Chapter 13, Problem 110P

(a)

To determine

Find the relation for the radius of the metallic strips.

Answer to Problem 110P

Thus proved that the radius of the metallic strips is $R\approx \frac{s}{\left({\alpha }_2-{\alpha }_1\right)\Delta T}$.

Explanation of Solution

Write the equation for linear expansion of the one metallic strip.

$\frac{\Delta L_1}{L_0}={\alpha }_{\text{1}}\Delta T$ (I)

Here, $\Delta L_1$ is the change in length of the first metallic strip, $L_0$ is the original length, ${\alpha }_{\text{1}}$ is the coefficient of linear expansion of the first metallic strip, and $\Delta T$ is the change in temperature.

Write the equation for linear expansion of the metallic strip.

$\frac{\Delta L_2}{L_0}={\alpha }_2\Delta T$ (II)

Here, $\Delta L_2$ is the change in length of the second metallic strip, $L_0$ is the original length, ${\alpha }_2$ is the coefficient of linear expansion of the second metallic strip, and $\Delta T$ is the change in temperature.

Rewrite the equation (I) for $\Delta L_1$.

$\Delta L_1=L_0{\alpha }_{\text{1}}\Delta T$ (III)

Rewrite the equation (II) for $\Delta L_2$.

$\Delta L_2=L_0{\alpha }_2\Delta T$ (IV)

If the expansion coefficient of the bimetallic coefficient ${\alpha }_2>{\alpha }_1$.

The relation for the radius of the one strip is.

$L_0+\Delta L_1=R\theta$ (V)

Here, $R$ is the radius of the circular arc and $\theta$ is the arc length

The relation for the radius of another strip is.

$L_0+\Delta L_2=\left(R+s\right)\theta$ (VI)

Here, $s$ is the thickness of the layer

Rewrite the equation (V) for $\theta$.

$\theta =\frac{L_0+\Delta L_1}{R}$ (VII)

Conclusion:

Substitute equation (VII) in equation (VI).

$L_0+\Delta L_2=\left(R+s\right)\frac{L_0+\Delta L_1}{R}$

Substitute the equation (III) and (IV) in above equation.

$\begin{array}{c}R\left(L_0+L_0{\alpha }_2\Delta T\right)=\left(R+s\right)\left(L_0+L_0{\alpha }_{\text{1}}\Delta T\right)\\ L_{0}R\left(1+{\alpha }_2\Delta T\right)=R{L}_0\left(1+{\alpha }_{\text{1}}\Delta T\right)+s{L}_0\left(1+{\alpha }_{\text{1}}\Delta T\right)\\ R\left(1+{\alpha }_2\Delta T\right)=R\left(1+{\alpha }_{\text{1}}\Delta T\right)+s\left(1+{\alpha }_{\text{1}}\Delta T\right)\\ R\left(1+{\alpha }_2\Delta T\right)-R\left(1+{\alpha }_{\text{1}}\Delta T\right)=s\left(1+{\alpha }_{\text{1}}\Delta T\right)\end{array}$

Apply the condition $\alpha \Delta T\ll 1$ and solve the above equation for $R$.

$\begin{array}{c}R\left({\alpha }_2-{\alpha }_{\text{1}}\right)\Delta T=s\left(1+{\alpha }_{\text{1}}\Delta T\right)\\ R\left({\alpha }_2-{\alpha }_{\text{1}}\right)\Delta T=s\\ R\approx \frac{s}{\left({\alpha }_2-{\alpha }_{\text{1}}\right)\Delta T}\end{array}$

Hence, proved the radius of the metallic strips is $R\approx \frac{s}{\left({\alpha }_2-{\alpha }_1\right)\Delta T}$.

(b)

To determine

The radius of the bimetallic strip.

Answer to Problem 110P

The radius of the bimetallic strip is $0.7\text{m}$.

Explanation of Solution

Write the equation for radius of the bimetallic strip from part (a).

$R=\frac{s}{\left({\alpha }_{\text{brass}}-{\alpha }_{\text{iron}}\right)\Delta T}$ (VIII)

Here, $R$ is the radius of the bimetallic strip, $s$ is the thickness of the each layer, ${\alpha }_{\text{iron}}$ is the coefficient of linear expansion of the iron, ${\alpha }_{\text{brass}}$ is the coefficient of linear expansion of the brass, and $\Delta T$ is the change in temperature.

Conclusion:

Substitute $0.1\text{mm}$ for $s$, $19\times {10}^{-6}{\text{K}}^{-1}$ for ${\alpha }_{\text{brass}}$, $12\times {10}^{-6}{\text{K}}^{-1}$ for ${\alpha }_{\text{iron}}$, and $20.0°\text{C}$ for $\Delta T$ in equation (VIII).

$\begin{array}{c}R=\frac{\left(0.1\text{mm}\right)\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}{\left[\left(19-12\right)\times {10}^{-6}{\text{K}}^{-1}\right]20.0\text{K}}\\ =7.1\times {10}^{-1}\text{m}\\ =0.71\text{m}\\ \simeq \text{0}\text{.7}\text{m}\end{array}$

Therefore, the radius of the bimetallic strip is $0.7\text{m}$.