154
4 Rotational Dynamics
As the axis about which the moment of inertia is calculated is common to both
the spheres, the moment of inertia of the hollow sphere will be
I = I 1 − I 2 =
2
5
(Ma
2
− mb
2
) = (M − m)k
2
(3)
where (M − m) is the mass of the hollow sphere and k is the radius of gyration.
Now M =
4
3
π a
3
ρ and m =
4
3
π b
3
ρ
(4)
Using (4) in (3) and simplifying we get
k =
2
5
(a 5 − b 5 )
(a 3 − b 3 )
4.6 (a) Let AB represent a thin rod of length L and mass M, Fig. 4.19. Choose the
x-axis along length of the rod and y-axis perpendicular to it and passing
through its centre of mass O. Consider a differential element of length dx
at a distance x from O. The mass associated with it is M (dx/L). The
contribution to moment of inertia about the y-axis by this element of length
will be M (dx/L) x 2 . The moment of inertia of the rod about y-axis passing
through the centre of mass is
I C =
dI C =
+L/2
−L/2
M
dx
L
x
2
=
M L 2
12
(b) Moment of inertia about y-axis passing through the end of the rod (A or B)
is given by the parallel axis theorem:
I A = I B = I C + M
L
2
2
=
M L 2
3
Fig. 4.19