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Homework Help: Rotational momentum (conservation problems)

  1. Jan 9, 2010 #1
    1.A rigid structure consisting of a circular hoop (on the right) of Radius R and mass m, and a square (on the left) made of four thin bars, each of length R and mass m. The rigid structure rotates at a constant speed about a verticle axis, with a period of rotation of 2.5 s. Assuming R = 0.50m and m = 2.0 kg, calculate (a) the structures rotational inertia about the axis of rotation and (b) its angular momentum about that axis

    There is a picture associated with the question but i cant upload it. Basically on the left is a square then right to the right of tht is a circle and inbetween the 2 there is a line drawn shoing the axis of rotation, the square is touching the circle.

    2. Rotational inertia for the hoop = 1/2 MR^2
    Parallel axis theorm = Icom + MH^2
    I dont know how to find the rotational inertia of the square that is composed of 4 thin bars each of length R. The rotational inertia equation for the hoop was already given to us in a previous table in the book, but nothing for a thin rod rotating about its axis type of thing.
    L = Iw w=(omega)

    3. i couldnt do anything for this question as i couldnt find the rotational inertia of the square, I dont know how to.
  2. jcsd
  3. Jan 9, 2010 #2
    I can't really picture what the system looks like, but you should be able to find the moment of inertia of the square as the combination of 4 bars (you should be able to look this up, or calculate it yourself) using the parallel axis theorem
  4. Jan 9, 2010 #3
  5. Jan 9, 2010 #4
    hi i did look those up but the problem with it is that the questions usually tell you what to assume the thing is. It says thin bars not rods and it doesnt say to assume rods. So i asnt sure if there was something i was missing.

    Also for all the moments of inertia of the rods that are parallel to the axis of rotation, there is no specified equation for the a thin rod with the axis of rotation down the rod; theres only perpendicular.

    ---- |
    | || This is kind of what the diagram looks like
    | || Except in this spot there is a circle that touches the axis of rotation
    ---- |

    I dont know i that helps.

    edit: seems it doesnt keep the formatting correclty. Ill try to get the image online so you can see it
  6. Jan 9, 2010 #5
    So they don't give dimensions of the bars (other than length)? My guess is the moment of inertia will be the same as for a rod since the other dimensions are very small so they can be ignored. This also means that the moment of inertia around the rod's axis is basically 0, but you can still apply the parallel axis theorem.
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