Question

A machinist wishes to insert a copper rod with a diameter of 4 mm into a hole with a diameter of 3.998 mm. By how much would the machinist have to lower the temperature (in °C) of the rod to make it fit the hole?

Answer 1

Actually, robot is wrong again...
It wasn't noted the starting temperature so let's assume it is 20 C

Then,

Initial diameter of the rod D1 = 4 mm

Final diameter of the rod D2 = 3.993 (at least)

Initial Temperature of the rod T1 = 20 C

Final Temperature of the rod T2 = ?

pi = 3.1415926...

Assume the coefficient of expansion (a) for brass is 19 x 10-6 m/m-C

The formula is

dL/L1 = a dT

But the L in this case is the circumference of the rod = pi*D

So,

d(pi*D)/(pi*D1) = a dT or (D2 - D1)/D1 = a(T2 - T1)

or (3.993 - 4)/4 = (19 x 10-6)(T2 - 20)

-92.1 = T2 - 20

So, T2 = -72.1 C (to get to approx. 3.993 mm)

Answer 2

The machinist would not need to lower the temperature of the rod to make it fit the hole. The difference in diameter between the rod and the hole is only 0.002 mm, which is too small to require a change in temperature.