm, respectively.
Z
x
30⬚
f ⫽ 15 rad/s
G
y
Since the only force that acts on the projectile is its own weight, the projectile
undergoes
torque-free
Iz = 1.5 A 0.0652 B = 6.3375 A 10 - 3 B kg # m2,
motion.
I = Ix = Iy = 1.5 A 0.1252 B = 0.0234375 kg # m2, and u = 30°. Thus,
#
HG
;
f =
I
#
HG = If = 0.0234375(15) = 0.352 kg # m2>s
Ans.
I - Iz #
#
f cos u
c =
Iz
=
0.0234375 - 6.3375 A 10 - 3 B
6.3375 A 10 - 3 B
(15) cos 30°
= 35.1 rad>s
Ans.
21–79. The satellite has a mass of 100 kg and radii of gyration
about its axis of symmetry (z axis) and its transverse axes (x or
y axis) of kz = 300 mm and kx = ky = 900 mm, respectively.
If# the satellite spins about the z axis at a constant rate of
c = 200 rad>s
# , and precesses about the Z axis, determine the
precession f and the magnitude of its angular momentum H G.
z
Z
c ⫽ 200 rad/s
15⬚
Since the weight is the only force acting on the satellite, it undergoes torque-free
motion.
Here, Iz = 100 A 0.32 B = 9 kg # m2,
u = 15°. Then,
I = Ix = Iy = 100 A 0.92 B = 81 kg # m2,
and
y
I - Iz #
#
f cos u
c =
Iz
200 = a
81 - 9 #
bf cos 15°
9
x
#
f = 25.88 rad>s
Using this result,
#
HG
f =
I
25.88 =
HG
81
HG = 2096 kg # m2>s = 2.10 Mg # m2>s
Ans.
985
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*21–80. The football has a mass of 450 g and radii of
gyration about its axis of symmetry (z axis) and its transverse
axes (x or y axis) of kz = 30 mm and kx = ky = 50 mm,
respectively. If the football has an angular momentum
of
#
#
HG = 0.02 kg # m2>s, determine its precession f and spin c.
Also, find the angle b that the angular velocity vector
makes with the z axis.
HG ⫽ 0.02 kg ⭈ m2/s
z
V
y
45⬚
x
G
Since the weight is the only force acting on the football, it undergoes torque-free
motion.
Iz = 0.45 A 0.032 B = 0.405 A 10 - 3 B kg # m2,
= 1.125 A 10 - 3 B kg # m2, and u = 45°.
I = Ix = Iy = 0.45 A 0.052 B
Thus,
#
HG
0.02
= 17.78 rad>s = 17.8 rad>s
=
f =
I
1.125 A 10 - 3 B
Ans.
1.125 A 10 - 3 B - 0.405 A 10 - 3 B
I - Iz
#
c =
HG cos u =
(0.02) cos 45°
IIz
1.125 A 10 - 3 B (0.405) A 10 - 3 B
= 22.35 rad>s = 22.3 rad>s
Ans.
Also,
vy =
HG sin u
0.02 sin 45°
=
= 12.57 rad>s
I
1.125 A 10 - 3 B
vz =
HG cos u
0.02 cos 45°
=
= 34.92 rad>s
Iz
0.405 A 10 - 3 B
Thus,
b = tan - 1 ¢
vy
vz
≤ = tan - 1 a
12.57
b = 19.8°
34.92
Ans.
986
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•21–81. The space capsule has a mass of 2 Mg, center of
mass at G, and radii of gyration about its axis of symmetry
(z axis) and its transverse axes (x or y axis) of kz = 2.75 m
and kx = ky = 5.5 m, respectively. If the capsule has the
#
angular velocity shown, determine its precession f and spin
#
c. Indicate whether the precession is regular or retrograde.
Also, draw the space cone and body cone for the motion.
y
G
30⬚
v ⫽ 150 rad/s
The only force acting on the space capsule is its own weight. Thus, it undergoes
torque-free motion. Iz = 2000 A 2.752 B = 15 125 kg # m2, I = Ix = Iy = 2000 A 5.52 B
= 60 500 kg # m2. Thus,
vy =
HG sin u
I
150 sin 30° =
HG sin u
60 500
HG sin u = 4 537 500
vz =
(1)
HG cos u
Iz
150 cos 30° =
HG cos u
15 125
HG cos u = 1 964 795.13
(2)
Solving Eqs. (1) and (2),
HG = 4.9446 A 106 B kg # m2>s
u = 66.59°
Using these results,
4.9446 A 106 B
#
HG
HG
=
=
= 81.7 rad>s
f =
I
60 500
60 500
Ans.
I - Iz
#
60 500 - 15 125
HG cos u = B
c =
R 4.9446 A 106 B cos 30°
IIz
60 500(15125)
= 212 rad>s
Ans.
Since I 7 Iz , the motion is regular precession.
Ans.
987
z
x