Reversing angular and linear formulas to find rpm.

  • Thread starter jrjack
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  • #1
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Homework Statement



A simple model of the core of a tornado is a right circular cylinder that rotates about its axis. If a tornado has a core diameter of 200 feet and maximum wind speed of 240 mi/hr (or 352 ft/sec) at the perimeter of the core, approximate the number of revolutions the core makes each minute. (Round your answer to one decimal place.)

Homework Equations



s=r(theta) and V=r(omega)

The Attempt at a Solution



The practice problems had a radius and rpm given and I could find the angular and linear speeds, where I'm struggling is I'm not sure how to reverse the example in order to convert the speed and radius into rpm.

V=r(omega) then (omega)=V/r ???is that right? then when I plug the numbers:
omega=(352ft/sec)/(100ft/rad)....shoud I multipy the reciprocal being (1rad/100ft)
then get: omega=3.52rad/sec...
somewhere I'm sure pi has to come in, 2pi=1rpm...so do I multiply by 1/2pi ??? this is where I'm getting lost.
 

Answers and Replies

  • #2
BruceW
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omega=3.52rad/sec is correct.
Now you must remember what the definition of omega is - it is the angular frequency, and it is given by the equation:
[tex] \omega = 2 \pi f [/tex]
(where f is frequency). So using this you can calculate the frequency in units of 1/seconds. Then, the question asks for it in units of 1/minutes, so you do the unit conversion to get that.
 
  • #3
BruceW
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It looks like my latex isn't working, but you can see what the equaton is anyway.
 
  • #4
BruceW
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just to make sure: omega = 2pi times f
 
  • #5
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so 3.52rad/sec times 1/2pi rev/rad = 3.52rev/2pi sec
then times 60= 211.2 rev/ 2pi min
which rounds to: 33.613 or to 1 decimal place of 33.6 rpm


edit: I fat-fingered my calculator and got the wrong answer the first time.

Thanks for your help.
 
Last edited:

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