© 2022 by R.C. Hibbeler. Published by Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws
as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 the
its
origin.
SOLUTION
Ixx + Iyy + Izz = (y2 + z2)dm + (x2 + z2)dm + (x2 + y2)dm
Lm Lm Lm
= 2 (x2 + y2 + z2)dm
Lm
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 independent
indepenent ofofthe
theorientation
orientationofofthe
thex,x,y,y,z zaxis.
axis. Q.E.D.
1077
,© 2022 by R.C. Hibbeler. Published by Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws
as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21–2.
Determine the moment of inertia of the cone with respect to y –y y¿
a vertical y axis passing 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.
a
x
SOLUTION
rpa2 h
The mass of the differential element is dm = rdV = r(py2) dx = x2dx.
h2
1
dIy = dmy2 + dmx2
4
1 rpa2 2 r pa2
B 2 x dx R a x b + ¢ 2 x2 ≤ x2 dx
a 2
=
4 h h h
r pa2
= (4h2 + a2) x4 dx
4h4
h
rpa2 r pa2h
Iy = dIy = 4
(4h2 + a2) x4dx = (4h2 + a2)
L 4h L0 20
However,
h
r pa2 r pa2h
m = dm = 2
x2 dx =
Lm h L0 3
Hence,
3m
Iy = (4h2 + a2)
20
Using the parallel axis theorem:
Iy = Iy + md2
3m 3h 2
(4h2 + a2) = Iy + m a b
20 4
3m 2
Iy = (h + 4a2) Ans.
80
Iy' = Iy + md2
3m 2 h 2
= (h + 4a2) + m a b
80 4
m
= (2h2 + 3a2) Ans.
20
Ans:
3m 2
Iy = ( h + 4a2 )
80
m
Iy′ = ( 2h2 + 3a2 )
20
1078
,© 2022 by R.C. Hibbeler. Published by Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws
as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21–3.
product of
Determine the product of inertia
inertia IIxy for the homogeneous
homogeneous z
material is
tetrahedron. The density of the material is rr.. Express
Express the
the
result in terms of the total mass m of of the
the solid.
soild. Suggestion:
Suggestion:
thickness dz
Use a triangular element of thickness and then
dz and then express
express
xy in
dIxy
dI interms
termsofofthe size
the andand
size mass of the
mass of element
the element.usingThe
the
result ofof
product Prob. 21–6.
inertia of the triangular prism shown with respect a
ra4h
to the xy and yz planes is Ixy = . y
24
a
SOLUTION
a
x
1
dm = r dV = r c (a - z)(a - z) ddz = (a - z)2 dz
r
2 2
a
r ra3
m = (a2 - 2az + z2)dz =
2 L0 6
The product of inertia of a triangular prism with respect to the xz and yz planes is
ra 4h rdz
Ixy = . For the element above, dIxy = (a - z)4. Hence,
24 24
a
r
Ixy = (a4 - 4a3z + 6z 2a 2 - 4az 3 + z 4)dz
24 L0
ra 5
Ixy =
120
or,
ma 2
Ixy = Ans.
20
Ans:
ma2
Ixy = 20
1079
, © 2022 by R.C. Hibbeler. Published by Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws
as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*21–4.
Determine the moments of inertia for the homogeneous r z¿
cylinder of mass m about the x¿ , y¿ , z¿ axes. x¿
y r
z
SOLUTION
y¿
x
Due to symmetry
Ixy = Iyz = Izx = 0
1 r 2 7mr 2 1 2
Iy = Ix = m(3r 2 + r 2) + m a b = Iz = mr
12 2 12 2
For x¿ ,
1 1
ux = cos 135° = - , uy = cos 90° = 0, uz = cos 135° = -
22 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 22 2 22
13
= mr 2 Ans.
24
For y¿ ,
7mr 2
Iy¿ = Iy = Ans.
12
For z¿ ,
1 1
ux = cos 135° = - , uy = cos 90° = 0, uz = cos 45° =
22 22
Iz¿ = 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 22 2 22
13
= mr 2 Ans.
24
Ans:
13 2
Ix′ = 24 mr
2
7mr
Iy′ = 12
13 2
Iz′ = 24 mr
1080
as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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 the
its
origin.
