Can interior pressure of a tornado be calculated?

AI Thread Summary
The discussion focuses on calculating the interior pressure of a tornado using centrifugal speed and radius. It suggests that while a simplistic model assumes uniform angular velocity and density, this approach is flawed. The pressure gradient can be derived from the air density and centrifugal acceleration at a specific distance from the tornado's axis. By calculating the centrifugal acceleration at the tornado's rim and integrating the pressure gradient inward, one can estimate the interior pressure. This method provides a theoretical framework for understanding tornado dynamics.
skeptic2
Messages
1,776
Reaction score
60
Can the interior pressure of a tornado be calculated from the centrifugal speed of the tornado and its radius (centripetal force)? For example, given a tornado with a 1/2 mile radius and a centrifugal speed of 200 mph, how much would the interior air pressure be?
 
Physics news on Phys.org
A simplistic calculation would assume that the air mass is rotating as a unit with a fixed angular velocity and a fixed density. [That's wrong, but it's a start].

You could then get the pressure gradient based on the density of air and the centrifugal acceleration at a given distance from the axis. Add up the pressure gradient along a path working from the outside to the center and you have an answer.

Why don't you begin by calculating the centrifugal acceleration at the rim?
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...

Similar threads

Replies
15
Views
4K
Replies
5
Views
2K
Replies
10
Views
5K
Replies
5
Views
2K
Replies
21
Views
3K
Replies
3
Views
2K
Back
Top