# Atwood pulley and calculating inertia based on slope (m)

1. May 6, 2014

### mack2014

I am working on a lab that is meant to test the validity of (m1

2. May 6, 2014

### mack2014

Sorry. For some reason I keep trying to post and it only posts part of the first line. Trying again

I am working on a lab that is meant to test the validity of (m1 m2)g=(m1 +m2 + I/R^2 )a . I created an Atwood pulley system. The system is accelerated due to the difference in weight on both sides of the pulley. After completing all the time trails for various weights I was asked to calculate acceleration using first a = [(m1 – m2)g] / m1 + m2 and then linearize to find slope and y intercept. I found the slope to equal m1 + m2 and the y intercept was 0.

I was then asked to use the averaging time for each trial and calculate a based on the equation
h = y - y0 = 0.5at^2, then graph [m1 – m2)g versus a, and perform a linear fit to find the slope and y intercept. My equation was y = 0.2109x + 0.03274 . Using the actual measurements, and neglecting moment of inertia, the slope was not equal to the mass because the pulley's inertia is actually there, although small enough that it is "lost" inside my measurement errors.

My question is, how do I calculate the moment of inertia from y = 0.2109x + 0.03274 , in particular, from the slope, which is what we were asked to do, keeping in mind the whole point of the lab is to test (m1 m2)g=(m1 +m2 + I/R^2 )a . I feel like it is something simple that I am missing.

Any help would be greatly appreciated!

3. May 6, 2014

### paisiello2

Doesn't (m1 +m2 + I/R^2 )= 0.2109 ?

You know m1, m2, and R, so solve for I.

4. May 6, 2014

### mack2014

I thought that at first but my mass for m1 is 0.11975kg and m2 is 0.11975kg. They add to 0.23385kg which is much greater then .2109..... I thought my graph may have been backwards, but the directions indicate that I am to plat (m1 - m2)g on the Y axis and acceleration on the x. Also, we aren't given R and although I can measure it, there is no mention of measuring the pulley to find it

5. May 7, 2014

### paisiello2

I would have thought there were varying m1 and m2 values for different "a" values to get a linear plot.

And I am not sure that you can determine "I" without knowing "R".

6. May 7, 2014

### Simon Bridge

I can see you got confused... lets recap:
How did you get this relation? Is it a typo?

If you plotted $y=(m_2-m_1)g$ vs $x=a$ and you get line $y=bx+c$, you compare that to the relation you are testing which was (assuming typo):
$(m_1 - m_2)g=(m_1 +m_2)a + Ia/R^2$

This is $y=(m_1+m_2)x +Ia/R^2$ so compare with your $y=bx+c$

See why the slope was expected to be $b=m_1+m_2$?

But That is not actually valid since $a$ is a variable - you cannot have it in the constant c. You have to be a bit more canny than that. Which is where you have got stuck yeah?

The actual theoretical line should be: $y=bx: b=m_1+m_2+I/R^2$
You know $b$ so you can rearrange the equation ... which is the suggestion in post #2.

You are still stuck though because:
... which shows a great deal of confusion.

Lets take this one at a time:

While $m_1$ and $m_2$ are not held constant, $M=m_1+m_2$ is a constant if you varied the masses by moving a weight from one side of the machine to the other.
Is that what you did?

Recheck the value of $M$.

You should always make sure you understand the reasons for the instructions. If you just follow them blindly in the lab, then you will make mistakes.

The lab instruction do not always tell you absolutely everything you must do - you are expected to be able to figure out the basic stuff yourself. Notice that your theoretical line needs a value for R ... unless you know another relation involving R that you can use to cancel it out of course?

Further notes:

Don't forget that measurements usually have units.

Don't forget to use your uncertainties to tell you how many decimal places you should keep.
(You may not know about uncertainties... but you may know some rules-of-thumb for rounding off.)

You may need to comment on the fact you got a non-zero y-intercept when the theory predicts it should be zero.
What does this mean, if anything?

7. May 8, 2014

### mack2014

Thank you for the response, I am working through it now. We are actially given
(m1 m2)g=(m1 +m2 + I/R^2 )a in the lab. The whole point of the lab is to derive this equation..

i just realize i posted m1 and m2 as the same number which they are not. The total mass (M) was 0.23385kg though. I did vary m1 and m2 to produce different acceleration rates. My value for b (slope) though is only 0.2019. This number is smaller than M (0.23385), it doesn't make sense to me. So yes, I am confused. My c value is 0.03264. If I measure the radius it is 0.02m.

8. May 8, 2014

### mack2014

So I think I have figured it out.... my acceleration is off, probably due to my time calculations. Therefore everything is off as Inertia should be a positive value not a negative value.

Thank you both for your help. :)

9. May 8, 2014

### Simon Bridge

Were you able to derive this equation algebraically - i.e. via a free-body diagram?

I ask because the equation does not seem to correspond to anything you have been asked to do.

... good, how did you vary m1 and m2. That was the important part of the question.

This all may be moot now anyway - but for the future, this sort of detail is usually important ;)

10. May 20, 2015

### SteveS

I'm doing the same lab and he is making a mistake in the formula. The formula given in the lab manual is:

(m1 - m2)g = (m1 + m2 + I/R^2) a

This maybe a source of his problem.

In my lab, I had one of those folding metal clasp type paper clips attached to each end of a piece of dental floss. Mass 1, a 100g mass, had 6 Canadian nickels taped to the bottom and was then hooked into one arm of the folding paper clip. Mass 2, also 100g mass, had 6 Canadian dimes held in the clasp paper clip. After running 5 trials one of the dimes from mass 2 as moved to the folding clip on mass 1. The masses of the coins were obtained from the Canadian mint. Every dime moved from mass 2 to 1 decreased mass 2 by 1.75g and increased mass 1 by the same.

Last edited: May 20, 2015
11. May 20, 2015

### SteveS

I'm actually wondering about this point. At when the masses are balanced m1 - m2 = 0, acceleration should also be 0. However, I'm not totally sure why the graph of mine shows a y-intercept greater than zero.

Does it have to do with error introduced by margins of error in my measurements?

12. May 22, 2015

### Simon Bridge

How it comes about will depend on the detals of the experiment. The theory you have used assumes ideal circumstances - like the pulley having zero moment of inertia and everything being frictionless. You should have a think about how non-ideal circumstances would affect the outcome of the experiment.

Statistical uncertainties will mean that the "best fit" line will not go through the origin - this is normal: it would be surprising if it did - but the origin should lie within reasonable uncertainty limits of the line. So you want to decide if the y intercept is a statistically significant distance from the origin.

I suspect that, at your level, you won't be expected to do much - just noticing and showing you have thought about it would be enough for now.