Different formulae for moment of inertia

[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!

 

[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.

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

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.

 

[Zynoakib]

Thanks! Nice and clear.

 

[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