1. Oct 28, 2012

### BlueCardBird

During a lab, we had to plot graph between Centripetal Force vs Radius of rubber stopper (some uniform acceleration lab) the result i got was a horizontal line. Then a question asks Based on your graph of centripetal force vs. radius of rubber stopper's motion, can you conclude that centripetal force is directly proportional to the radius of the rubber stoppers motion? Explain your answer. I'm just wondering if my answer makes sense.

Based on our graph, it states that the theoretical centripetal force is not proportional to the radius of the rubber stoppers motion. Since Fc=m.a and a=4π²r/T² then Fc=m (4π²r/T²), basically we are only graphing the Fc’s relationship to r. This statement can only be held true if the mass is constant, and T is constant. In this case, though m stays constant, T does not, and that is why the graph does not display a direct proportionality between Fc and r.

2. Oct 28, 2012

### sophiecentaur

What were the axes of the graph and what variables did you keep constant? What were the details of the experiment. You haven't told us enough to be able to help you yet. This is a Physics Forum, not a Psychic one.

3. Oct 28, 2012

### BlueCardBird

4. Oct 28, 2012

### sophiecentaur

Think about this. What was the tension in the string balanced by? Could that change? What would happen if T were greater - or less?

5. Oct 28, 2012

### BlueCardBird

The T is dependent upon the mass at the bottom, in our case we used washers, Fg=Ft=Fc

6. Oct 28, 2012

### sophiecentaur

So what dependent variable was staying constant during the experiment (i.e. the vertical axis)? You haven't said what were the axes on your graph.

7. Oct 28, 2012

### BlueCardBird

the vertical axis was the Centripetal Force and Horizontal axis was the Radius

8. Oct 28, 2012

### sophiecentaur

This does not compute for me. The centripetal force is the same as the tension in the string - which is the weight hanging on the bottom. You said that varies - so how do you get a horizontal line if the y co-ordinates (i.e. weights hanging on the string) are not all the same?

9. Oct 28, 2012

### BlueCardBird

the y values are the same, it is that radius that is changing

10. Oct 28, 2012

### sophiecentaur

That makes no sense if you are plotting tension (y) against radius (x) and you say the tension is varying.

11. Oct 28, 2012

### BlueCardBird

no, what I'm saying is that Tension is the same but the radius is changing; also period is changing aswell.

12. Oct 29, 2012

### sophiecentaur

Of course you can arrange for many combinations of radius and speed which will give the same tension. The instructions you quoted included varying the tension. It would have helped if you had bothered to describe your particular experiment instead of leaving us to read that doc (as I asked at the start).

Last edited: Oct 29, 2012
13. Oct 29, 2012

### sophiecentaur

The formula for the tension in the string is
T = mv2/r

If you are keeping T constant, shouldn't you be plotting r against v, rather than r against T, which you are keeping constant by adjusting v?

To yield a useful result, an experiment needs to reveal the relationship between two variables whilst all others are kept constant. Isn't that the basis of the scientific approach?