### Derivation of moment of inertia of an uniform rigid rod

Calculate/derive the moment of inertia of an uniform rigid rod of length L and mass M about an axis perpendicular to the rod and passing through O, at an arbitrary distance h from one end.

**GUIDE: (I’ll be guiding you through this as it may be the first time that you are seeing this)**

- Imagining the rod is cut into infinitesimally many pieces of infinitesimally thin slices. (“slicing” the object)
- Each of these slices have a mass of dm (remember your calculus lessons?) and length of dx.
- Choose a variable to sum. E.g. In this problem, we are summing from left of the axis to right of axis. The variable is x. In other problem, the variable can be theta or r or …

Now, we show our formula for the calculation for moment of inertia first:

$$dI = dm \: x^{2}$$

Hey, there is a dm in the equation! Recall that we’re using x to sum. Hence, we have to force a dx into the equation for moment of inertia. Now, lets find an expression for dm. Since the rod is uniform, the mass varies linearly with distance.

$$dm = \frac{M}{L} dx$$

Using the equation for dm, we substitute it into the first equation. Hence, we have:

$$dI = \frac{M}{L} x^{2} dx$$

Now,

$$I = \int \: dI$$

Substituting dI, (write the appropriate limits)

$$I = \frac{M}{L} \int\limits_{-h}^{L – h} \: x^{2} \: dx$$

Note: The lower limit is -h because the left side of the rod is -h units away from the axis of rotation. (We take right as positive)

Solving the integration, we have:

**We’re done!**

**Now, for the special cases,**

**When the rotation axis is at one end of the rod (h = 0), we have:**

$$I = \frac{1}{3} ML^{2}$$

**However, if the rotation axis is through the centre of mass of the rod. (remember rod is uniform, hence $h = \frac{L}{2}$)**

**We have:**

$$I = \frac{1}{12} M L^{2}$$

**Note: **The moment of inertia is expected to be highest when the axis is at one end since the mass are now furthest away from the axis of rotation. Lowest is when axis is at the center.

– Always check your expression after deriving them.

**Derivation Of Moment Of Inertia Of Common Shapes:**

Gamerboy scribbled

Helped me a ton! Thank you

Archa scribbled

Thank u so much.

Amateur Academic scribbled

Thanks! This was the only place I could find an easy-to-follow derivation of the moment of inertia for a rod about an arbitrary axis. Helped a ton!

John Solomon Nyanga scribbled

its good

nic scribbled

good work

Adam scribbled

Thanks alots

Bob Boyd scribbled

Good stuff!