Mathematical Methods for Physics and Engineering : A Comprehensive Guide

VECTOR ALGEBRA

7.22 In subsection 7.6.2 we showed how the moment or torque of a force about an axis
could be represented by a vector in the direction of the axis. The magnitude of
the vector gives the size of the moment and the sign of the vector gives the sense.
Similar representations can be used for angular velocities and angular momenta.

(a) The magnitude of the angular momentum about the origin of a particle of

massmmoving with velocityvon a path that is a perpendicular distanced

from the origin is given bym|v|d. Show that ifris the position of the particle

then the vectorJ=r×mvrepresents the angular momentum.

(b) Now consider a rigid collection of particles (or a solid body) rotating about

an axis through the origin, the angular velocity of the collection being

represented byω.

(i) Show that the velocity of theith particle is

vi=ω×ri

and that the total angular momentumJis

J=

∑

i

mi[r^2 iω−(ri·ω)ri].

(ii) Show further that the component ofJalong the axis of rotation can

be written asIω,whereI, the moment of inertia of the collection

about the axis or rotation, is given by

I=

∑

i

miρ^2 i.

Interpretρigeometrically.

(iii) Prove that the total kinetic energy of the particles is^12 Iω^2.

7.23 By proceeding as indicated below, prove theparallel axis theorem,whichstates
that, for a body of massM, the moment of inertiaIabout any axis is related to
the corresponding moment of inertiaI 0 about a parallel axis that passes through
the centre of mass of the body by

I=I 0 +Ma^2 ⊥,

wherea⊥is the perpendicular distance between the two axes. Note thatI 0 can

be written as

∫

(nˆ×r)·(ˆn×r)dm,

whereris the vector position, relative to the centre of mass, of the infinitesimal

massdmandˆnis a unit vector in the direction of the axis of rotation. Write a

similar expression forIin whichris replaced byr′=r−a,whereais the vector

position of any point on the axis to whichIrefers. Use Lagrange’s identity and

the fact that

∫

rdm= 0 (by the definition of the centre of mass) to establish the
result.
7.24 Without carrying out any further integration, use the results of the previous
exercise, the worked example in subsection 6.3.4 and exercise 6.10 to prove that
the moment of inertia of a uniform rectangular lamina, of massMand sidesa
andb, about an axis perpendicular to its plane and passing through the point
(αa/ 2 ,βb/2), with− 1 ≤α, β≤ 1 ,is

M

12

[a^2 (1 + 3α^2 )+b^2 (1 + 3β^2 )].