A 5 m long cylindrical steel wire with radius 2 x 10^-3 m is suspended vertically from a rigid suppo

Q: Why do we use the mass of the wire and not the bob in the heat equation?

The heat is generated *within* the wire when it snaps back. The bob falls away and takes its kinetic energy with it, but the elastic energy was stored in the wire itself, so it heats the wire.

Q: Why does the wire heat up when the bob snaps?

When the wire is stretched, work is done on it, storing elastic potential energy. When the tension is suddenly released, the wire rapidly contracts. Internal friction (damping) during this rapid contraction converts the elastic energy into thermal energy.

Q: Can we ignore the change in length of the wire when calculating its volume?

Yes, the extension $\Delta l$ is extremely small compared to the original length $l$ (on the order of millimeters compared to 5 meters). Including it would make a negligible difference to the mass and volume.

Q: What happens to the potential energy of the bob?

The gravitational potential energy of the bob is converted into its kinetic energy as it falls. It does not contribute to the heating of the wire after the snap.

Q: Why is the temperature rise so small?

Metals like steel have a relatively high specific heat capacity, and the amount of elastic energy stored is quite small (less than 1 Joule) compared to the mass of the wire (about 0.5 kg). It takes a lot of energy to heat up half a kilogram of steel!

Question

A 5 m long cylindrical steel wire with radius $2 	imes 10^{-3}	ext{ m}$ is suspended vertically from a rigid support and carries a bob of mass 100 kg at the other end. If the bob gets snapped, calculate the change in temperature of the wire ignoring losses. (For the steel wire : Young's modulus $= 2.1 	imes 10^{11}	ext{ Pa}$; Density $= 7860	ext{ kg/m}^3$; Specific heat $= 420	ext{ J/kg-K}$).

Sigma Percentile · Accepted Answer

Welcome to a fascinating problem that beautifully bridges two seemingly unrelated worlds of physics: elasticity and thermodynamics! At first glance, we are dealing with a heavy mass stretching a steel wire. But by the end of it, we are calculating a change in temperature. How does a mechanical stretch turn into heat? Let's embark on this journey.

### The Setup: A Wire Under Tension

Imagine a robust steel wire, $5	ext{ m}$ long, hanging from a rigid ceiling. At the bottom of this wire, we attach a massive $100	ext{ kg}$ bob. Now, steel is incredibly strong, but it is not perfectly rigid. Under the immense weight of the $100	ext{ kg}$ bob, the wire stretches. It behaves exactly like a very stiff spring.

When you stretch a spring, you do work on it. This work is stored within the material as **elastic potential energy**. The atoms in the steel lattice are pulled slightly apart from their equilibrium positions, and they want to snap back.

### Step 1: The Mass of the Wire

Before we calculate the energy, we need to know how much steel we are actually heating up. The problem gives us the density of the steel, its length, and its radius.

The volume of a cylindrical wire is given by the area of its cross-section multiplied by its length:
$$V = \pi r^2 l$$

The mass $m$ is simply the density $ho$ multiplied by this volume:
$$m = ho \pi r^2 l$$

Let's plug in the numbers. The density $ho$ is $7860 	ext{ kg/m}^3$, the radius $r$ is $2 	imes 10^{-3} 	ext{ m}$, and the length $l$ is $5 	ext{ m}$.
$$m = (7860) (3.14) (2 	imes 10^{-3})^2 (5)$$
$$m = 0.494 	ext{ kg}$$

So, the wire itself has a mass of nearly half a kilogram.

