find KE as a function of radius slice dr
from r = 10^10^3 to r = 150*10^3
208 km/hr /3.6 = 57.8 m/s
Velocity = m r + b linear
57.8 = m (10*10^3) + b
0 = m( 150*10^3) + b
so solve for slope and intercept
m = -.413 * 10^-3
b = 61.9
so
v = 61.9 - .413*10^-3 r
the volume at radius r = 2 pi r dr(5550)
so the mass at radius r = .891*5550* 2 pi r dr
so the KE at radius r = (1/2) (.891*5550* 2 pi r dr) (61.9-.413*10^-3 r)^2
when you square that and multiply by r dr you will get terms like
a r + b r^2 + c r^3
which will integrate to
a r^2/2 + b r^2/3 + c r^4/4
evaluate that integral at r = 10*10^3 and at 150*10^3 and you have it
In a simple model of the wind speed associated with hurricane Emily, we assume there is calm eye 10.0 km in radius. The winds, which extend to a height of 5550 m, begin with a speed of 208.0 km/hr at the eye wall and decrease linearly with radial distance down to 0 km/hr at a distance of 150.0 km from the center. Assume the average density of the air from sea level to an altitude of 5550 m is 0.891 kg/m3. Calculate the total kinetic energy of the winds.
2 answers
Where did 3.6 come from?
2 answers