Question

How do you estimate the energy stored in the rotational motion of a hurricane if you assume the hurricane is a uniform cylinder?

Answer 1

Oops, sorry, made a typo. Hold on.

Answer 2

The moment of inertia of a uniform cylinder of radius R, mass M, and height h is
\[\frac{MR^2h}{2} \]
then the kinetic energy is equal to
\[KE = \frac{1}{2} I \omega^2 = \frac{1}{4}MR^2\omega^2h\]
or, recognizing that
\[R\omega = v\]
is the velocity of a point on the outer edge of a cylinder,
\[KE = \frac{1}{4}Mv^2h\]
where v is then the speed of a point on the outer edge. Of course, this applies only to rigid bodies, and hurricanes are not rigid, so the inside will be spinning faster than we'd expect, and this will give an underestimate of the true kinetic energy.

Answer 3

how do you know the moment of Inertia is .5MR^2H? My book doesn't account for the height of a cylinder when giving the equation for moment of inertia.