# Answer to Question #1508 in Mechanics | Relativity for mike

Question #1508
At 21C, an aluminum ring has an inner diameter of 8 cm and a brass rod has a diameter of 8.06 cm. If the temperature coefficient of expansion for brass is 1.9 &times; 10[sup]&minus;5[/sup] (degrees C)[sup]&minus;1[/sup] and the temperature coefficient of expansion for aluminum is 2.4 &times; 10[sup]&minus;5[/sup] (degrees C)[sup]&minus;1[/sup], to what temperature must the ring be heated so that it will just slip overthe rod?
Answer in units of degrees C.
1
2011-02-08T06:11:53-0500
Denote the diameter of the ring as D1 and the the diameter of the rod as D2, the temperature coefficient of expansion for aluminium as &alpha;1, for the brass as &alpha;2.
The change in the linear dimension can be estimated by the formula:
&Delta; L = &alpha; * L * &Delta; T

In our case to slip over the rod the ring must have minimum diameter equal to the diameter of the rod:
D1+&Delta;D1 = D2 + &Delta;D2
D1 + D1* &alpha;1 * &Delta;T = D2 + D2* &alpha;2* &Delta;T
&Delta; T = (D1-D2)/(&alpha;2D2-&alpha;1D1)= (8-8.06)/(1.9*8.06-2.4*8) x105 = 154.4 C.
&Delta; T = T - Tinitial
T = &Delta; T +Tinitial = 154.4 + 21 = 175.4 C.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!