Rotations - Biomechanics of the Human Body - page 61

4.3 Moment of Inertia of Regularly Shaped Uniform Solids

Many times, a body of complex shape can be simulated by the composition of
uniform and geometric regular-shaped bodies. We have already adopted this to
evaluate the center of gravity of a human body (Fig. 3.9 ). Therefore, it is very useful
to know the moments of inertia of some of these bodies. The moments of inertia about
a predefined axis of rotation for some solids of total mass M can be calculated with
the mathematical technique of integrals. As such calculations are outside the scope of
this topic, here we will present the results of such calculations and discuss them.
In the figures, I y means moment of inertia about the y -axis and so on. Observe that
we are using the same symbol M for the torque and for the total mass of a solid body.

4.3.1 Radius of Gyration

For bodies of any shape, we can always find a point where the total mass M of the
body as a whole is considered to be concentrated, without changing its moment of
inertia about a given axis. This point is at a distance k from the axis of rotation and
is called the radius of gyration of the body about this axis. Note here that, in general,
the mass of a body cannot be considered as being concentrated at the center of
gravity for the purpose of calculating the moment of inertia. The introduction of the
radius of gyration is simply to facilitate the visualization since the values of k are
given in tables for solids of uniform density and commonly encountered shapes.

(a) Figure 4.3 shows a disc of radius R and mass M that rotates about the y -axis
passing through its center of gravity. In this case,

M R 2

Mk 2 I y ¼

4 ¼

from which we find that k ¼ R /2 ¼ 0.5 R . For a disc with radius R ¼ 0.20 m

0.01 kg m 2 . Note that if the figure is rotated 90 ,
the disc will be horizontal and the axis of rotation now will be also horizontal
and called x , but the situation is equivalent and, hence, the formula for the
moment of inertia is the same.
(b) Figure 4.4 shows a disc of radius R and mass M that rotates about the x -axis
passing through the center of gravity. In this case,

and M ¼ 1 kg, we obtain I y ¼

2

M R
2

Mk 2 I x ¼ ¼

R

from which we find that k 71 R . Observe that the body is the same as in

the previous example; only the axis of rotation has been changed, and the moment

¼

p

0 :