Question

At its peak, a tornado is 55.4 m in diameter and carrits 586 km/h winds. What is its angular velocity in revolutions per second?

Answer 1

0.935 rev/s

Explanation:

You want the angular velocity in revolutions per second of a tornado 55.4 m in diameter that has 586 km/h winds.

Revolutions

Each revolution is the circumference of the circle 55.4 m in diameter:

C = πd = π(55.4 m) ≈ 174.044 m

Velocity

Then the angular velocity is ...

`\frac{586\text{ km}}{1\text{ h}}*\frac{1000\text{ m}}{1\text{ km}}*\frac{1\text{ h}}{3600\text{ s}}*\frac{1\text{ rev}}{174.044\text{ m}}\\\\=(586*1000)/(3600*174.044)\text{ rev/s}\approx\boxed{0.935\text{ rev/s}}`