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

16771720740.20324 J

Step-by-step explanation:

`\rh0` = Density = 0.97 kg/m³

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

d = Diameter of cylinder = 230 m

r = Radius =
`(d)/(2)=(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* \pi 115^2* 640\\\Rightarrow m=25792727.013\ kg`

Moment of inertia of a cylinder

`I=(1)/(2)mr^2\\\Rightarrow I=(1)/(2)* 25792727.013* 115^2\\\Rightarrow I=170554407373.4625\ kgm^2`

Angular speed

`\omega=(v)/(r)\\\Rightarrow \omega=(51)/(115)\ rad/s`

Kinetic energy is given by

`K=(1)/(2)I\omega^2\\\Rightarrow K=(1)/(2)* 170554407373.4625* \left((51)/(115)\right)^2\\\Rightarrow K=16771720740.20324\ J`

The kinetic energy contained by the tornado is 16771720740.20324 J