A steel rod of length $2l$, cross sectional area $A$ and mass $M$ is set rotating in a horizontal plane about an axis passing through the centre. If $Y$ is the Young's modulus for steel, find the extension in the length of the rod. (Assume the rod is uniform.)

A steel rod of length 2l, cross sectional area A and mass M is set rotating in a horizontal plane about an axi

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Concept understanding — Youngs Modulus

Young’s Modulus: The Stretchiness of a Solid

When you pull on a rubber band, it stretches easily. When you pull on a steel rod of the same size, it barely moves. Both are elastic — they return to their original shape when you let go — but they resist stretching very differently. Young’s modulus is the number that tells you exactly how much a material resists being stretched or compressed lengthwise.

The Intuition: Stiffness per Unit Size

Think of a spring. A stiff spring requires a large force to stretch it a little. A soft spring stretches a lot with a small force. Young’s modulus is like the “stiffness” of a material, but it’s cleverly designed to be independent of the object’s shape and size.

If you take a thick steel rod and a thin steel wire of the same length, the rod is harder to stretch. That’s because you’re pulling on more material. Young’s modulus removes this size effect — it tells you the stiffness of the material itself, not the particular piece you’re holding.

The Precise Definition

Young’s modulus ( $E$ or $Y$ ) is defined as the ratio of tensile stress to tensile strain, as long as the material obeys Hooke’s law (the deformation is reversible and proportional to the force).

Y = \frac{\text{Tensile Stress}}{\text{Tensile Strain}}

Let’s break down the two parts.

Tensile Stress ( $\sigma$ ) is the force per unit area. If you pull with a force $F$ on a rod of cross-sectional area $A$ , the stress is:

\sigma = \frac{F}{A}

Stress has units of pressure — pascals (Pa) or $\text{N/m}^2$ . It tells you how “intense” the pulling is, regardless of the rod’s thickness.

Tensile Strain ( $\varepsilon$ ) is the fractional change in length. If the original length is $L_0$ and it stretches by $\Delta L$ , the strain is:

\varepsilon = \frac{\Delta L}{L_0}

Strain is a pure number — it has no units. A strain of 0.01 means the rod stretched by 1% of its original length.

Putting it together:

Y = \frac{F/A}{\Delta L / L_0} = \frac{F L_0}{A \Delta L}

What the Number Tells You

A high Young’s modulus means the material is very stiff — it takes a huge stress to produce even a tiny strain. Steel has $Y \approx 200 \times 10^9$ Pa. A low Young’s modulus means the material is easily stretched. Rubber has $Y \approx 0.01 \times 10^9$ Pa — about 20,000 times smaller than steel.

Note

Young’s modulus is only valid in the elastic region — where the material returns to its original shape after the force is removed. If you stretch too far (past the elastic limit), the material deforms permanently or breaks, and Young’s modulus no longer applies.

A Worked Example

A steel wire of length 2.0 m and cross-sectional area $1.0 \times 10^{-6} \text{ m}^2$ is pulled by a force of 100 N. How much does it stretch? (Young’s modulus of steel = $2.0 \times 10^{11}$ Pa)

From $Y = \frac{F L_0}{A \Delta L}$ , rearrange:

$\Delta L = \frac{F L_0}{A Y} = \frac{100 \times 2.0}{(1.0 \times 10^{-6}) \times (2.0 \times 10^{11})} = \frac{200}{2.0 \times 10^5} = 1.0 \times 10^{-3} \text{ m} = 1.0 \text{ mm}$ …

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The rod stretches because each element experiences a centrifugal force that creates tension. The total extension is $\boxed{\frac{M \omega^2 l^2}{3 A Y}}$ .

Why Young’s modulus applies here

When a rod rotates, every tiny segment of it wants to fly outward. The material resists this stretching through internal elastic forces. Young’s modulus $Y$ relates stress (force per area) to strain (fractional change in length):

Y = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L / L}

The trick is that the tension is not constant along the rod — it’s zero at the free ends and maximum at the centre. So we cannot just use $F = M \omega^2 R$ for the whole rod. We must integrate the stretching of each infinitesimal piece.

Step-by-step solution

1. Set up coordinates and consider a small element

Take the rod’s centre as the origin. The rod rotates with angular speed $\omega$ . Consider an element of length $dx$ at a distance $x$ from the centre (so $0 \le x \le l$ for one half). Its mass is:

dm = \frac{M}{2l} \, dx

2. Find the tension at a distance $x$ from the centre

The part of the rod beyond $x$ (from $x$ to $l$ ) is being pulled outward by centrifugal force. That outward pull is the tension $T(x)$ at position $x$ .

The centrifugal force on a small mass $dm$ at distance $r$ is $dm \, \omega^2 r$ . So the tension at $x$ equals the sum of centrifugal forces on all elements from $x$ to $l$ :

T(x) = \int_{x}^{l} \omega^2 r \, dm = \int_{x}^{l} \omega^2 r \left( \frac{M}{2l} dr \right)

T(x) = \frac{M \omega^2}{2l} \int_{x}^{l} r \, dr = \frac{M \omega^2}{2l} \left[ \frac{r^2}{2} \right]_{x}^{l}

T(x) = \frac{M \omega^2}{4l} (l^2 - x^2)

Tip

Notice $T(x)$ is maximum at $x=0$ (centre) and zero at $x=l$ (free end) — exactly what we expect.

3. Relate tension to extension of an infinitesimal segment

For a small segment of original length $dx$ at position $x$ , the tension $T(x)$ causes a small extension $d(\Delta L)$ . Using Young’s modulus: …

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Q.A steel rod of length 2l2l2l, cross sectional area AAA and mass MMM is set rotating in a horizontal plane about an axis passing through the centre. If YYY is the Young's modulus for steel, find the extension in the length of the rod. (Assume the rod is uniform.)

Why Young’s modulus applies here

Step-by-step solution

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Q.A steel rod of length $2l$ , cross sectional area $A$ and mass $M$ is set rotating in a horizontal plane about an axis passing through the centre. If $Y$ is the Young's modulus for steel, find the extension in the length of the rod. (Assume the rod is uniform.)

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