Question

linear expansion help

Answer 1

A steel ball bearing is 5 cm in diameter
at 20◦C. A bronze plate has a hole that is
4.995 cm in diameter at 20◦C.
What common temperature must they have
in order for the ball to just squeeze through
the hole?

Answer 2

note, values for thermal expansion may differ for you.

if\[L = L_0 + \Delta L\]and\[\Delta L = \alpha \Delta T L_0\]then\[L = \alpha \Delta T L_0 + L_0\]



For the steel ball:\[L_{St} = L_0 + \Delta L\]For the bronze plate:\[L_{Br} = L_0 + \Delta L\]Final 'lengths' are equal\[L_{st} = L_{Br}\]Thus: \[ \alpha_{St} \Delta T L_{0St} + L_{0St} =\alpha_{Br} \Delta T L_{0Br} + L_{0Br}\ \]\[ \ 13X10^{-6} \Delta T (0.05) + 0.05 =\ 19X10^{-6}\Delta T (0.04995) + 0.04995\ \] Solve for Change in Temperature



*hint change in temperature will be near 160 C