A Classical Approach of Newtonian Mechanics

8 ROTATIONAL MOTION 8.6 Moment of inertia

̧ ̧ ̧+d M.

The above equation can be expanded to give

IJ = M

[(x^2 + y^2 ) − 2 d x + d^2 ] dx dy dz

̧ ̧ ̧

dx dy dz

̧ ̧ ̧^

(x^2 + y^2 ) dx dy dz

̧ ̧ ̧

x dx dy dz

= M ̧ ̧ ̧^

dx dy dz

− 2 d M ̧ ̧ ̧^

dx dy dz

2 dx^ dy^ dz^

(8.46)

dx dy dz

It follows from Eqs. (8.43) and (8.44) that

IJ = I + M d^2 , (8.47)

which proves the theorem.

Let us use the parallel axis theorem to calculate the moment of inertia, IJ, of

a thin ring about an axis which runs perpendicular to the plane of the ring, and

passes through the circumference of the ring. We know that the moment of inertia

of a ring of mass M and radius b about an axis which runs perpendicular to the

plane of the ring, and passes through the centre of the ring—which coincides

with the centre of mass of the ring—is I = M b^2. Our new axis is parallel to this
original axis, but shifted sideways by the perpendicular distance b. Hence, the

parallel axis theorem tells us that

IJ = I + M b^2 = 2 M b^2. (8.48)

See Fig. 78.

As an illustration of the direct application of formula (8.34), let us calculate

the moment of inertia of a thin circular disk, of mass M and radius b, about an

axis which passes through the centre of the disk, and runs perpendicular to the

plane of the disk. Let us choose our coordinate system such that the disk lies

in the x-y plane with its centre at the origin. The axis of rotation is, therefore,

coincident with the z-axis. Hence, formula (8.34) reduces to

I = M

(x^2 + y^2 ) dx dy

̧ ̧

dx dy

,

(8.49)

̧ ̧ ̧

̧ ̧