## Physics Friday 148

A thin bimetallic strip of thickness d is straight at a temperature T0, with length L0. The two metals have coefficients of linear thermal expansion α1 and α2, with α2>α1. If the temperature is raised to a temperature T not significantly greater than T0, what is the angle θ through which the strip bends?

We should expect the curvature to be uniform, leaving the strip to form a circular arc of angle θ. If the radius of curvature for the inner half of the strip is r, then (with θ in radians), the length of that metal is L1=rθ.
We can ignore any expansion in the direction of the thickness, so the length of the other metal is
L2=(r+d)θ.
Taking the difference, we see
L2-L1=dθ,
so $\theta=\frac{L_2-L_1}{d}$.

Now, the formula for thermal expansion is
$\frac{\Delta{L}}{L}=\alpha\Delta{T}$ (see this previous post),
so for the inner side,
$\frac{L_1-L_0}{L_0}=\alpha_1\left(T-T_0\right)$,
and solving for L1,
$L_1=L_0\left(1+\alpha_1\left(T-T_0\right)\right)$.
Analogously,
$L_2=L_0\left(1+\alpha_2\left(T-T_0\right)\right)$.
Thus, the difference is
$L_2-L_1=L_0\left(\alpha_2-\alpha_1\right)\left(T-T_0\right)$,
and so the angle θ is
$\theta=\frac{L_0}d\left(\alpha_2-\alpha_1\right)\left(T-T_0\right)$.