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Angular acceleration and Steiners theorem

  1. Jul 20, 2011 #1
    Hey all!

    I'm currently doing some classical mechanics re-work and i've stumbled upon something that I don't quite understand.

    Calculating the angular velocity for, say a disc rolling on the ground, that has a tangential foce acting on it, I would view the angular acceleration from the middle of the disc.


    I(c) is Inertia in centre of disc.
    M = alpha*I(c)
    M = F x R = FR
    alpha = FR/I(c)

    For some reason, this is the wrong way of viewing the problem. I need to use steiners and move the I to the edge of my disc for what I have gathered from my failed attempts on many problems like the one presented.

    Why do I have to use steiners theorem and move my point of inertia from the centre to the edge the disc? My general assumption would be that the angular acceleration (and velocity) would be the same in the middle as on any edge as it should be seen from the centre. What I think is it covers just as much angle in the centre as it does on the edge.

    Would someone please explain to me why this is the case? I do have a few general ideas and I've done some reading but I can't find anything definitive, just math heavy explenations (which are definitive, but I dont understand them :) )

    Any help on this matter would be greatly appreciated!

    EDIT: I have no clue if this is coursework related or not. If it is move it...
     
    Last edited: Jul 20, 2011
  2. jcsd
  3. Jul 20, 2011 #2
    The way you're solving the problem, you're not taking into account the translational component of the wheel. There are really two ways to solve the problem and it has to do with your point of reference:

    1) The point of reference of the center of the wheel. You started this way. A disc traveling on the ground has two components of motion with respect to the ground, the rotational component and the translational component. Your solution only takes into the rotational and that would be correct is the disc isn't moving forward and just spinning. You have to somehow take into consideration that linear acceleration also depends on the angular acceleration (like a_x = R*alpha). If you add that term in there, your way can work.

    2) The point of reference on the ground. Here, we're taking advantage of the fact that the point of contact for the wheel and the ground are traveling the same speed: zero. At any moment in time, you could say that the wheel is "rotating" about that contact point. That "instant center or rotation" changes as the wheel rotates, but you can still analyze it as if the wheel is in pure rotation about that instant center. This is why we want to move the moment of inertia to this instant center. This eliminates the need to throw in a translational term and we can just use equations of rotational motion.

    Hope that helps!
     
  4. Jul 20, 2011 #3

    Doc Al

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    Staff: Mentor

    You should always be able to find the angular acceleration by finding torques about the center of mass. Can you give a specific example where that didn't work?
     
  5. Jul 20, 2011 #4
    @ timthereaper

    Thank you very much, it helped alot and im on a good way of understanding this. I'm not quite there yet though :)

    @Doc Al

    Yes I do have an example!

    We have a cylindrical disc which lies on its side with a cord wrapped around it. You pull the cord with F tangentially forward (no angle for F that is relative the cylinder). Find the angular acceleration of the disc.

    m = 100kg
    static friction = kinetic friction = 0.2 (prob won't be)
    F = 500N
    R = 0.3m
    I(c) = 1/2*m*R^2 (Given in question)

    M = alpha*I(c)
    M = F x R = FR
    alpha = FR/I(c)

    I(c) = (100*0.3^2)/2 = 4.5
    alpha = 500*0.3/4.5 = 33.33 rad/s^2

    This is when i'm using I(c) in the middle. Now instead if I'd use steiners and move to a parallel axis R meters from center i'd get

    I(o) = I(c) +mR^2 = 3/2 mR^2
    I(o) = 3/2(100*0.3^2) = 13.5, also im moving down 1 R from F so M = 2RF
    alpha = 2*500*0.3/13.5 = 22.22 rad/s^2

    The difference is huge not only in answer but also understanding.
     

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  6. Jul 20, 2011 #5

    Doc Al

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    Staff: Mentor

    The problem here is that you have failed to consider all the forces. You forgot about the friction at the surface.

    That's not an issue with the second method, since for that case the torque due to friction will be zero.
     
  7. Jul 20, 2011 #6
    Yes, that is true!

    I tried calculating it using M = F*R + 0.2*m*g*R as I assume friction is what it does, it hinders the attempted spin of the plate and creating a force helping my initial pull force. Though using this fact I get a much much higher angular acceleration of 46.42 rad/s^2. Even tried using friction as a negative force but that only got me 20.22 rad/s^2. Can you see where my calculations are wrong?

    Back on the topic.

    Steiners seems useful for cases where I want to ignore any force of friction, such as in this case. If I have no force of friction, say for a hanging JoJo, would it be smart to use steiners again and calculate angular acceleration in I(o) rather than I(c)?
     
  8. Jul 20, 2011 #7

    Doc Al

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    Staff: Mentor

    Two problems:
    (1) You assume that the friction is μmg, which is incorrect. That's the maximum possible value of the static friction force--the actual friction force needed to prevent slipping may be less. Use Newton's laws to calculate the needed friction.
    (2) What direction is the friction acting? As always, draw yourself a free body diagram.

    Why? I'd stick to I(c).
     
  9. Jul 20, 2011 #8
    Wow, you are so correct!
    I've just re-calculated everything and this time used my brain and, low and behold, I got alpha in the exact form I wanted. This without using nifty tricks, just accounting for all forces in all directions.

    I learned a great deal about friction and how to approach and think when we have objects of mass with a momentum.

    I thank you all for helping and I must say this is the best forum i've ever been to :) I'm gonna help around as much as I can!
     
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