# Solution manual engineering mechanics dynamics 12th edition chapter 21

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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 ϭ — 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,
= C 0 + 12(1)(0) D + C 0 + 12(2)(1) D + C 0 + 12(2)(2) D
= 72 kg # m2

Ans.

= C 0 + 12(0)(0) D + C 0 + 12(1)(0) D + C 0 + 12(2)(-1) D
= -24 kg # m2

Ans.

= 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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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
= C 0 + 125.6(0.2)(0.2) D - C 0 + 31.4(0.3)(0.1) D
= 4.08 kg # m2

Ans.

= C 0 + 125.6(0.2)(0.05) D - C 0 + 31.4(0.1)(0.05) D
= 1.10 kg # m2

Ans.

= 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
60Њ

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

60Њ
0.3 m

0.2 m
0.2 m

y

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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

= C 0 + 8(0.2)(0.2) D + C 0 + 4(0)(0.2) D

x

= 0.32 kg # m2

y

400 mm

Ans.

= C 0 + 8(0.2)(0) D + C 0 + 4(0.2)(0.1) D
= 0.08 kg # m2

Ans.

= 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
=
= -0.667 ft
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

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

45Њ

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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Page 936

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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

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.

= 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.

= 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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

z

21–19. 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

120Њ
120Њ
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
120Њ
C

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Page 939

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

z

*21–20. 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
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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

•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

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

z

21–22. 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):
vA = 0.667 ft>s
v is perpendicular to the rod.

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

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

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

z

21–23. 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

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):
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

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

z

*21–24. 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
(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),

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

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

z

•21–25. 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
from rest. The disk rolls without slipping.

Mϭ5Nиm
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

Ans.

944

1.5 m

B
0.2 m
y

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Page 945

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

z

21–26. 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.

Mϭ5Nиm
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¿
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

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

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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Page 947

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

z

*21–28. 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

Hz = c

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

y

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Page 948

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

z

•21–29. 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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Page 949

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

z

21–30. 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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