How Much Energy is Stored in a Hurricane's Rotation?

[sinclair18]

Rotational Motion Question -- please help

Homework Statement

Estimate the energy stored in the rotational motion of a hurricane. Model the hurricane as a uniform cylinder 300 km and 5 km high, made of air whose mass is 1.3 kg/m^3. Estimate the outer edge of the hurricane to move at a speed of 200 km/h.

Homework Equations

KE = (1/2)*I*w^2

v= wR

For a uniform cylinder, I=(1/2)MR^2 (but it's hollow so would that make a difference, even though you're only given one radius)

The Attempt at a Solution

So here's how I started out:

KE = (1/2)*I*w^2

v= wR

--> Therefore KE = (1/2)[(1/2)MR^2][v/R]^2

The answer is 4E17 J, but I can't seem to get that. Can someone please tell me where I'm going wrong? I've been doing this problem for so long and I'm just not getting what I keep doing wrong. I feel like it might have something to do with the mass I'm using (M=1.3 kg/m^3) or the fact that I'm neglecting the height (h=5km)? It's not supposed to be a difficult problem...please help!

 

[Curious3141]

Homework Statement

Estimate the energy stored in the rotational motion of a hurricane. Model the hurricane as a uniform cylinder 300 km and 5 km high, made of air whose mass is 1.3 kg/m^3. Estimate the outer edge of the hurricane to move at a speed of 200 km/h.

Homework Equations

KE = (1/2)*I*w^2

v= wR

For a uniform cylinder, I=(1/2)MR^2 (but it's hollow so would that make a difference, even though you're only given one radius)

The Attempt at a Solution

So here's how I started out:

KE = (1/2)*I*w^2

v= wR

--> Therefore KE = (1/2)[(1/2)MR^2][v/R]^2

The answer is 4E17 J, but I can't seem to get that. Can someone please tell me where I'm going wrong? I've been doing this problem for so long and I'm just not getting what I keep doing wrong. I feel like it might have something to do with the mass I'm using (M=1.3 kg/m^3) or the fact that I'm neglecting the height (h=5km)? It's not supposed to be a difficult problem...please help!

What you're given is not a "mass" of air - it's the density of air. Units are mass per unit volume.

What's the volume of a cylinder? Hence, what is its mass? Work in symbols throughout (try to use LaTex, if possible), as there's less chance of error.

You definitely need the height that's given. Also, I'm assuming that 300km is the diameter, not the radius?

 

[sinclair18]

Yeah I just figured it out actually thanks soooo much!

 

[Curious3141]

Yeah I just figured it out actually thanks soooo much!

No problem. Glad to help (if I did).

 

[Nickie]

First, it's important to note that the given dimensions for the hurricane are not sufficient to accurately estimate the energy stored in its rotational motion. In reality, the hurricane is a complex system with varying wind speeds and air densities throughout its structure. Therefore, the estimate provided may not be entirely accurate.

That being said, let's work through the problem with the given information. You are correct in using the formula KE = (1/2)*I*w^2, where I is the moment of inertia and w is the angular velocity.

To find the moment of inertia for a cylinder, we use the formula I = (1/2)MR^2. Since we are given the mass per unit volume, we can calculate the mass of the cylinder by multiplying the volume by the density. The volume of a cylinder is given by V = πr^2h, where r is the radius and h is the height. However, since we are only given one radius, we will assume that the cylinder is solid and use the formula V = πr^2h to calculate the volume. Therefore, the mass of the cylinder is:

M = (1.3 kg/m^3)(π)(300 km)^2(5 km) = 6.54 x 10^14 kg

Now we can plug in the values for I and w into the formula for kinetic energy:

KE = (1/2)*(1/2)(6.54 x 10^14 kg)(300 km)^2*(200 km/h)^2

= 4.91 x 10^22 J

This is a very large amount of energy, but keep in mind that this is just an estimate and the actual value may be different due to the simplifications we made in our calculations. Also, the height of the hurricane may have some impact on the energy stored in its rotational motion, but it is likely to be negligible compared to the other factors.

In conclusion, the key to solving this problem is to use the correct formula for moment of inertia and to be careful with units. I hope this helps!