Question

Prove the theorem of parallel axes.

(Hint: If the centre of mass is chosen to be the origin ${\displaystyle \sum m_i{r}_i=0}$

Answer 1

Theorem of parallel axes: According to this theorem, moment of inertia of a rigid body about any axis AB is equal to moment of inertia of the body about another axis KL passing through centre of mass C of the body in a direction parallel to AB, plus the product of total mass M of the body and square of the perpendicular distance between the two parallel axes. If h is perpendicular distance between the axes AB and KL, then Suppose the rigid body is made up of n particles m1, m2, …. mn, mn at perpendicular distances r₁, r₂, r_i…. r_n. respectively from the axis KL passing through the centre of mass C of the body.

If h is the perpendicular distance of the particle of mass m{ from KL, then

The perpendicular distance of ith particular from the axis

${\displaystyle AB=(r_i+n)}$

or ${\displaystyle I_{AB}=\sum _i{m}_i{(r_i+h)}^2}$

${\displaystyle =\sum _i{m}_i{r}_i^2+\sum _i{m}_i{h}^2+2h\sum _i{m}_i{r}_i}$ ....(ii)

As the body is balanced about the centre of mass the algebraic sum of the moments of the weights of all particles about an axis passing through C must be zero

${\displaystyle \sum _i\left(m_{i}g\right)r_i=0\text{or}g\sum _i{m}_i{r}_i}$ or ${\displaystyle \sum _i{m}_i{r}_i=0}$  ...(iii)

From equation (ii) we have

${\displaystyle I_{AB}=\sum _i{m}_i{r}_i^2+(\sum m_i)h^2+0}$

or ${\displaystyle I_{AB}=I_{KL}+M{h}^2}$

Where ${\displaystyle I_{\text{KL}}=\sum _i{m}_i{r}_i^2}$  and ${\displaystyle M=\sum m_i}$