Question

An object with mass M1 lies on a friction-less surface. It is tied, through a pulley, to an object with mass M2, which is hanging from the pulley, vertically downward. How can you calculate the acceleration of the first object?

Answer 1

As I couldn't draw in my original question, here is the freebody diagram of the question above: The second object is affected by gravity and I have to calculate the acceleration of the first object.

Answer 2

Well, both objects must have the same acceleration, because they're connected by a string, right? So what's the acceleration of the first object?

Answer 3

I'm not sure. The acceleration of M2 is gravity, but the movement of the entire system can't be equal to gravity, because when you increase the mass of M2, M1 moves faster.

Answer 4

The acceleration of M2 is not -9.8 m/s^2. If I consider down to be positive,
\[m_2 a = m_2 g - T\]
And for my first mass,
\[m_1 a = T\]
So I can certainly plug the second into the first to get
\[m_2 a = m_2 g - m_1 a\]
so
\[(m_1+m_2)a = m_2 g\]
and
\[a = \frac{m_2}{m_1 + m_2} g\]
downward.

Answer 5

The problem is that I am doing a lab report based on the problem above, where my group timed "M1" for various masses in M2. M1 is a constant mass, and is suspended in an "air track" to essentially eliminate friction. My teacher told us to find acceleration using \[s=(1/2)a t^{2}+ut\] where u is 0 (starts at rest). These are the experimental values. But when I find the "literature acceleration" (with your formula), I get percentage error of -58% and such. Such large error worries me, and it can't be possible we were THAT bad at timing to get that error. I've double checked both my expermiental and literature values, so now I'm not sure what to do! I can give you the excel document I was working with if you'd like.

Answer 6

Sure, I'll take a look

Answer 7

Here it is:

Answer 8

Okay, I've gone through this several times and I have a few guesses as to what is going on here. I promise, the equation I provided you is correct. So, I started searching for sources of error. I assume the track length and the masses are known to ~1% precision or better, so I did not take those into account. That leaves only your timing. Now, you said the timing couldn't have been bad enough to do this much damage, but I'd like to draw your attention to several things.

Answer 9

Notice that your "experimental error" gets larger and larger. It's not randomly distributed about some average error, it increases. Secondly, note that except for the first time, it's always negative, which means your experimental value for the acceleration is too large. This implies a systematic error, not random fluctuations.

Answer 10

The fact that your error increases as the time measured decreases suggests that the timer had a tendency to anticipate the end of the run by a constant amount (no more than a quarter of a second or so) but as the total run time decreased, the proportional error got larger and larger.

Answer 11

Thanks alot! I was cracking my head thinking that it couldn't be possible to be THAT off. I converted the literature acceleration to time in seconds and realized that, as you said, timing was off by 0.15-0.2 seconds (except the first one). This is reasonable for a human, but once the formulas are applied these values got way off. Thanks again, you have saved my day.

Answer 12

No problem! If you ever get the opportunity to do it again, I would advise trying to do it with a longer track or heavier glider to increase the total time of the runs and decrease your fractional error. As you can see your first one is quite close!

Answer 13

Thanks! I'll keep that in mind and tell it to my teacher for next year.

Answer 14

Something I just noticed, but I'm not sure if it is true, if we re-write your formula as: \[y=(x/(0.383+x))*9.81\] we get a horizontal asymptote at 9.81. Does this mean that if we place an infinite mass as M2, its maximum acceleration will be gravity?