Question

A typical tornado can be envisioned as a cylinder of height 640 m, with a diameter of about 230 m (note: the visible funnel cloud is usually somewhat less than the actual diameter). The rotation in the tornado is approximately a `solid body rotation', meaning that it rotates very much like a solid cylinder would, even though air is a gas. If the outer edge of the cylinder has a speed of 51 m/s, and the air has a density of 0.97 kg/m3, determine the kinetic energy contained by the tornado.

Answer 1

Answer:

16771720740.20324 J

Explanation:

$\rh0$ = Density = 0.97 kg/m³

V = Volume = $\pi r^2h$

d = Diameter of cylinder = 230 m

r = Radius = $\frac{d}{2}=\frac{230}{2}=115\ m$

h = Height of the cylinder = 640 m

v = Velocity of cylinder = 51 m/s

Mass of object is given by

$m=\rho V\\\Rightarrow m=0.97\times \pi 115^2\times 640\\\Rightarrow m=25792727.013\ kg$

Moment of inertia of a cylinder

$I=\frac{1}{2}mr^2\\\Rightarrow I=\frac{1}{2}\times 25792727.013\times 115^2\\\Rightarrow I=170554407373.4625\ kgm^2$

Angular speed

$\omega=\frac{v}{r}\\\Rightarrow \omega=\frac{51}{115}\ rad/s$

Kinetic energy is given by

$K=\frac{1}{2}I\omega^2\\\Rightarrow K=\frac{1}{2}\times 170554407373.4625\times \left(\frac{51}{115}\right)^2\\\Rightarrow K=16771720740.20324\ J$

The kinetic energy contained by the tornado is 16771720740.20324 J