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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 515
y2+z2 = r2 sin2 ϕ+z2.�e moment is inertia is therefore found from the
integral
I = ∫
cyl.
(r2 sin2 ϕ + z2) × ρ dV = ρ∫
cyl.
r2 sin2 ϕ dV
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
A
+ ρ∫
cyl.
z2 dV
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
B
where the integration is over the complete cylinder. �e integrals A and
B are conveniently evaluated separately.
A =ρ∫
z=+l/2
z=−l/2
∫
r=a
r=0
∫
ϕ=2π
ϕ=0
r2 sin2 ϕ × r dz dr dϕ
=ρ∫
z=+l/2
z=−l/2
dz∫
r=a
r=0
r3 dr∫
ϕ=2π
ϕ=0
sin2 ϕ dϕ
=ρ × l × a
4
4
× π = ρl a4π/4 = mtota2/4
where the integral over ϕ is found using Integral T.2 with k = 1 and a =
2π.
B =ρ∫
z=+l/2
z=−l/2
∫
r=a
r=0
∫
ϕ=2π
ϕ=0
z2 × r dz dr dϕ
=ρ∫
z=+l/2
z=−l/2
z2 dz∫
r=a
r=0
r dr∫
ϕ=2π
ϕ=0
dϕ
=ρ × l
3
12
× a
2
2
× 2π = ρl 3a2π/12 = mtot l 2/12
�e moment of inertia about the perpendicular axis is therefore
I� = A+ B = mtota2/4 +mtot l 2/12 = mtot(a2/4 + l 2/12)
A rigid rotor with the same totalmass hasmoment of inertia I = mtotR2g,�,
hence Rg,� = (a2/4 + l 2/12)1/2 .
(c) Consider the moment of inertia about the z-axis passing through the
centre of a solid sphere. �e square of the perpendicular distance of a
point to this axis is x2 + y2, hence I = ∫sphere(x
2 + y2)ρ dV . In spherical
polar coordinates
x2 + y2 = r2 sin2 θ cos2 ϕ + r2 sin2 θ sin2 ϕ = r2 sin2 θ
�e integral is therefore evaluated as
I =ρ∫
r=a
r=0
∫
θ=π
θ=0
∫
ϕ=2π
ϕ=0
(r2 sin2 θ) × r2 sin θ dr dθ dϕ
=ρ∫
r=a
r=0
r4 dr∫
θ=π
θ=0
sin3 θ dθ ∫
ϕ=2π
ϕ=0
dϕ
=ρ × a
5
5
× 4
3
× 2π