Answer:
[tex]2.7\cdot 10^{16} J[/tex]
Explanation:
We can approximate the hurricane as a rotating uniform cylinder, so its energy is the rotational kinetic energy, given by:
[tex]E=\frac{1}{2}I\omega^2[/tex] (1)
where
I is the moment of inertia
[tex]\omega[/tex] is the angular velocity
The moment of inertia of a cylinder rotating about its axis is
[tex]I=\frac{1}{2}MR^2[/tex]
where
M is the mass
R is the radius
So formula (1) can be written as
[tex]E=\frac{1}{2}(\frac{1}{2}MR^2)\omega^2=\frac{1}{4}MR^2\omega^2[/tex] (2)
For an object in rotation, the linear speed at the edge is related to the angular velocity by
[tex]v=\omega R[/tex]
So we can rewrite (2) as
[tex]E=\frac{1}{4}Mv^2[/tex]
where we have:
[tex]v=100 km/h = 27.8 m/s[/tex] is the speed at the edge of the hurricane
We have to calculate the mass of the cylinder. We have:
[tex]R=88 km = 88,000 m[/tex] (radius)
[tex]h=4.4 km = 4400 m[/tex] (height)
So the volume is
[tex]V=\pi R^2 h = \pi (88,000)^2 (4400)=1.07\cdot 10^{14} m^3[/tex]
The density is
[tex]\rho = 1.3 kg/m^3[/tex]
So the mass is
[tex]M=\rho V=(1.3)(1.07\cdot 10^{14})=1.39\cdot 10^{14} kg[/tex]
Therefore, the energy is
[tex]E=\frac{1}{4}(1.39\cdot 10^{14})(27.8)^2=2.7\cdot 10^{16} J[/tex]
