Calculating tangential velocity of an air parcel circulating a tornado

[Timebomb3750]

Homework Statement

If an air parcel 2000m from a tornado center has a tangential velocity of 10 m/s, what is the resultant tangential velocity if the parcel is 100m from the tornado center.

Homework Equations

v = rω, where r = radius, and ω = angular velocity
ω = v/r

The Attempt at a Solution

I understand that the tangential velocity of the air parcel should be much faster as radius drops. But how am I supposed to calculate tangential velocity of it requires the calculation of angular velocity which requires velocity. I'm really lost.

 

[NewtonsHead]

Angular Velocity is not dependant on radius, therefore, it will be the same for each radius. You can find angular velocity with the equation ω = v/r. Use that value to find the tangential velocity at the new radius.

 

[Timebomb3750]

Angular Velocity is not dependant on radius, therefore, it will be the same for each radius. You can find angular velocity with the equation ω = v/r. Use that value to find the tangential velocity at the new radius.

Okay, here's my work.

ω = 10 (m/s)/ 2000m = .005 radians/sec

Then I put that into the tangential velocity question:

v = rω = 100m*.005 radians/sec = .5 m/s

Why is it that the tangential velocity got slower as radius decreased? I thought it was the other way around as I mentioned in my first post.

 

[NewtonsHead]

The tangential velocity is slower near the center of the disk because it doesn't have to circle around as big of a radius as the outside. Imagine running around a circular track... the people on the outside have to run faster to keep up with the people on the inside lanes. This is not true for planets though, because the inner planets are affected by the sun's gravity more and travel faster in the tangential direction to maintain their orbits.

 

[Nickie]

Calculating the tangential velocity of an air parcel circulating a tornado requires an understanding of the relationship between tangential velocity, radius, and angular velocity. As the air parcel moves closer to the tornado center, its radius decreases, resulting in an increase in tangential velocity. This can be calculated using the equation v = rω, where v is the tangential velocity, r is the radius, and ω is the angular velocity. In this case, we are given the tangential velocity of 10 m/s at a radius of 2000m. To calculate the resultant tangential velocity at a radius of 100m, we can use the equation ω = v/r to find the angular velocity, and then plug that value into the original equation to find the new tangential velocity. It is important to note that the angular velocity of a tornado is not constant and can vary greatly, so this calculation will only provide an estimate of the resultant tangential velocity. Additionally, other factors such as air pressure and temperature can also affect the tangential velocity of an air parcel in a tornado.