Question

A boy is twirling a model airplane on a string 4 feet long. If he twirls the plane at 0.5 revolutions per minute, how far does the plane travel in 4 minutes

Answer 1

Airplane covers a distance of approximately 50.265 feet in 4 minutes.

Explanation:

From Rotation Physics we know that angular speed is the number of revolutions done by a particle at a given time. If the boy is twirling a model airplane at constant speed, then we calculate such variable as follows:

`\dot n = (n)/(t)` (Eq. 1)

Where:

`\dot n` - Angular speed, measured in revolutions per minute.

`n` - Number of revolutions done by airplane, measured in revolutions.

`t` - Time, measured in minutes.

Now we clear the number of revolutions done by airplane within expression:

`n = \dot n \cdot t`

If we know that
`\dot n = 0.5\,(rev)/(min)` and
`t = 4\,min`, then we get that number of revolutions is:

`n = \left(0.5\,(rev)/(min) \right)\cdot (4\,min)`

`n = 2\,rev`

The airplane made two revolutions in four minutes.

Now, the distance covered by the airplane (
`s`), measured in feet, is calculated by means of the equation below:

`s = 2\pi\cdot l\cdot n`

Where:

`l` - Length of the string, measured in meters.

`n` - Number of revolutions done by airplane, measured in revolutions (dimensionless).

If we get that
`n = 2\,rev` and
`l = 4\,ft`, then the distance covered by the airplane in 4 minutes is:

`s = 2\pi\cdot (4\,ft)\cdot (2\,rev)`

`s \approx 50.265\,ft`

Airplane covers a distance of approximately 50.265 feet in 4 minutes.