# Homework Help: Centripetal force lab: meaning of slope of radius vs Fc graph

1. Jul 15, 2011

### 5.98e24

1. The problem statement, all variables and given/known data
I have a graph with the radius on the x axis and Fc on the y axis. I had to then calculate the slope of this linear relationship. I did that, and I got 0.17.

The problem is that I don't know what this represents. mass? acceleration??

2. Relevant equations
Fc = m4π^2rf^2

btw π = pi = 3.14

3. The attempt at a solution
I tried to isolate for the slope of the line, which would be making the left side equal to Fc/r.
This is what I got:

Fc/r = m4πf^2

I know that [STRIKE]Fc[/STRIKE] mass is kept constant.. then I have no idea where to go from there :S

Last edited: Jul 15, 2011
2. Jul 15, 2011

### cupid.callin

i dont think it represents something in general,
its just like change in Fc with unit change in radius ...

3. Jul 15, 2011

### gdbb

How can $F_c$ be kept constant if you got a slope of 0.17 for an $F_c vs. r$ graph?

4. Jul 15, 2011

### 5.98e24

Ah whoops! I made a mistake; Fc isn't constant. I was looking at a graph I extrapolated from. Apologies!

I mean that mass is constant.

So does it really not represent anything significant?

5. Jul 15, 2011

### cupid.callin

Oh!! I didnt notice OP wrote Fc is constant.
I Need lenses now ...

6. Jul 15, 2011

### gdbb

Well, force divided by radius gives units of $\frac{kg}{s^2}$, which, technically, is the unit for surface tension. However, I don't think you wanted to find that surface tension is equal to $0.17 \frac{kg}{s^2}$.

You could just note that, as the radius increased from r = a to r = b meters, the centripetal force increase by 0.17 Newtons per meter.

7. Jul 16, 2011

### cupid.callin

you mean $\frac{kg}{m^2}$.

8. Jul 16, 2011

### gdbb

I'm talking about surface tension in fluid dynamics. I was just trying to give an extreme example of how little "kilograms per second squared" means in a Circular Motion context, as $\frac{kg}{s^2}$ is also the unit of $k$, the spring constant, which is much more likely to be the case in the Introductory Physics sub-forum.

9. Jul 16, 2011

### RTW69

What is "f" in your equation, is it angular velocity? Is it 2*pi or 2*pi^2, you show it both ways? Does your graph have a "y' intercept? What is it? You say the slope is .17 what is the rest of the equation in the form Y=mx+b? Y is Fc, x is the radius in meters, the slope has units of Newtons/meters or kg/s^2. I am guessing the slope is mass* angular velocity^2

10. Jul 16, 2011

### cupid.callin

Oh yes ... i got a bit confused :shy:

11. Jul 17, 2011

### 5.98e24

The y-intercept is 0.

f is the frequency.

The only thing is that we didn't learn about angular velocity. Is that the only thing that it can represent?

Last edited: Jul 17, 2011
12. Jul 17, 2011

### Redbelly98

Staff Emeritus
Welcome back to PF.

Okay, so the slope equals $4 \pi^2 m f^2$. If you know m and the value of the slope, you can calculate what the constant f is. Or, if you know f and the slope, you can calculate what m is. But -- was frequency f really a constant throughout your experiment?

By the way, does your lab apparatus resemble something like the following?

13. Jul 17, 2011

### 5.98e24

frequency f was not constant. The only constant values were Fc (centripetal force), which was 0.5N, and mass, which was 0.03kg.

are you saying that I can't use the equation if f isn't constant?

Yes, it is similar.

14. Jul 17, 2011

### Redbelly98

Staff Emeritus
Thanks, that helps explain better what is going on.

Something is wrong here. If Fc is constant, and you plot it on the y-axis, you should get a horizontal line, so slope=0.

It looks like the variables here are f and r, so the graph should involve those quantities instead. But, just graphing f vs. r will not give a straight line. Did you get any instruction in what you are supposed to actually graph?

15. Jul 17, 2011

### 5.98e24

This was similar to our procedure starting on page 4: http://schools.hwdsb.on.ca/highland/files/2011/01/centripetal_force.pdf

So basically the question I'm trying to answer now is #23.

The graph of radius and Fc was made after doing #20 and #21 in the PDF. The values of Fc were extrapolated from a graph of f^2 on the x axis and Fc on the y axis, seen in #20.

Last edited by a moderator: Apr 26, 2017
16. Jul 17, 2011

### Redbelly98

Staff Emeritus
Got it.

Looks like you have already seen that the graph is linear or a direct proportion, and that the slope should equal $4 \pi^2 m f^2$. You can measure the slope from your graph and, since you know m and f, compare the measured-from-the-graph slope value with what you get by calculating $4 \pi^2 m f^2$.