Derivation of r (radius) in terms of f (frequency)? Help please.

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SUMMARY

The discussion centers on deriving the radius (r) of a rotor machine in terms of frequency (f). Key equations include centripetal force (F = mv²/r), linear velocity (v = 2πr/T), and angular velocity (ω = 2πf). Participants emphasize the necessity of centripetal force to keep riders pressed against the wall, indicating that the derivation involves manipulating these equations to isolate r. The consensus is that while the task appears complex, it can be simplified by correctly applying the relevant equations.

PREREQUISITES
  • Centripetal force and its role in rotational motion
  • Understanding of angular velocity and frequency relationships
  • Basic algebraic manipulation of equations
  • Familiarity with the concept of linear velocity in circular motion
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  • Study the derivation of centripetal force in relation to circular motion
  • Learn how to manipulate equations involving angular velocity and frequency
  • Explore practical applications of centripetal force in amusement park rides
  • Investigate the relationship between radius and speed in circular motion scenarios
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Physics students, mechanical engineers, and anyone interested in the dynamics of rotational systems, particularly in the context of amusement rides and centripetal force applications.

BLZ
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Okay, so the job I need to do is derive an equation for the radius of an object in terms of its frequency.

These are the equations that we are allowed to use:

v(Linear velocity) = rω
v=2πr/T
ω (angular velocity)=2πf
f (frequency)= 1/T (time period)
T= 2πr/v
a (centripetal acceleration)=v^2/r
F (centripetal force)=mv^2/r

So now.. my attempt:

r= v/ω
= √ar/2πf
= (√(a*(vT/2π))/2πf
2πfr=√(a*(vT/2π)
2πfr^2= a * (vT/2π)
2π * 2πfr^2 = a * vT

... I'm lost I have so many variables that I need to eliminate and I'm not sure what else to do from here. Help would be very appreciated.
 
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Hi
I am not sure what you are trying to do. Radius and frequency are entirely different quantities and they are not directly related except when you state the situation that you are studying. You have written down some equations that relate them and some other quantities. What are you trying to achieve by jiggling around the original equations in the second half? I think you are trying for something that is impossible or has no meaning. Could you explain more fully, please?
 
sophiecentaur said:
Hi
I am not sure what you are trying to do. Radius and frequency are entirely different quantities and they are not directly related except when you state the situation that you are studying. You have written down some equations that relate them and some other quantities. What are you trying to achieve by jiggling around the original equations in the second half? I think you are trying for something that is impossible or has no meaning. Could you explain more fully, please?

okay, so the whole situation from what I've been told by my teacher is that we are deriving an equation for the radius in terms of frequency for a "rotor machine" aka the type of amusement rides that spin so fast that people are pushed against the walls and can't move. That is all the info that we've really been given for the equation. My teacher says this is a simple derivation but it seems very complicated to me. :(
 
Well, which of the equations you listed in the OP would be useful in figuring out how a person can be pushed up against the outer wall of the ride as it turns? What is keeping the person pressed up against the wall?
 
BLZ said:
okay, so the whole situation from what I've been told by my teacher is that we are deriving an equation for the radius in terms of frequency for a "rotor machine" aka the type of amusement rides that spin so fast that people are pushed against the walls and can't move. That is all the info that we've really been given for the equation. My teacher says this is a simple derivation but it seems very complicated to me. :(

So, now we know the actual scenario, the solution is fairly straightforward. You need a centripetal force to push your body inwards so that it doesn't slip. What sort of value could that involve? (Imagine how hard you would need to push a body against a rough wall, to stop it sliding down.) All you need now is to look in your equation tool box and find an equation that contains centripetal Force, speed and radius. Then arrange it so that F is one side and r is the other.

What did you expect, for such a sophisticated problem? But 'complication' is a very relative thing. :wink:
 
2nd attempt
Okay so I did rearranged an equation first to get...
√ar = v

and inserted it into the equation: F=mv^2/r
and got... F=m*(√ar^2)/r
= m*a*r/r
= ma

3rd attempt:
v = 2πr/T
√ar = 2πr/(2πr/v)
√ar = v ? (Cancelled itself out)

The wrong variables keep cancelling out for some reason. I've done quite the number of attempts and "r" keeps getting canceled out and that's one of the variables that I need. I'm not sure what I'm doing wrong? This equation has force, speed, and radius so did I pick the right equation to use or...?
 
Last edited:
Why are you surprised when the variables to cancel out? You can get all sorts of results when you substitute and reduce.
You start off with the right equation ( F=mv^2/r) and then shoot yourself in the foot by 'tidying it up' too much. Put in the value you think is appropriate for the Force and then you can rearrange to get how V and r are related. Job done, bish bash bosh!
 

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