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

In summary, the conversation involves a student attempting to derive an equation for the radius of an object in terms of its frequency, specifically for a "rotor machine" amusement ride. The equations they are allowed to use are v=2πr/T, ω=2πf, f=1/T, T=2πr/v, a=v^2/r, and F=mv^2/r. The student is struggling to eliminate variables and is unsure if they are using the correct equation, but it is suggested to start with F=mv^2/r and substitute in a value for force to rearrange and find the relationship between v and r.
  • #1
BLZ
3
0
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.
 
Physics news on Phys.org
  • #2
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
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. :(
 
  • #4
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
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:
 
  • #6
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:
  • #7
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!
 

1. What is the formula for finding the radius in terms of frequency?

The formula for finding the radius (r) in terms of frequency (f) is r = c / (2πf), where c is the speed of light.

2. How do you derive the formula for radius in terms of frequency?

The formula for radius in terms of frequency can be derived by using the formula for the circumference of a circle, which is C = 2πr, and the formula for the speed of light, c = fλ, where λ is the wavelength. By substituting c = fλ into the formula for circumference, we get C = 2πrf. Solving for r gives us r = c / (2πf).

3. Can the formula for radius in terms of frequency be used for any type of wave?

Yes, the formula for radius in terms of frequency can be used for any type of wave as long as the speed of the wave (c) is known.

4. How does the radius change as the frequency increases?

As the frequency increases, the radius decreases. This is because the wavelength (λ) is inversely proportional to frequency, so as frequency increases, the wavelength decreases. Since the radius is half of the wavelength, it also decreases.

5. Is there a limit to how small the radius can be in terms of frequency?

There is no limit to how small the radius can be in terms of frequency. However, if the frequency is extremely high, the radius may reach a point where it becomes too small to measure accurately.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
618
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
82
Replies
10
Views
312
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
23
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
884
Back
Top