Find out the increase in moment of inertia $I$ of a uniform rod (coefficient of linear expansion $\alpha$) about its perpendicular bisector when its temperature is slightly increased by $\Delta T$.

> [!TLDR] > When a uniform rod expands due to a temperature increase, its length increases, which in turn increases its moment of inertia about its perpendicular bisector. The increase in moment of inertia is $\boxed{\frac{1}{6}ML^2 \alpha \Delta T}$. The moment of inertia of an object depends on its mass distribution and its dimensions. When the temperature of a material changes, its dimensions typically change due to thermal expansion or contraction. For a rod, this means its length will change. Since the moment of inertia of a rod about its perpendicular bisector is proportional to the square of its length, any change in length will directly affect its moment of inertia. Our approach will be to first write down the initial moment of inertia, then determine the new length after thermal expansion, and finally calculate the new moment of inertia. The difference between the new and initial moments of inertia will give us the increase. 1. **Initial Moment of Inertia** Let the original length of the uniform rod be $L$ and its mass be $M$. The moment of inertia of a uniform rod about an axis passing through its center and perpendicular to its length (its perpendicular bisector) is given by: $$I = \frac{1}{12}ML^2$$ 2. **Change in Length due to Thermal Expansion** When the temperature of the rod is increased by $\Delta T$, its length changes due to linear thermal expansion. The new length, $L'$, can be expressed in terms of the original length $L$ and the coefficient of linear expansion $\alpha$ as: $$L' = L(1 + \alpha \Delta T)$$ Here, $\alpha \Delta T$ represents the fractional change in length. Since the temperature is "slightly increased", $\alpha \Delta T$ will be a very small quantity. 3. **New Moment of Inertia** The mass $M$ of the rod remains constant. Only its length changes. We can find the new moment of inertia, $I'$, by substituting the new length $L'$ into the moment of inertia formula: $$I' = \frac{1}{12}M(L')^2$$ Su

Find out the increase in moment of inertia I of a uniform rod (coefficient of linear expansion α) about its pe

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Concept understanding — Thermal Expansion Coefficient

Thermal Expansion Coefficient: From Intuition to Precision

The Intuition: What Happens When Things Get Hot?

Think about a metal railway track on a summer day. The track is laid in sections with small gaps between them. On a hot afternoon, those gaps get smaller — sometimes the track even buckles. Why? Because the metal expands when heated.

Or consider a mercury thermometer. The liquid mercury sits in a bulb at the bottom. When your body warms the bulb, the mercury expands and rises up the narrow tube. The hotter you are, the higher it climbs.

This is thermal expansion: most materials get bigger when heated and smaller when cooled. The atoms inside vibrate more vigorously as temperature rises, pushing their neighbours slightly farther apart. The entire object grows in every direction.

But different materials expand by different amounts. A steel rod and an aluminium rod of the same length, heated by the same amount, will not end up the same length. Aluminium expands more. So we need a number that tells us how much a given material expands per degree of temperature change. That number is the thermal expansion coefficient.

The Precise Statement: Defining the Coefficient

There are actually three coefficients, depending on whether we care about length, area, or volume. For a first meeting, we focus on the most common one: the linear thermal expansion coefficient, denoted by the Greek letter $\alpha$ (alpha).

\alpha = \frac{1}{L_0} \cdot \frac{\Delta L}{\Delta T}

Where:

$L_0$ is the original length of the object (at some starting temperature)
$\Delta L$ is the change in length (final length minus original length)
$\Delta T$ is the change in temperature (final temperature minus initial temperature)

What this formula says in plain English: The coefficient $\alpha$ is the fractional change in length per degree of temperature change. If $\alpha = 2.5 \times 10^{-5} \, \text{per} \, ^\circ\text{C}$ , it means that for every $1^\circ\text{C}$ rise in temperature, the material expands by $0.0025\%$ of its original length.

How to Use It: The Working Formula

From the definition, we can rearrange to get the practical formula:

\Delta L = \alpha \, L_0 \, \Delta T

So the final length $L$ after a temperature change is:

L = L_0 + \Delta L = L_0 (1 + \alpha \, \Delta T)

Tip

For small temperature changes (say, less than $100^\circ\text{C}$ ), this linear approximation is excellent. For very large changes, the coefficient itself may change slightly with temperature, but at the introductory level we treat $\alpha$ as constant.

A Concrete Example

A steel bridge girder is $50.00 \, \text{m}$ long at $20^\circ\text{C}$ . The linear expansion coefficient of steel is $\alpha = 1.2 \times 10^{-5} \, /^\circ\text{C}$ . How much longer is it on a $40^\circ\text{C}$ day?

