# Different formulae for moment of inertia

1. Aug 22, 2015

### Zynoakib

I know the formula for moment of inertia is but there are I = MR^2 but there are also formulae for different objects as shown in the picture.

So, how and when do you use I = MR^2 ? Just in case of (a)?

Thanks!

2. Aug 22, 2015

### brainpushups

The moment of inertia is a sum of all $mr^2$ for the particles in the system (or an integral for a continuous system of masses). The formulas you posted can all be derived by integrating.

For an object rotating about its center the uniform ring is the only one for which the moment of inertia is the $MR^2$ This is because each particle has the same distance from the axis of rotation so the sum essentially amounts to summing over all of the masses.

3. Aug 22, 2015

### Zynoakib

Thanks! Nice and clear.

4. Aug 22, 2015

### Noctisdark

I just want to get a little deep into this and present what brainpushups said mathematically, any object can be represented at a set of points, each of these point have a moment of inertia $\delta I = \delta m\cdot r^2$ (this is true because they are points each of mass delta m), and we know that $I_{net} = \sum \limits_i \delta I$, if we want to sum infinetely small quantities, our best option would be an integral so $I_{net} = \int \delta I = \int r^2 dm$ .
For example if you want to determine to moment of intertia of a solid cylinder about its central axis, you start by defining $\rho = \frac{m}{V} = \frac {m}{\pi R^2L}$ so that $dm = \rho dV = \rho r\cdot dr\cdot d\theta\cdot dz$ and then set the boundaries, for example $0 \leftarrow r \rightarrow R, 0 \leftarrow \theta \rightarrow 2\pi$ and $0 \leftarrow z \rightarrow L$ and finally integrate $I_{net} = \rho \int_0^L \int_0^{2\pi} \int_0^R r^3 \cdot dr\cdot d\theta\cdot dz = 2\pi L\rho \int_0^R r^3 \cdot dr = 2\pi L\cdot \frac {m}{\pi R^2L} \cdot \frac{R^4}{4} = \frac{1}{2} mR^2$ Cheers :D