Chapter 8, Problem 64P

(a)

To determine

The energy of a hurricane.

Answer to Problem 64P

The energy of a hurricane is $4.53\times {10}^{16}J$

Explanation of Solution

Given:

Speed of wind in hurricane, $v=120\text{km/h =120}\times \frac{5}{18}=33.33\text{m/s}$

Density of air, $\rho =1.3{\text{kg/m}}^{\text{3}}$

Radius of hurricane, $r=100km={10}^{5}m$

Height of hurricane, $h=4km=4000m$

Formula used:

Volume of cylinder, $V=\pi r^{2}h$

Mass, $M=\rho V=\rho \pi r^{2}h$

Moment of inertia of cylinder, $I=\frac{M{r}^2}{2}$

Angular velocity, $\omega =\frac{v}{r}$

Kinetic energy of rotating body, $K=\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(\frac{M{r}^2}{2}\right){\left(\frac{v}{r}\right)}^2=\frac{M{v}^2}{4}$

Calculation:

Substituting values in formula of mass,

  $\begin{array}{l}$ M=(1.3)(\pi ){({10}^5)}^2(4000)$M=1.63\times {10}^{14}kg\end{array}$

Substituting values in formula of kinetic energy,

  $\begin{array}{l}K=\frac{(1.63\times {10}^{14}){(33.33)}^2}{4}\\ K=4.53\times {10}^{16}J\end{array}$

Conclusion:

The energy of a hurricane is $4.53\times {10}^{16}J$

(b)

To determine

The angular momentum of the hurricane.

Answer to Problem 64P

The angular momentum of the hurricane is $5.43\times {10}^{20}{\text{kg-m}}^{\text{2}}\text{/s}$

Explanation of Solution

Given:

Mass of the hurricane, $M=1.63\times {10}^{14}kg$

Speed of wind in hurricane, $v=120\text{km/h =120}\times \frac{5}{18}=33.33\text{m/s}$

Radius of hurricane, $r=100km={10}^{5}m$

Formula used:

Angular momentum, $L=Mvr$

Calculation:

Substituting values in formula of angular momentum,

  $\begin{array}{l}L=(1.63\times {10}^{14})(33.33)({10}^5)\\ L=5.43\times {10}^{20}{\text{kg-m}}^{\text{2}}\text{/s}\end{array}$

Conclusion:

The angular momentum of the hurricane is $5.43\times {10}^{20}{\text{kg-m}}^{\text{2}}\text{/s}$