Question

At its peak, a tornado is 56.0 m in diameter and carries 400 km/h winds. What is its angular velocity in revolutions per second?

rev/s

Answer 1

Given the diameter of the tornado d = 56m. So its radius r = 28m. Also the velocity of the tornado v = 400km/h = 111.11m/s. The angular velocity of the tornado is given by,

$\omega = \frac{v}{r}=\frac{111.11}{28}=3.97rad/s$

Now the angular velocity in revolutions per second is,

$\omega =\frac{3.97}{2\pi}=0.63rps$

So the angular velocity is 0.63 revolutions per second.