## 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
L2L1=dθ,
so .

Now, the formula for thermal expansion is
 (see this previous post),
so for the inner side,
,
and solving for L1,
.
Analogously,
.
Thus, the difference is
,
and so the angle θ is
.