Question

Find the angular and linear speed for a wheel that rotates 32 rpm

Answer 1

One revolution corresponds to an angular displacement of
`2\pi`, so its angular speed is

`32\,\mathrm{rpm}=\left(32(\rm rev)/(\rm min)\right)\left(2\pi(\rm rad)/(\rm rev)\right)=64\pi(\rm rad)/(\rm min)`

`\pi` radians is equivalent to 180 degrees, so the angular speed could also be

`\left(64\pi(\rm rad)/(\rm min)\right)\left(\frac{180}\pi(\rm deg)/(\rm rad)\right)=11,520(\rm deg)/(\rm min)`

The linear speed depends on the wheel's radius. Suppose its radius is
`r`. Then the wheel has circumference
`2\pi r` units. A point on the edge of the wheel travels this distance in one revolution, so its linear speed is

`\left(32(\rm rev)/(\rm min)\right)\left(2\pi r(\rm u)/(\rm rev)\right)=64\pi r(\rm u)/(\rm min)`

(where
`u` stands for units of length)