## Ring on a Rod

An aluminum ring is to fit snugly on a cylinder aluminum rod. At 21°C, the diameter of the rod is 6.019 cm and the inside diameter of the ring is 5.988 cm. To slip over the rod, the ring must be slightly larger than the rod diameter by about 0.005 cm. To what temperature must the ring be brought if its hole is large enough so it will slip over the rod?

Material Coefficient of Linear

Expansion, α (C°)-1 Aluminum 2.5 × 10-5

Material Coefficient of Linear

Expansion, α (C°)-1 Aluminum 2.5 × 10-5

**262 degC**## Approach

The inner diameter of the ring must expand to at least 0.005 cm larger than the diameter of the rod before the ring will fit into the rod. Expansion of material is given b:

change in L = alphaLo*dT

change in L = alphaLo*dT

## Solution

Given quantities:

dT=dL/alphaLo=0.36/2.5e-5*5.988=240.5

The final temperature is then

Tf=Ti+dT=262 deg C

- diameter of ring = 5.988
- diameter of rod = 6.019 cm

dT=dL/alphaLo=0.36/2.5e-5*5.988=240.5

The final temperature is then

Tf=Ti+dT=262 deg C