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Angular velocity - simple question.

  1. May 19, 2006 #1
    Alright, I am doing an experiment that involves rolling a marble down a curved railing in order to find its kinetic energy, both in terms of translation and rotation. My teacher recommended I set up two photogates in order to find the time taken for the ball to travel in between two points on the railing, and then from the length of the railing, the circumference of the marble and the time taken, I figure out the angular velocity and hence the rotational energy .
    I'm sure I'm missing something here (late at night and I'm tired), but wouldn't it be easier to use the translational velocity (which I already know) and just divide that by the circumference?

    Note: I actually never learnt about rotational kinetic energy OR angular velocity.
     
  2. jcsd
  3. May 19, 2006 #2
    Well, it's not that simple.
    Have you taken the moment of inertia of the marble into acount? (You could say it's roughly spherical)
    If the railing is y meters tall, then at the very bottom, by energy conservation:
    mgy = 1/2 mv^2 + 1/2 I w^2
    Where w is the angular speed.
    Because w = v/r you can obtain an expression for the angular speed with a bit of algebra...
    I is the moment of inertia of the object which is = 2/5 mR^2....
    This is, however, assuming that the marble doesn't slip!
     
    Last edited: May 19, 2006
  4. May 19, 2006 #3
    Yes, I'm assuming no slippage. Also, I am taking the moment of inertia into account. I must know though, if we know the translational velocity at a point, it just seems to make sense that I could figure out the frequency of rotation from that (again, assuming no slippage) and then from that obtain the angular velocity and eventually kinetic energy.
     
  5. May 20, 2006 #4
    So what you want to know is whether to use the translational velocity or the velocity you get from the photogates? They're pretty much the same thing, though I have no idea how think you can experimentally get instantaneous velocity.

    Also, if you divide translational velocity by the RADIUS you should get angular velocity in rads per second.
     
  6. May 20, 2006 #5
    Getting the instantaneous translational velocity is quite a nifty trick actually. The railing is set up on the table, curving upwards from left to right. I place a marble at the top of the curve and then release it, it flies off the railing (and table) at a level (horizontal) spot of known height. As the height is known - rearranging s = 1/2 at^2 - I can find out the time taken for it to hit the ground. I then measure the horizontal distance travelled during this time, and assuming no loss of speed just use a simple distance/time equation to calculate the velocity at take off (and travel).
     
  7. May 20, 2006 #6

    Hootenanny

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    Further to what Phoenix said, angular velocity ([itex]\omega[/itex]) is given by the instantateous tangental velocity over the radius. Then you can use the expression;

    [tex]E_{kr} = \frac{1}{2}I\omega^2[/tex]

    Your total kinetic energy would be given by the sum of the rotational and linear kinetic energies.

    ~H
     
  8. May 20, 2006 #7

    Curious3141

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    I would advise a little caution since aerodynamic drag can throw your results off more than you'd expect. Your way is fine if you can neglect air resistance (and you probably can for low final speeds), so I'm probably just being overcautious.

    If you want greater accuracy, I would suggest working in an evacuated chamber if you can manage it. Also, use smoothened soft modelling clay or play dough as the landing surface, so you can accurately capture the imprint of the landing marble with minimal rolling after it lands.

    EDIT : I have a couple of other suggestions if you're interested. When you're choosing the marbles, go for the ones which look homogeneous. A lot of marbles have weird designs inside them, and poorly constructed ones even have air bubbles, etc. Try to go for the very plain, smooth ones.

    In fact, if you just want a spherical object and you're not restricted to using glass marbles, go with steel ball bearings of a decent size. Not only is their construction much more quality controlled, they are smoother and less fragile. Seriously, use ball bearings. The greater density will also mean greater mass for the same volume and a bigger imprint on the clay at the end, meaning easier measurement.

    I don't know what exactly you're hoping to study here, but you can study some nice things here, like get the moment of inertia of that particular projectile, and if you want, even study the phenomenon using spheres of different mass and radius, to work out a relationship between moment of inertia, mass and radius for a sphere in general and see how it compares to the theoretical expression. If you can use spheres of different materials (glass and metal), then you can vary radius and mass independently.
     
    Last edited: May 20, 2006
  9. May 26, 2006 #8
    Yes thanks for your advice. I, of course, realise that air resistance has an effect, but due to the constraints on the equipment available to me (a school lab) I can't do much about that. Much of what you said I'm already doing, however. I'm using a box full of sand to identify where the ball lands. Also, coincidentally, I moved on to ball bearings when I began to expand my experiments to spheres of different and radii. I found that when I was trying to see if marbles were similar (comparing volumes and masses), in some cases the densities were quite different.

    Besides, a less than perfect experiment gives me more to write about ; ). The ball-bearings ought to travel the same distance, but they didn't - the larger ones travelled further, so that gives me a bit to write about concerning air resistance and friction etc...
     
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