Dot Physics

Playing with a merry go round

Thanks to reader Cleon for notifying me of this video on youtube. Check it out.

First, some notes.

  • I am sure you noticed that the aspect ratio is incorrect (at least that was the first thing I noticed). The boys must have made a 16:9 video, but then uploaded it to youtube as a 4:3. This doesn’t really affect the analysis, but I had to fix it. I used the awesome firefox plugin NetVideoHunter to download the video and then used MPEG Streamclip to resize the video.
  • Other than that, they did follow my suggestions for making videos. The camera doesn’t move and is mostly perpendicular to the motion (well, kind of).
  • The only thing I could think of to improve the analysisability of the video would be to add some marker on the merry-go-round to easily see the rotation rate and also include something for scale. But really, overall a very good video.

Using Tracker Video Analysis, I could get the horizontal position of one spot of the merry go round. From this, there is a way to get the angular data. However, there is a trick. For an object moving in a circle, I can get a relationship between the angle and the horizontal (I call x) position:

i-001207072c505cf37ce6c5b6e291399f-la_te_xi_t_13.jpg

The problem is that the inverse cosine function is kinda stupid. Look at the following two x-positions (let me assume that r = 1 meter):

i-62a585be068be525f71290ef99fde81b-untitled1.jpg

These two (different) angles have the same x-value. So, I had to go and adjust the angle for each half-turn of the merry-go-round. Here is the angular data I get:

i-8d1beadd3751c0cdab8afc12e6a631d1-untitled_21.jpg

Just looking at the data, it is difficult to see a change in angular speed. The little bumps are there because the uncertainty in the position is not linear with respect to the uncertainty for angle points – that is, small changes around the two ends of the circle can make a big difference.

If you look the video, the boys move in closer to the center around the 1:00 minute mark. This corresponds to the arrow in the graph above (the time is different because I only used a small part of the whole clip). And yes, the merry-go-round did speed up. From the slope before and after this, I can say:

i-5d60157f43c5110ea6bd502634ba95b6-la_te_xi_t_1_13.jpg

So, the big question. Why did the merry-go-round speed up? Cleon made a great point. It would be wrong to say it sped up because angular momentum was conserved. This would be the same as saying that when a bullet is shot from a rifle, it speeds up because linear momentum is conserved. For the bullet, momentum was conserved since the rifle moved backwards. But why did the bullet speed up? It was because there was an energy transfer. Energy was stored in the gun powder and as it exploded, this gas did work on the bullet (and on the rifle).

Ok, it is a little more complicated with the merry-go-round with boys on it. First let me look at the angular momentum before and after they moved. Here is a diagram of the two situations:

i-44df0202aa78a3a8e3228176643dface-untitled_3.jpg

How do you calculate the angular momentum? I don’t think I have talked about this much before – well I need to. Anyway, for a rigid object rotating about a fixed axis, I can write the angular momentum as: (it has to be a about a fixed axis for this to work – see this post about CD players in space for more details)

i-a8a32674346a22ef127156a0fd76fa2b-la_te_xi_t_1_22.jpg

L is the angular momentum vector and ? is the angular momentum vector. I is normally called the moment of inertia, I like to call it the rotational mass. If you call it that, you can see the connection with this formula:

i-230bee5bd0272b44d046b0ee7b00a07a-la_te_xi_t_1_32.jpg

The linear mass is what makes things difficult to change their linear motion and rotational mass is what makes things difficult to change their rotational motion. Get it? The rotational mass depends on two things (when the axis of rotation is fixed) – the mass, and how far this mass is from the axis of rotation. For a discrete set of masses, this could be written as:

i-c6e49002cfcd2be4c567c4601a76925a-la_te_xi_t_1_42.jpg

For the moving boys, they can be treated as point masses. The merry-go-round is essentially a disk. To find its rotational mass, you would have to pretend like it is a whole bunch of discrete masses and add them up – essentially integrate. So, you can see that as the boys move closer to the center, they decrease their rotational mass. The merry-go-round speeds up. In the end, the angular momentum before and after should be about the same (although there is some friction that makes it not conserved). If I estimated the mass and position of the boys, I could get a value for the rotational mass of the merry-go-round. Maybe I will save that for homework.

Comments

  1. #1 Fran
    July 28, 2009

    This is nice, thanks! I need to get better at teaching exactly this (among other things) and I will plan to use this!

  2. #2 chong
    January 1, 2011

    I was doing a question on If somebody on the merry go round just falls off without giving any tangential force on it, will the angular speed of the merry go round change? I thought it should change since the moment of inertia of the merry go round has changed, but the answer for that question claimed that the person did not exert any tangential force of the merry go round and so the angular speed stay the same

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