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

1. Apr 25, 2014

### BLZ

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.

2. Apr 25, 2014

### sophiecentaur

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?

3. Apr 25, 2014

### BLZ

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. :(

4. Apr 25, 2014

### SteamKing

Staff Emeritus
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?

5. Apr 26, 2014

### sophiecentaur

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.

6. Apr 26, 2014

### BLZ

2nd attempt
Okay so I did rearranged an equation first to get...
√ar = v

and inserted it in to 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 cancelled 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: Apr 26, 2014
7. Apr 26, 2014

### sophiecentaur

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!