Question

A wheel of radius 25.0 cm has an angular speed of 44.0 rpm. What is the linear speed of a point on its circumference? [Hint: use the equation v=rw]

Answer 1

Approximately
`\rm 1.15\; m \cdot s^(-1)`.

Step-by-step explanation:

The equation in the hint
`v = r \cdot \omega` relates the linear speed
`v` of an object to its angular speed
`\omega`and its radius
`r`.

Both
`\omega` and
`r` are given. However,
`\omega` is in the non-standard unit of (RPM) rotations per minutes. To make sure that this equation gives the linear velocity in its standard SI unit
`\rm m \cdot s^(-1)`, convert
`\omega` to the its standard unit radians-per-second.

The wheel rotates
`44` times each minute. Divide that by
`60` (number of seconds in each minute) to get the number of rotations in each second. The wheel turns
`11/15` of a rotation in each second on average.

The angular distance in each rotation is
`2\pi` radians. Multiply the number of rotations in each second by
`2\pi` to get the number of radians in each second. The angular velocity
`\omega` of the wheel is
`1.15` radians per second.

Convert the radius of the circle to standard units:
`r = \rm 25.0\; cm = 0.250 \; m`.

Apply the equation. Note that the "radians" part of the unit "radians per second" is ignored in this calculation.

`v = r \cdot \omega = \rm 0.250\; m \cdot 1.15\; s^(-1) \approx 1.15\; m \cdot s^(-1)`.