4.3 Solutions
157
I =
dI = 2π R 4 σ
π
0
sin
3
θ d θ
= 2π R 4 σ
4
3
But σ =
M
4π R 2
∴ I =
2
3
MR 2
4.11 By prob. (4.5), the radius of gyration of a hollow sphere of external radius a
and internal radius b is
k =
2
5
(a 5 − b 5 )
(a 3 − b 3 )
(1)
The derivation of (1) is based on the assumed value of moment of inertia for
a solid sphere about its diameter
I =
2
5
MR 2
. Squaring (1) and multiplying
by M, the mass of the hollow cylinder is
I = Mk
2
=
2
5
M
(a 5 − b 5 )
(a 3 − b 3 )
(2)
Let a = b +
(3)
where is a small quantity. Then (2) becomes
I =
2
5
M
(b + ) 5 − b 5
(b + ) 3 − b 3
=
2
5
M
b 5 + 5b 4 + · · · − b 5
b 3 + 3b 3 + · · · − b 3
where we have neglected higher order terms in . Thus
I =
2
5
M
5b 4
3b 2
=
2
3
Mb
2
=
2
3
MR
2
where b = a = R is the radius of the hollow sphere.
4.3.2 Rotational Motion
4.12 Potential energy at height h is mgh and kinetic energy is zero. At the bottom
the potential energy is assumed to be zero. The kinetic energy (K ) consists of
translational energy
1
2
mv 2
+ rotational energy
1
2
I ω 2
: