Question

A cylinder 15.0 cm in diameter rotates at 1000 rpm. (a) what is its angular velocity in rad/s? (b) what is the tangential velocity of a point on the rim of the cylinder?

Answer 1

(a) Approximately
`105\; {\rm rad \cdot s^(-1)}`.

(b) Approximately
`15.7\; {\rm m \cdot s^(-1)}`, assuming that this cylinder is rotating along the axis that goes through the center.

Step-by-step explanation:

The unit
`{\rm rpm}` stands for "revolutions per minute", where each revolution is
`2\, \pi` radians.

With an angular velocity of
`1000\; {\rm rpm}`, this cylinder would turn
`1000* 2\, \pi = 2000\, \pi` radians every minute (
`60\; {\rm s}`). Thus, the angular velocity of this cylinder would be:

`\begin{aligned} \omega &= \frac{2000\, \pi\; {\rm rad}}{60\; {\rm s}} \\ &\approx 104.720\; {\rm rad \cdot s^(-1)}\end{aligned}`.

A point at the rim of this cylinder would be at a distance of
`r = 15.0\; {\rm cm} = 0.150\; {\rm m}` from the axis of revolution of this cylinder. If the angular velocity of this cylinder is
`\omega`, the tangential velocity of this point would be:

`\begin{aligned} v&= \omega\, r \\ &= \frac{2000\, \pi}{60\; {\rm s}} * 0.150\; {\rm m} \\ &\approx 15.7 \; {\rm m\cdot s^(-1)}\end{aligned}`.