#### Question

Prove the theorem of parallel axes.

(Hint: If the centre of mass is chosen to be the origin `summ_ir_i = 0`

#### Solution

**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 r_{1}, r_{2}, 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

`AB = (r_i + n)`

or `I_(AB) = sum_(i)m_i(r_i + h)^2`

`= sum_(i) m_ir_i^2 + sum_(i) m_ih^2 + 2h sum_(i) m_ir_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

`sum_(i)(m_ig)r_i = 0 or g sum_(i) m_ir_i` or `sum_(i) m_ir_i = 0` ...(iii)

From equation (ii) we have

`I_(AB) = sum_(i) m_ir_i^2 + (summ_i)h^2 + 0`

or `I_(AB) = I_(KL) + Mh^2`

Where `I_"KL" = sum_(i) m_ir_i^2` and `M = sum m_i`