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 × 10[sup]−5[/sup] (degrees C)[sup]−1[/sup] and the temperature coefficient of expansion for aluminum is 2.4 × 10[sup]−5[/sup] (degrees C)[sup]−1[/sup], to what temperature must the ring be heated so that it will just slip overthe rod?
Answer in units of degrees C.

Expert's answer

Denote the diameter of the ring as D_{1} and the the diameter of the rod as D_{2}, the temperature coefficient of expansion for aluminium as α_{1}, for the brass as α_{2}.

The change in the linear dimension can be estimated by the formula:

Δ L = α * L * Δ T

In our case to slip over the rod the ring must have minimum diameter equal to the diameter of the rod:

D_{1}+ΔD_{1} = D_{2} + ΔD_{2}

D_{1} + D_{1}* α_{1} * ΔT = D_{2} + D_{2}* α_{2}* ΔT

Δ T = (D_{1}-D_{2})/(α_{2}D_{2}-α_{1}D_{1})= (8-8.06)/(1.9*8.06-2.4*8) x10^{5} = 154.4 C.

Δ T = T - T_{initial}

T = Δ T +T_{initial} = 154.4 + 21 = 175.4 C.

The change in the linear dimension can be estimated by the formula:

Δ L = α * L * Δ T

In our case to slip over the rod the ring must have minimum diameter equal to the diameter of the rod:

D

D

Δ T = (D

Δ T = T - T

T = Δ T +T

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