(II) Hurricanes can involve winds in excess of 120km/h at the ...

Question

(II) Hurricanes can involve winds in excess of 120km/h at the outer edge. Make a rough estimate of

(a) the energy, of such a hurricane, approximating it as a rigidly rotating uniform cylinder of air (density 1.3 kg/m³) of radius 85 km and height 4.5 km.

Accepted Answer

Welcome back. Everyone. In this problem, we want to calculate the energy linked with a tornado. Hypothetically represented as a uniformly rotating cylindrical mass of air with a density of 1.2 kg per cubic meter, a radius of 55 m and a height of two kilometers. Consider the wind velocity is 300 kilometers per hour in the most intense zone which is at the outer edge of the cylinder. For our answer choices. A says that it's 2.6 multiplied by 10 to the 10th jewels. B four multiplied by 10 to the 10th jewels. C 5.2 multiplied by 10 to the 10th jewels and D 7.9 multiplied by 10 to the 10th jewels. Now, if we're going to figure out the energy linked with the tornado, let's first make sure we understand all of the information that we're given here. So we're talking about a cylindrical tornado, ok. Hypothetically where it has a radius of 55 m. Ok. So let's call that R it has a height of two kilometers or 2000 m, ok? And we know that its wind velocity is 300 kilometers per hour. So if we think about the information we have here R equals 55 m, our height edge equals 2000 m. Let's put it in SI units. Our density raw is 1.2 kg per cubic meter. OK? And the wind's linear velocity V, OK is 300 kilometers per hour. And if we were to write that in SI units, that's gonna be 300 multiplied by 1000 m divided by 3600 seconds, which is equal to 83 and a third meters per second. OK. And now we want to use all of that information to figure out the energy linked with our tornado, the energy of our tornado. No, let's ask ourselves what type of energy is at play here. What do we know? Well, since it's rotating, we can find its rotational kinetic energy. OK. And recall that rotational kinetic energy is equal to a half of I multiplied by omega squared. OK. Where I is a moment of inertia for our object and omega is its angular velocity. Now, we don't know any of these off the bat, but how can we use it to, to how can we figure that out? What can help us? Well, we also know that our moment of inertia for our cylinder, OK, is equal to a half of Mr R squared. OK. We know that our angular velocity is actually equal to our linear velocity V divided by our radius R and we have that information as well. So now if we were to substitute that into our formula, then we will be able to find the energy for our tornado because no substitute in those values to solve for our energy. OK. Then that means our energy, the kinetic energy is going to be equal to a half of a half Mr squared multiplied by omega V divided by R all squared. No, we're almost there. OK? But we don't know what our mass is. We don't know the mass of the tornado. But is there anything else we know that can help us? Well, if we think about it, we know that it's a cylindrical mass of air and we have its density. What relates density to mass? Well, recall as well. OK. Recall that density is actually defined as the amount of mass that we have per unit of volume for a cylinder. Its volume is going to be pi R squared H. So really, OK, its density equals its mass divided by pi R squared H which means its mass is going to be equal to its density multiplied by pi R squared H. So we can use that idea to substitute our value for M into our formula because no, that means it's going to be a half. Well, we can start simplifying from no a half multiplied by a half. That's 1/4. Our mass is raw pi R squared inch and that's being multiplied by R squared. And all of that is being multiplied by V squared divided by R squared. Here, we can cancel R squared O from one of our terms. And that leaves us then our energy to be a half of raw pi R squared H multiplied by V squared. We know what all of these values are. So now let's substitute to figure out the energy linked to our tornado. Sorry, this should be 1/4 not a half my bad. So no, that's going to be 1/4 of rule which we said is 1.2 kilograms per cubic meter multiplied by pi OK, multiplied by our radius 55 m or squared multiplied by the height of the tornado 2000 m. OK? Multiplied by its speed squared. So that's 83.33 m per second squared. Now we can put this expression to solve for our energy. So let's put this into our calculator. Now, when we do that, OK, when we do that, and when we write our answer to 1 to, to 2 significant figures like the rest of our answers, then we should get that. The energy is going to be equal to 4.0 multiplied by 10 to the 10th jewels. Thus, this is the energy linked to our tornado. If we look back at our answer choices, that means option B is the correct answer. Thanks a lot for watching everyone. I hope this video helped

Struggling with Physics?

(II) Hurricanes can involve winds in excess of 120km/h at the outer edge. Make a rough estimate of
(a) the energy, of such a hurricane, approximating it as a rigidly rotating uniform cylinder of air (density 1.3 kg/m³) of radius 85 km and height 4.5 km.

Watch next

Intro to Rotational Kinetic Energy practice set

Struggling with Physics?

(II) Hurricanes can involve winds in excess of 120km/h at the outer edge. Make a rough estimate of(a) the energy, of such a hurricane, approximating it as a rigidly rotating uniform cylinder of air (density 1.3 kg/m³) of radius 85 km and height 4.5 km.

Watch next

Intro to Rotational Kinetic Energy practice set

(II) Hurricanes can involve winds in excess of 120km/h at the outer edge. Make a rough estimate of
(a) the energy, of such a hurricane, approximating it as a rigidly rotating uniform cylinder of air (density 1.3 kg/m³) of radius 85 km and height 4.5 km.