9–3.
Locate the distance x to the center of gravity of the
homogeneous rod bent into the parabolic shape. If the rod
has a weight per unit length of 0.5 lb>ft, determine the
reactions at the fixed support O.
y
0.5 ft
SOLUTION
O
dL = 2dx 2 + dy 2
dy = x dx
1
'
x =
L
x dL
=
L
dL
L0
x2dx2 + x2 dx2
1
L0
2dx2 + x2 dx2
Let x = tan u
dx = sec2 u du
p
4
x =
L0
tan u21 + tan2 u sec2 ud u
p
4
L0
21 + tan2 u sec 2 udu
p
C sec3 u D 04
3
=
p
C sec u2tan u + 12 {ln | sec u + tan u|} D 04
'
x = 0.531 ft
Ans.
Also,
dL = 2dx2 + dy 2
L =
LB
1 + a
dy 2
b dx
dx
1
=
L0
21 + x 2 dx
= 1.148 ft
L
'
xdL =
1
L0
x 21 + x 2 dx
= 0.6095
x =
0.6095
= 0.531 ft
1.148
Ans.
+ ©F = 0;
:
x
Ox = 0
+ c ©Fy = 0;
Oy - 0.5(1.148) = 0
Oy = 0.574 lb
a + ©MO = 0;
Ans.
Ans.
MO - 0.5(1.148)(0.531) = 0
MO = 0.305 lb # ft
Ans.
y
0.5x2
x
1 ft
9–4.
Locate the distance y to the center of gravity of the
homogeneous rod bent into the parabolic shape.
y
0.5 ft
SOLUTION
O
dL = 2dx2 + dy2
L =
LA
1 + a
dy 2
b dx
dx
1
=
L0
21 + x 2 dx
= 1.148 ft
L
'
ydL =
1
L0
0.5x2 21 + x2 dx
= 0.2101 ft
y =
0.2101
= 0.183 ft
1.148
Ans.
y
0.5x2
x
1 ft
9–7.
Determine the area and the centroid x of the parabolic area.
y
h
SOLUTION
y
Differential Element:The area element parallel to the x axis shown shaded in Fig. a
will be considered. The area of the element is
dA = x dy =
a
h1>2
y 1>2 dy
x
a
'
'
Centroid: The centroid of the element is located at x =
y 1>2 and y = y.
=
2
2h1>2
Area: Integrating,
h
A =
LA
LA
=
dA
LA
h
'
xdA
x =
dA =
L0
¢
a
1>2
2h
a
L0 h
1>2
y1>2 ≤ ¢
2
ah
3
y 1>2 dy =
a
1>2
h
h
2a
A y3>2 B 2 =
1>2
3h
y 1>2 dy ≤
0
2
ah
3
Ans.
a2 y2 h
a
¢ ≤`
y dy
2h 2 0
3
L0 2h
=
=
= a Ans.
2
2
8
ah
ah
3
3
h 2
h x2
––
a2
x
a
9–11.
y
Locate the centroid x of the area.
h
y ⫽ —2 x2
b
SOLUTION
h
dA = y dx
'
x = x
x
b
x =
LA
h 3
x dx
2
b
L0
'
x dA
=
LA
dA
b
h 2
x dx
2
b
L0
B
b
h 4 b
x R
4b2
0
=
=
b
h
B 2 x3 R
3b
0
3
b
4
Ans.
9–12.
y
Locate the centroid y of the shaded area.
h
y ⫽ —2 x2
b
SOLUTION
dA = y dx
y
'
y =
2
x
b
y =
LA
=
dA
2
h 4
x dx
4
2b
L0
'
y dA
LA
h
b
h 2
x dx
2
L0 b
B
2
b
b
h
x5 R
10b4
0
=
=
b
h
B 2 x3 R
3b
0
3
h
10
Ans.
*9–28.
y
Locate the centroid x of the area.