Step 1: Identify the quantities.

$L_0 = 50.00 \, \text{m}$
$\Delta T = 40 - 20 = 20^\circ\text{C}$
$\alpha = 1.2 \times 10^{-5} \, /^\circ\text{C}$

Step 2: Apply the formula.

\Delta L = \alpha \, L_0 \, \Delta T = (1.2 \times 10^{-5})(50.00)(20)

Step 3: Calculate.

\Delta L = 1.2 \times 10^{-5} \times 1000 = 0.012 \, \text{m} = 1.2 \, \text{cm}

So the girder expands by $1.2 \, \text{cm}$ . That is why bridges have expansion joints — without them, the structure would buckle.

Two Important Cousins: Area and Volume Expansion

For a thin sheet (like a metal plate), we care about area expansion. The area expansion coefficient is approximately $2\alpha$ . For a solid object, the volume expansion coefficient is approximately $3\alpha$ . These come from the same idea: if every linear dimension grows by a factor $(1 + \alpha \Delta T)$ , then area grows by $(1 + \alpha \Delta T)^2 \approx 1 + 2\alpha \Delta T$ , and volume by $(1 + \alpha \Delta T)^3 \approx 1 + 3\alpha \Delta T$ .

Watch out

These approximations ( $2\alpha$ and $3\alpha$ ) are valid only when $\alpha \Delta T$ is small compared to 1. For most solids and modest temperature changes, this is true. For gases, the expansion is much larger and a different treatment is needed.

What the Coefficient Tells Us About Materials

| Material | $\alpha$ (per $^\circ\text{C}$ ) | Behaviour |

|----------|--------------------------------|-----------| …

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When a uniform rod expands due to a temperature increase, its length increases, which in turn increases its moment of inertia about its perpendicular bisector. The increase in moment of inertia is $\boxed{\frac{1}{6}ML^2 \alpha \Delta T}$ .

The moment of inertia of an object depends on its mass distribution and its dimensions. When the temperature of a material changes, its dimensions typically change due to thermal expansion or contraction. For a rod, this means its length will change. Since the moment of inertia of a rod about its perpendicular bisector is proportional to the square of its length, any change in length will directly affect its moment of inertia.

Our approach will be to first write down the initial moment of inertia, then determine the new length after thermal expansion, and finally calculate the new moment of inertia. The difference between the new and initial moments of inertia will give us the increase.

Initial Moment of Inertia Let the original length of the uniform rod be $L$ and its mass be $M$ . The moment of inertia of a uniform rod about an axis passing through its center and perpendicular to its length (its perpendicular bisector) is given by:

I = \frac{1}{12}ML^2

Change in Length due to Thermal Expansion When the temperature of the rod is increased by $\Delta T$ , its length changes due to linear thermal expansion. The new length, $L'$ , can be expressed in terms of the original length $L$ and the coefficient of linear expansion $\alpha$ as:

L' = L(1 + \alpha \Delta T)

Here, $\alpha \Delta T$ represents the fractional change in length. Since the temperature is "slightly increased", $\alpha \Delta T$ will be a very small quantity.

3. New Moment of Inertia

The mass $M$ of the rod remains constant. Only its length changes. We can find the new moment of inertia, $I'$ , by substituting the new length $L'$ into the moment of inertia formula:

I' = \frac{1}{12}M(L')^2

Substitute the expression for $L'$:

I' = \frac{1}{12}M[L(1 + \alpha \Delta T)]^2

I' = \frac{1}{12}ML^2(1 + \alpha \Delta T)^2

Calculate the Increase in Moment of Inertia The increase in moment of inertia, $\Delta I$ , is the difference between the new moment of inertia $I'$ and the initial moment of inertia $I$ :

\Delta I = I' - I

\Delta I = \frac{1}{12}ML^2(1 + \alpha \Delta T)^2 - \frac{1}{12}ML^2

Factor out $\frac{1}{12}ML^2$:

\Delta I = \frac{1}{12}ML^2[(1 + \alpha \Delta T)^2 - 1]

Now, expand the term $(1 + \alpha \Delta T)^2$: …

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Q.Find out the increase in moment of inertia III of a uniform rod (coefficient of linear expansion α\alphaα) about its perpendicular bisector when its temperature is slightly increased by ΔT\Delta TΔT.

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Q.Find out the increase in moment of inertia $I$ of a uniform rod (coefficient of linear expansion $\alpha$ ) about its perpendicular bisector when its temperature is slightly increased by $\Delta T$ .

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