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Homework Help: Angular Velocity Problem

  1. Dec 12, 2011 #1
    1. The problem statement, all variables and given/known data
    A large turnable shaped like a wooden board with sizes A and B and mass M rotates. A drop of water with mass m drops on 80cm from the axis of rotation. Find the final angular velocity of the system if the initial angular speed is ω.

    2. Relevant equations
    I know how to work this question out if it were to land on the edge of the turntable. However, I am not suree what to do if it lands 80cm away from the axis of rotation. I am not really looking for the answer, just the theory to work it out. I don't have any values... I just made the question up. Thanks!

    3. The attempt at a solution
    he angular momentum before the drop drops equals the angular momentum after the drop drops, or

    Lb = La (b, a refer to before and after)

    since angular momentum, L, = I w where I is the moment of inertia and w the angular velocity,we need to find the moment of inertia of a rectangle of sides, A, B

    the moment of inertia of a rectangle around an axis perpendicular to the plane and passing throught the middle of the plane is

    I=1/2 M(A^2+B^2)

    so we have:

    1/2 M(A^2+B^2) w = Ia wa

    the moment of inertia after is the original moment of inertia + the moment due to the drop of water; for point masses (and we consider a drop a point mass), this contribution is mr^2 where r is the distance from the rotation axis

    the pythagorean theorem tells us that the distance of m from the rotation axis is

    r^2=(A/2)^2 +(B/2)^2 = 1/4(A^2+B^2)

    so we have:

    1/12 M(A^2+B^2) wb = [1/12M(A^2+B^2)+1/4 m(A^2+B^2)]wa

    collect terms and solve for wa (notice, interestingly, that the (A^2+B^2) terms drop out)
  2. jcsd
  3. Dec 12, 2011 #2

    Simon Bridge

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    Science Advisor
    Homework Helper

    Welcome to PF.
    Is there a question in there?

    I have one though:
    I thought A and B were the lengths of the sides of the rectangular turntable?
    Surely this radius is, therefore, the distance from the center to each corner?
    Isn't the radius to the drop supplied to you in the question (80cm)?

    Note: if I wrap "tex" tags around that, I get:
    [tex]r^2=(A/2)^2 +(B/2)^2 = (A^2+B^2)/4[/tex]
    ... just saying.
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