### Step 2: The Stored Elastic Energy

Now, let's find out exactly how much energy is trapped in this stretched wire. The formula for the elastic potential energy $U$ stored in a stretched wire is:
$$U = \frac{1}{2} 	imes 	ext{stress} 	imes 	ext{strain} 	imes 	ext{Volume}$$

We know that **stress** is the force applied per unit area. The force is the weight of the bob, $Mg$, and the area is $\pi r^2$.
$$	ext{stress} = \frac{Mg}{\pi r^2}$$

**Strain** is the fractional change in length, $\frac{\Delta l}{l}$. According to Hooke's Law, Young's modulus $Y$ is the ratio of stress to strain. Therefore, we can express strain as:
$$	ext{strain} = \frac{	ext{stress}}{Y} = \frac{Mg}{\pi r^2 Y}$$

Substituting these into our energy equation, along with the volume $V = \pi r^2 l$:
$$U = \frac{1}{2} \left( \frac{Mg}{\pi r^2} ight) \left( \frac{Mg}{\pi r^2 Y} ight) (\pi r^2 l)$$

Notice how one of the $\pi r^2$ terms beautifully cancels out! We are left with a highly elegant master equation for the stored energy:
$$U = \frac{M^2 g^2 l}{2 \pi r^2 Y}$$

Let's substitute the given values. The mass of the bob $M = 100 	ext{ kg}$, $g = 10 	ext{ m/s}^2$, $l = 5 	ext{ m}$, $r = 2 	imes 10^{-3} 	ext{ m}$, and $Y = 2.1 	imes 10^{11} 	ext{ Pa}$.
$$U = \frac{(100)^2 (10)^2 (5)}{2 (3.14) (2 	imes 10^{-3})^2 (2.1 	imes 10^{11})}$$
$$U = 0.9478 	ext{ J}$$

Almost $1	ext{ Joule}$ of energy is stored in the atomic lattice of the steel wire.

### Step 3: The Snap and The Heat

Here is where the physics gets incredibly exciting. The problem states: "the bob gets snapped."

Imagine holding a stretched rubber band and suddenly letting go of one end. It snaps back violently. When the bob detaches, the tension at the bottom of the wire instantly drops to zero. The stretched steel wire rapidly contracts to its original length.

But where does that $0.9478	ext{ Joules}$ of stored elastic energy go? It cannot just disappear—the law of conservation of energy forbids it. Because the wire snaps back and vibrates, internal friction (damping) within the steel quickly dissipates this macroscopic kinetic energy into microscopic kinetic energy. In other words, the organized potential energy turns into chaotic thermal energy. **The wire heats up!**

We can equate the stored elastic energy to the heat energy $Q$ gained by the wire:
$$Q = U = 0.9478 	ext{ J}$$

### The Final Calculation

To find the temperature rise, we turn to the fundamental equation of calorimetry. The heat $Q$ required to raise the temperature of a mass $m$ with specific heat $s$ by an amount $\Delta 	heta$ is:
$$Q = m s \Delta 	heta$$

We want to find $\Delta 	heta$, so we rearrange the formula:
$$\Delta 	heta = \frac{Q}{ms}$$

We already calculated the mass of the wire $m = 0.494 	ext{ kg}$, and we are given the specific heat of steel $s = 420 	ext{ J/kg-K}$. Let's plug everything in:
$$\Delta 	heta = \frac{0.9478}{0.494 	imes 420}$$
$$\Delta 	heta = \frac{0.9478}{207.48}$$
$$\Delta 	heta = 4.568 	imes 10^{-3} \, ^\circ	ext{C}$$

And there we have it! The temperature of the wire increases by a tiny fraction of a degree. While you wouldn't be able to feel this heat with your hand, it is a profound testament to the interconnectedness of mechanics and thermodynamics. Every time materials deform and snap back, heat is born.

Visualized Solution

$Visualizing the Setup$

$Mass of the Wire$

$Calculating Mass of the Wire$

$Elastic Potential Energy$

$Energy in terms of knowns$

$Calculating Stored Energy$

$The Snap and Heat Conversion$

$Temperature Rise$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Young's Modulus, Bulk Modulus and Modulus of Rigidity

The Setup

Step 1

Step 2

Step 3

The Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mass of the Wire

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Mass of the Wire

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Elastic Potential Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy in terms of knowns

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Stored Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Snap and Heat Conversion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Temperature Rise

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward