# Mass of a rubber stopper being swung in a horizontal circle

1. Jul 20, 2014

### CharlesDerpwin

1. The problem statement, all variables and given/known data
The problem statement, all variables and given/known data[/b]
The purpose of this lab is to determine the mass of a rubber stopper being swung in a horizontal circle.

I made an apparatus that resembles the picture attached, with a string (with a rubber stopper on one end, and a weight on the other end) put through a tube. I whirl the tube above my head so that a horizontal circle is made by the stopper.

The radius was increased from 0.25m, to 0.45m, to 0.65m, to 0.85m.
This increased the period of the stopper's circular motion, 0.330s, 0.37575s, 0.45925s, 0.50625s, respectively.

The lab requires me to draw graph of my results, with period squared in order to straighten the graph, and calculate the mass of the stopper using the slope of my graph

2. Relevant equations
Fc=4(pi)2mr/T2

3. The attempt at a solution
using the equation 4(pi)2mr/T2, I rearranged it and found that r is proportional to t2, as everything else in this experiment is supposed to be constant.

Graphing the radius and period2, the slope of the line of best fit was 0.25555555s2/m.

Now this is where I'm not sure what to do, using the slope to find the mass of the stopper.

Thanks for reading this and I hope you can help me out

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Last edited: Jul 20, 2014
2. Jul 20, 2014

### yeahgene

I assume you know the value of the weight you used that is at the end of the string. The radial force, is of course, equal to this weight $w$. So you can write $$r = \frac{w}{4\pi^2 m} T^2$$. The slope you calculated is equal to $\frac{w}{4\pi^2 m}$ where you can get the mass you're looking for.

3. Jul 21, 2014

### CharlesDerpwin

So, in your equations, W is the force of gravity on the hanging weight, which is also the centripetal force on the stopper?

I was, in fact, given the mass of the hanging weight, 200g

In that case, 0.200kg x 9.81m/s2 = 1.962N
Now 1.962N is my Fg and also Fc, I can put it in this as W in $\frac{w}{4\pi^2 m}$, this expression is equal to my slope of 0.255555555, and i solve from there
$0.255555555=\frac{1.962}{4\pi^2 m}$
giving me a mass of 0.194470594kg?

4. Jul 21, 2014

### yeahgene

To be clear, the centripetal acceleration is equal to the tension on the string, the tension in the string is equal to the force due to the gravity on the hanging weight.

Does a mass of 0.1945 kg or 194.5 g for the rubber stopper makes sense? Which implies it is as massive as the weight.

5. Jul 21, 2014

### CharlesDerpwin

ahh, of course that mass makes no sense. The equation should be more like
$\frac{1}{0.2555555555}=\frac{1.962}{4\pi^2 m}$
I forgot that the slope is really $\frac{0.2555555555s^2}{1m}$
Now I can manipulate the equations:
$m=\frac{(1.962)(0.2555555555)}{4\pi^2}$
so the mass of the stopper = 0.0127006104kg, or 12.7 grams

Much more reasonable, and as a matter of fact, I weighed the stopper and it was 12.67g!

Thanks so much for the help!