A wooden wheel of radius R is made of two semi-circular parts (see figure). The two parts are held t

Q: Why do we use $\Delta l / L = \alpha \Delta T$ if the length of the ring doesn't actually change?

Great question! The *actual* length doesn't change, but its *natural* (unstressed) length decreases as it cools. The difference between its actual forced length and its new natural length creates the thermal strain.

Q: Why is the force $F = 2T$ and not just $T$?

Look at the free body diagram of one half of the wheel. The ring is cut at *two* places (the left and right ends). At each cut, the ring exerts a tension $T$ inwards. So, the total downward force from the ring is $T + T = 2T$.

Q: Does the expansion of the wooden wheel matter?

In this problem, we assume the wooden wheel is perfectly rigid and its thermal expansion is negligible compared to the metal ring, or that the $\Delta T$ only applies to the ring (as it was heated and then cooled to the wheel's temperature).

Q: What if the ring was heated instead of cooled while on the wheel?

If it were heated further, it would expand and its natural length would become greater than the wheel's circumference. It would become loose and fall off, so the normal force would be zero!

Q: Why is the cross-sectional area $S$ important?

Tension is a force, and force depends on how thick the material is. A thicker ring (larger $S$) requires more force to stretch it by the same amount, hence it will exert a greater crushing force on the wheel.

Question

A wooden wheel of radius $R$ is made of two semi-circular parts (see figure). The two parts are held together by a ring made of a metal strip of cross-sectional area $S$ and length $L$. $L$ is slightly less than $2\pi R$. To fit the ring on the wheel, it is heated so that its temperature rises by $\Delta T$ and it just steps over the wheel. As it cools down to surrounding temperature, it presses the semi-circular parts together. If the coefficient of linear expansion of the metal is $\alpha$ and its Young's modulus is $Y$, the force that one part of the wheel applies on the other part is

Options:

(A) $2\pi SY\alpha\Delta T$

(B) $SY\alpha\Delta T$

(C) $\pi SY\alpha\Delta T$

(D) $2SY\alpha\Delta T$

Sigma Percentile · Accepted Answer

### The Physics of the Shrinking Ring

Imagine a wooden wheel split perfectly down the middle into two semi-circular halves. To hold them together, we use a metal ring. But there is a catch—the ring is intentionally made slightly smaller than the circumference of the wheel. To fit it on, we heat the ring. As its temperature rises by $\Delta T$, it expands just enough to slip over the wooden halves.

The real magic happens when the ring cools back down to room temperature. As it cools, it desperately wants to shrink back to its original, smaller length. However, the rigid wooden wheel prevents it from doing so. This forced stretching creates a massive internal tension within the metal ring, which in turn crushes the two wooden halves together.

### Calculating the Thermal Tension

Let's quantify this crushing force. If the ring were free to contract, its length would decrease by an amount $\Delta l = L \alpha \Delta T$, where $\alpha$ is the coefficient of linear expansion.

Because the wheel forces the ring to stay stretched, the ring experiences a thermal strain. We can write this strain as:

$$ 	ext{Strain} = \frac{\Delta l}{L} = \alpha \Delta T $$

According to Hooke's Law, the stress developed in the material is directly proportional to the strain, governed by Young's modulus $Y$. Since stress is defined as the internal tension $T$ divided by the cross-sectional area $S$, we have:

$$ \frac{T}{S} = Y 	imes 	ext{Strain} $$

Substituting our expression for strain, we get:

$$ \frac{T}{S} = Y \alpha \Delta T $$

Rearranging this gives us the tension pulling through every cross-section of the metal ring:

$$ T = Y S \alpha \Delta T $$

### The Free Body Diagram

Now, we need to find the force that one half of the wheel applies on the other. To do this, we isolate the upper semi-circular half of the wheel and draw a free body diagram.

The metal ring wraps around the outside, but if we cut our system in half horizontally, we slice through the ring at two places: the far left and the far right. At each of these cuts, the ring is pulling downwards with the tension $T$ we just calculated.

To keep this top half of the wheel from accelerating downwards, the bottom half must be pushing up on it with a normal force $F$.

### The Final Force Balance

Since the top half is in perfect static equilibrium, the net vertical force must be zero. The single upward normal force $F$ must exactly balance the two downward tension forces from the ring.

$$ F = T + T = 2T $$

Finally, we substitute our expression for the tension $T$ into this force balance equation:

$$ F = 2(Y S \alpha \Delta T) $$

$$ F = 2 S Y \alpha \Delta T $$

This elegant result shows how thermal properties (like $\alpha$) and mechanical properties (like $Y$) combine to create immense structural forces. It is the exact same principle used by blacksmiths for centuries to fit iron tires onto wooden cartwheels!

Select Answer:

Visualized Solution

Analyzing the Setup

Thermal Strain

Tension in the Ring

Calculating Tension

Free Body Diagram

Force Balance

Final Answer

The Way Forward

The Sigma Insight: Thermal Expansion

The Physics of the Shrinking Ring

Calculating the Thermal Tension

The Free Body Diagram

The Final Force Balance

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