Considering the Fig. 8.3,

Suppose AB and CD are two parallel lines. AB passes through any point P of a rigid body and CD passes through its centre of mass, G.

From trigonometry, we get the calculation as-

r^{2}_{G} + d^{2} +r^{2}/ 2r_{G} d

= r^{2}_{G} + d^{2} +r^{2}/ 2r_{G} d

2r^{2}_{G}+ d = r^{2}_{G} + d^{2} – r^{2}

r^{2} = r^{2}_{G} + d^{2} + 2r_{G}d

Now,

I = ^{2}dm

= ^{2}_{G} + d^{2} + 2r_{G}d } dm

=^{2}_{G}dm + d^{2}^{2}_{G}dm

= I_{G} + d2m + 2d dm

= I_{G} + md^{2} + 0 (Since, dm = 0)

Hence,

I = I_{G} + md^{2}

Similarly, we can calculate,

K^{2} = K^{2}_{G} + d^{2}

Where,

K_{G} is the Radius of Gyration

**Links of Previous Main Topic:-**

- Introduction concept of equilibrium of rigid body
- Friction introduction
- Introduction about distributed forces
- Area moments of inertia in rectangular and polar coordinates
- Mass moment of inertia introduction
- Radius of gyration in mass moments of inertia second moment of mass

**Links of Next Mechanical Engineering Topics:-**