Solutions. - UC Davis Mathematics

LECTURE 5 SOLUTIONS TO EXERCISES
15.6.6 Centroid of region between x-axis and y = sin x, 0 ≤ x ≤ π.
It’s symmetric
R π arond x = π/2, so only need the y-coordinate. From previous computations, know
that M = 0 sin x = 2. The moment about the x-axis is then
π
Z
Z π Z sin x
1
1
1 π 2
1
sin xdx =
ydydx =
x − sin 2x = (π − 0 − 0 + 0) = π/4
Mx =
2 0
4
2
4
0
0
0
Therefore the center of mass has coordinates (¯
x, y¯ = (π/2, My /M ) = (π/2, π/8).
y-axis with δ = 1, bounded by curve y = (sin2 x)/x2 and
15.6.8 Moment of inertia w.r.t.
π ≤ x ≤ 2π.
Z
2π
sin2 x/x2
Z
2
Z
2π
x dydx =
π
0
sin2 xdx =
π
1
1
1
(x − sin(2x))|2π
π = (2π − 0 − π + 0) = π/2.
2
2
2
15.6.14 Center of mass and moment of intertia about x-axis of thin plate bdd by curves x = y 2
and x = 2y − y 2 if density is δ(x, y) = y + 1.
The curves intersect at y = 0, 1, so to find the center of mass:
Z 1 Z 2y−y2
1
M=
y + 1dxdy = ... =
2
2
0
y
x
¯=
y¯ =
1
M
Z
1
M
Z
1
Z
2y−y 2
x(y + 1)dxdy = ... = (2)(
4
)
15
y(y + 1)dxdy = ... = (2)(
4
)
15
y2
0
1
2y−y 2
Z
y2
0
And to find the moment of inertia about the x axis,
Z 1 Z 2y−y2
1
Ix =
y 2 (y + 1)dxdy = ... = .
6
2
0
y
15.6.16 Find c.o.m. and Iy of thin plate bdd by y = 1 and y = x2 if δ(x, y) = y + 1.
1
Z
Z
1
M=
(y + 1)dydx = ... =
x2
0
x
¯=
1
M
Z
1
Z
1
x(y + 1)dydx = (
0
x2
16
15
15 5
)( )
16 12
1Z 1
1
15 24
y¯ =
y(y + 1)dydx = ( )( )
M 0 x2
16 35
Z 1Z 1
8
Iy =
x2 (y + 1)dydx =
.
35
2
0
x
Z
1
2
LECTURE 5 SOLUTIONS TO EXERCISES
15.6.22 Wedge thing.
The top of the wedge has equation z = 4−2y
(it’s independent of x, and then use point-slope
3
form to figure out the equation). So
Z 3 Z 4 Z (4−2y)/3
(y 2 + z 2 )dzdydx = ... = 208.
Ix =
−4/3
−2
−3
3
Z
4
Z
Z
(4−2y)/3
(x2 + z 2 )dzdydx = ... = 280.
Iy =
3
Z
−4/3
−2
−3
4
Z
Z
(4−2y)/3
(x2 + y 2 )dzdydx = ... = 360.
Iz =
−4/3
−2
−3
15.6.24 Solid, const density, bdd below by z = 0, on sides by x2 + 4y 2 = 4, above by z = 2 − x.
(a)
Z 2 Z √4−x2 /2 Z 2−x
M=
dzdydx = ... = 4π
√
− 4−x2 /2
−2
0
(b)
2
Z
√
4−x2 /2
Z
Myz =
xdzdydx = ... = −2π
√
− 4−x2 /2
−2
2−x
Z
0
(c)
Z
2
√
Z
Mxz =
4−x2 /2
Z
−2
2−x
ydzdydx = ... = 0
√
− 4−x2 /2
0
So x
¯ = −1/2 and y¯ = 0.
(d)
2
Z
√
Z
Mxy
−2
4−x2 /2
2−x
Z
zdzdydx = ... = 5π
√
− 4−x2 /2
0
so z¯ = 54 .
15.6.30 Solid, first octant, bdd by 4 − x2 and x = y 2 , δ(x, y) = kxy.
2
Z
√
Z
x
Z
4−x2
M=
kxydzdydx = ... =
0
Z
0
0
√ Z
2Z
x
4−x2
32k
15
5
8k
=⇒ x
¯=
3
4
0
0
0
√
√
Z 2 Z √x Z 4−x2
256 2k
40 2
Mxz =
kxy 2 dzdydx = ... =
=⇒ y¯ =
231
77
0
0
0
Z 2 Z √x Z 4−x2
256k
8
Mxy =
kxyzdzdydx = ... =
=⇒ z¯ = .
105
7
0
0
0
15.6.32 Same wedge as in 22 with different a, b, c, and δ(x, y, z) = x + 1. ALL of the integrals are
R 1 R 4 R (2−y)/2
dzdydx. Here are the integrands and the answer you get:
−1 −2 −1
Myz =
M : δ, 18
Myz : xδ, 6
Mxz : yδ, 0
Mxy : zδ, 0
Ix : δ(y 2 + z 2 ), 45
Iy : δ(x2 + z 2 ), 15
kx2 ydzdydx = ... =
LECTURE 5 SOLUTIONS TO EXERCISES
3
Iz : δ(x2 + y 2 ), 42
15.6.35 Proof of parallel axis theorem.
(a) Since, say, the yz plane goes through the center of mass, we have that x
¯ = 0, so Myz = 0 as
well.
(b)
ZZZ
ZZZ
ZZZ
2
2
ˆ
ˆ
ˆ
IL =
|v − hi| dm =
|(x − h)i + y j| dm =
(x2 − 2xh + h2 + y 2 )dm =
D
D
D
ZZZ
ZZZ
ZZZ
=
(x2 + y 2 )dm − 2h
xdm + h2
dm = Ix − 0 + h2 m.
D
D
D
15.6.38 That wedge thing again, given that Ix = 208, find the moment of inertia about the line
y = 4, z = −4/3 (the narrow end of the wedge)
Z
3
Z
4
Z
(4−2y)/3
dzdydx = ... = 72
M=
−3
−2
−4/3
r
x
¯ = y¯ = z¯ = 0(f rom22) =⇒ Ix = Ic.m. + 72(0) = Ic.m. =⇒ IL = Ic.m. + 72
16
16 +
9
!2
= 1488.

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