In 1993 the radius of hurricane emily was about 350 km. the wind speed near the center ("eye") of the hurricane, whose radius was about 30 km, reached about 200 km/h. as air swirled in from the rim of the hurricane toward the eye, its angular momentum remained roughly constant. estimate
a. the wind speed at the rim of the hurricane.
b. the pressure difference at the earth's surface between the eye and the rim.
c. if the kinetic energy of the swirling air in the eye could be converted completely to gravitational potential energy, how high would the air go?
d. in fact, the air in the eye is lifted to heights of several kilometers. how can you reconcile this with your answer to part c?

a. v_r = 17.14 km/h

b. P_r - P_e = 1838.55 N/m²

c. h = 157.50 m

d. Air is compressible and does not meet the Bernoulli equation. There are more variables that change the behaviour of the air:

Explanation:

a. If angular momentum is constant: (I_r)(ω_r) = (I_e)(ω_e)

Since ω = v/r and I = mr²

Thus,

(m(r_r)²)(v_r/r_r) = (m(r_e)²)(v_e/r_e) ⇒ (v_r)(r_r) = (v_e)(r_e)

⇒ v_r = ((v_e)(r_e))/r_r

          = ((200)(30))/350

          = 17.14 km/h

b. v_e = 55.56 m/s      v_r = 4.76 m/s

   h_e  = 0                    h_r = 0 (Earth's surface)

By Bernoulli's equation:

P_r + ρg(h_r) + 1/2(ρ)((v_r)²) = P_e + ρg(h_e) + 1/2(ρ)((v_e)²)

P_r - P_e = 1/2(ρ)((v_e)²- (v_r)²) = 1/2(1.20)(55.56² - 4.76²)

                = 1835 N/m²

Since P_r - P_e > 0 ⇒P_r > P_e (pressure is higher in the rim)

c. Kinetic Energy = Gravitational Potential Energy

1/2(m)(v_e)² = mgh

⇒ h = (v_e)²/2g

      = (55.56)²/2(9.80)

      = 157.50 m

d. Air is compressible and does not meet the Bernoulli equation. There are more variables that change the behaviour of the air:

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