2
––
SOLUTION
y x3
Differential Element: The differential element parallel to the y axis shown shaded
in Fig. a will be considered. The area of the element is
1m
3
––
dA = (y1 - y2) dx = (x
2>3
- x
3>2
y x2
) dx
x
'
Centroid: The centroid of the element is located at x = x.
1m
1
3
2
1
(x2>3 - x3>2) dx = ¢ x5>3 - x5>2 ≤ ` = m2 Ans.
5
5
5
0
1m
A =
dA =
LA
L0
We have
1m
'
x dA
x =
LA
L0
=
dA
LA
3
8
¢ x8>3 =
1m
x(x2>3 - x3>2) dx
1
5
=
1
5
2 7>2 1m
x ≤`
7
0
=
L0
25
m
56
(x5>3 - x5>2) dx
1
5
Ans.
9–29.
y
Locate the centroid y of the area.
2
––
SOLUTION
y x3
Differential Element: The differential element parallel to the y axis shown shaded
in Fig. a will be considered. The area of the element is
1m
3
––
dA = (y1 - y2) dx = (x
2>3
- x
3>2
y x2
) dx
x
1
1
'
Centroid: The centroid of the element is located at y = (y1 + y2) = (x2>3 + x3>2).
2
2
1
3
2
1
(x2>3 - x3>2) dx = ¢ x5>3 - x5>2 ≤ ` = m2
5
5
5
0
1m
A =
dA =
LA
L0
Ans.
We have
1m
'
y dA
LA
y =
L0
=
dA
LA
1 2>3
(x + x3>2)(x2>3 - x3>2) dx
2
L0
=
1
5
1m
1 4>3
(x - x3) dx
2
1
5
x4 1 m
1 3 7>3
B x - R`
2 7
4 0
=
=
1
5
25
m
56
Ans.
1m
9–39.
Locate the centroid y of the paraboloid.
z
z2 = 4y
4m
y
SOLUTION
Volume and Moment Arm: The volume of the thin disk differential element is
'
dV = pz2dy = p14y2dy and its centroid y = y.
Centroid: Applying Eq. 9–3 and performing the integration, we have
y =
LV
4m
'
ydV
LV
=
dV
L0
y3p14y2dy4
4m
p14y2dy
L0
4p
=
y3
3
4m
2
4m
y
4p
2
0
0
= 2.67 m
Ans.
4m
9–40.
Locate the center of gravity of the volume. The material is
homogeneous.
z
2m
SOLUTION
y 2 = 2z
Volume and Moment Arm: The volume of the thin disk differential element is
'
dV = py2dz = p12z2dz = 2pzdz and its centroid z = z.
Centroid: Due to symmetry about z axis
' '
x = y = 0
Ans.
Applying Eq. 9–3 and performing the integration, we have
'
Lv
z =
Lv
2p
=
2m
'
z dV
=
dV
2m
L0
z3
3
2m
2
2m
z
2p
2
z12pzdz2
L0
0
0
=
2pzdz
4
m
3
Ans.
2m
y
9–46.
z
The hemisphere of radius r is made from a stack of very thin
plates such that the density varies with height r = kz,
where k is a constant. Determine its mass and the distance
to the center of mass G.
G
r
SOLUTION
y
Mass and Moment Arm: The density of the material is r = kz. The mass of the thin
disk differential element is dm = rdV = rpy2dz = kz3p(r2 - z2) dz4 and its
'
centroid z = z. Evaluating the integrals, we have
r
m =
Lm
dm =
L0
kz3p(r2 - z2) dz4
= pk ¢
z4 r
r2z2
pkr4
- ≤` =
2
4 0
4
Ans.
r
Lm
'
z dm =
L0
z5kz3p(r2 - z2) dz46
= pk ¢
r2z3
z5 r
2pkr5
- ≤` =
3
5 0
15
Centroid: Applying Eq. 9–3, we have
z =
_
z
Lm
'
z dm
=
Lm
dm
2pkr5>15
pkr4>4
=
8
r
15
Ans.
x