SOLUTION
Ixx + Iyy + Izz = (y2 + z2)dm + (x2 + z2)dm + (x2 + y2)dm
Lm Lm Lm
= 2 (x2 + y2 + z2)dm
Lm
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 independent
indepenent ofofthe
theorientation
orientationofofthe
thex,x,y,y,z zaxis.
axis. Q.E.D.
1077
,© 2022 by R.C. Hibbeler. Published by Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws
as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21–2.
Determine the moment of inertia of the cone with respect to y –y y¿
a vertical y axis passing 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.
a
x
SOLUTION
rpa2 h
The mass of the differential element is dm = rdV = r(py2) dx = x2dx.
h2
1
dIy = dmy2 + dmx2
4
1 rpa2 2 r pa2
B 2 x dx R a x b + ¢ 2 x2 ≤ x2 dx
a 2
=
4 h h h
r pa2
= (4h2 + a2) x4 dx
4h4
h
rpa2 r pa2h
Iy = dIy = 4
(4h2 + a2) x4dx = (4h2 + a2)
L 4h L0 20
However,
h
r pa2 r pa2h
m = dm = 2
x2 dx =
Lm h L0 3
Hence,
3m
Iy = (4h2 + a2)
20
Using the parallel axis theorem:
Iy = Iy + md2
3m 3h 2
(4h2 + a2) = Iy + m a b
20 4
3m 2
Iy = (h + 4a2) Ans.
80
Iy' = Iy + md2
3m 2 h 2
= (h + 4a2) + m a b
80 4
m
= (2h2 + 3a2) Ans.
20
Ans:
3m 2
Iy = ( h + 4a2 )
80
m
Iy′ = ( 2h2 + 3a2 )
20
1078
,© 2022 by R.C. Hibbeler. Published by Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws
as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21–3.
product of
Determine the product of inertia
inertia IIxy for the homogeneous
homogeneous z
material is
tetrahedron. The density of the material is rr.. Express
Express the
the
result in terms of the total mass m of of the
the solid.
soild. Suggestion:
Suggestion:
thickness dz
Use a triangular element of thickness and then
dz and then express
express
xy in
dIxy
dI interms
termsofofthe size
the andand
size mass of the
mass of element
the element.usingThe
the
result ofof
product Prob. 21–6.
inertia of the triangular prism shown with respect a
ra4h
to the xy and yz planes is Ixy = . y
24
a
SOLUTION
a
x
1
dm = r dV = r c (a - z)(a - z) ddz = (a - z)2 dz
r
2 2
a
r ra3
m = (a2 - 2az + z2)dz =
2 L0 6
The product of inertia of a triangular prism with respect to the xz and yz planes is
ra 4h rdz
Ixy = . For the element above, dIxy = (a - z)4. Hence,
24 24
a
r
Ixy = (a4 - 4a3z + 6z 2a 2 - 4az 3 + z 4)dz
24 L0
ra 5
Ixy =
120
or,
ma 2
Ixy = Ans.
20
Ans:
ma2
Ixy = 20
1079
, © 2022 by R.C. Hibbeler. Published by Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws
as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*21–4.
Determine the moments of inertia for the homogeneous r z¿
cylinder of mass m about the x¿ , y¿ , z¿ axes. x¿
y r
z
SOLUTION
y¿
x
Due to symmetry
Ixy = Iyz = Izx = 0
1 r 2 7mr 2 1 2
Iy = Ix = m(3r 2 + r 2) + m a b = Iz = mr
12 2 12 2
For x¿ ,
1 1
ux = cos 135° = - , uy = cos 90° = 0, uz = cos 135° = -
22 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 22 2 22
13
= mr 2 Ans.
24
For y¿ ,
7mr 2
Iy¿ = Iy = Ans.
12
For z¿ ,
1 1
ux = cos 135° = - , uy = cos 90° = 0, uz = cos 45° =
22 22
Iz¿ = 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 22 2 22
13
= mr 2 Ans.
24
Ans:
13 2
Ix′ = 24 mr
2
7mr
Iy′ = 12
13 2
Iz′ = 24 mr
1080
