## Table of Contents

## Step-By-Step Guide For 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.

Moment Of Inertia Of An Uniform Rigid Rod is:

$$I = \frac{1}{3} M (L^{2}-3Lh + 3h^{2})$$

## Outline Of Solution

This kind of solution might be your initial encounter with this concept, so we will walk through it together with clear steps and explanations.

### Step 1: Conceptualizing the Problem

**Understanding the Rod:**Consider the rod as a linear object where its mass distribution is uniform. This means that every segment of the rod has the same mass per unit length.**Axis of Rotation:**The axis about which we are calculating the moment of inertia is perpendicular to the length of the rod and passes through a point (O), which is (h) units away from one end.

### Step 2: Dividing the Rod into Elements

**Slicing the Rod:**Imagine slicing the rod into a vast number (infinitely many) of very thin slices. This method allows us to consider the mass and position of each tiny segment to calculate the total moment of inertia.**Mass of a Slice:**Each slice can be considered to have a mass $dm$. This is a small segment of the total mass $M$ of the rod, corresponding to a tiny length $dx$ of the rod.**Position Variable:**We need to choose a variable that represents the position of each slice relative to our axis of rotation. In this case, let’s use $x$, which measures the distance from the point $O$ to the location of each slice along the rod.

### Step 3: Mathematical Formulation

**Expression for $dm$:**Given the uniform mass distribution, the mass per unit length of the rod is $\frac{M}{L}$. Thus, for a small segment $dx$, the mass $dm$ is given by $dm = \frac{M}{L}dx$.**Moment of Inertia of a Slice:**The moment of inertia $dI$ for a small slice at a distance $x$ from the axis is given by $dI = dm \cdot x^2$. This formula captures the contribution of each small slice to the total moment of inertia.

### Step 4: Integrating Over the Rod

**Setting Up the Integral:**To find the total moment of inertia $I$, integrate $dI$ over the length of the rod. The integration limits depend on how you set the origin and the direction of $x$.**Integration Process:**You will perform the integration considering the entire length of the rod, from one end to the other, accounting for the position of the axis $O$ and the distance $h$.

### Step 5: Final Calculation

**Computing I:**The final step is to integrate the expression for $dI$ across the appropriate limits to cover the entire rod. This will yield the formula for the moment of inertia of the rod about the specified axis.

**Conclusion:** By slicing the rod into infinitesimally small segments, calculating the moment of inertia for each segment, and then integrating these contributions along the length of the rod, we can derive the total moment of inertia for the rod about the chosen axis. This method combines principles from physics (moment of inertia) with calculus (integration of infinitesimal contributions) to solve the problem.

## Full Derivation Of Moment Of Inertia Of An Uniform Rigid Rod

The moment of inertia of a small segment of mass $dm$ at a distance $x$ from the axis is given by:

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

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.

Given the rod’s uniform density, the mass distribution is linear, which allows us to express $dm$, the mass of an infinitesimal segment, in terms of its length $dx$. The total mass $M$ of the rod is uniformly distributed over its total length $L$, leading to:

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

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

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

To find the total moment of inertia $I$, we integrate $dI$ across the length of the rod. Considering the axis of rotation passes through a point $O$ at a distance $h$ from one end, we define our coordinate system such that the axis of rotation is at $x = 0$, and the rod extends from $-h$ to $L-h$ along the $x$-axis.

$$I = \int \: dI$$

Substituting dI, (write the appropriate limits)

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

**Reminder:** 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)

Calculating the integral, we have:

The integration of $x^2$ across the specified limits yields:

$$I = \frac{M}{L} \left[ \frac{1}{3} x^3 \right]_{-h}^{L-h}$$

By performing the integration and simplifying, we find:

$$I = \frac{M}{L} \left[ \frac{1}{3} (L-h)^3-\frac{1}{3} (-h)^3 \right]$$

Further simplifying, we get:

$$I = \frac{M}{3L} \left[ (L^3-3L^2h + 3Lh^2-h^3) + h^3 \right]$$

$$I = \frac{M}{3L} \left[ L^3-3L^2h + 3Lh^2 \right]$$

This simplifies to the final formula for the moment of inertia of the rod about the axis through (O):

$$I = \frac{1}{3} M (L^{2}-3Lh + 3h^{2})$$

**We’re done!**

## Special Cases: Moment Of Inertia For Different Axis Position Of An Uniform Rigid Rod

### Axis At One End Of The Rod

**In the scenario where the rotation axis is located at one end of the rod (h = 0), the moment of inertia is calculated as follows:**

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

This configuration places all segments of the rod at the maximum possible distance from the axis of rotation, resulting in a relatively high moment of inertia.

### Axis Through Centre Of Mass Of The Rod

**Conversely, when the rotation axis passes through the rod’s centre of mass (with the rod being uniform, this implies $h = \frac{L}{2}$), we find that:**

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

Positioning the axis at the rod’s center of mass minimizes the moment of inertia since the distribution of mass around the axis is symmetrical, leading to a lower value.

### Observations

**Impact of Axis Position:**The moment of inertia reaches its maximum when the axis is situated at one end of the rod, as the mass elements are furthest from the axis of rotation. Conversely, it attains its minimum when the axis intersects the rod’s center of mass.**Verification:**It is crucial to verify your expressions for accuracy after derivation, ensuring they logically reflect the physical configuration and principles involved.