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Moment of Inertia/Angular Speed/Torque

  1. Sep 13, 2007 #1
    1. The problem statement, all variables and given/known data
    A flat construction of two circular rings that have a common center and are held together by three rods of negligible mass. The construction, which is initially at rest, can rotate around the common center (like a merry-go-round) through which another rod of negligible mass extends. Mass 1 is 0.12 kg, its inside and outside radii are 0.016 m and 0.045 m, respectively. Mass 2 is 0.24 kg, its inside and outside radii are 0.090 m and 0.140 m, respectively. A tangential force of magnitude 13.0 N is applied to the outer edge of the outer ring for 0.300 s. What is the change in the angular speed of the construction during that time interval?

    2. Relevant equations
    I believe the following is the correct moment of inertia formula for this problem:

    I = (1/2)M(R1^2 + R2^2)

    3. The attempt at a solution
    What I have done so far is calculate each ring's moment of inertia.

    Ring 1:
    I1 = (1/2)M(R1^2 + R2^2)
    I1 = (1/2)(0.12 kg)((0.016 m)^2 + (0.045 m)^2)
    I1 = 1.3686e-4 (kg*m^2)

    Ring 2:
    I2 = (1/2)M(R1^2 + R2^2)
    I2 = (1/2)(0.24 kg)((0.090 m)^2 + (0.140 m)^2)
    I2 = .003324 (kg*m^2)

    This is as far as I have gotten and I am not even sure if the work I have done thus far is relevant. I am not sure what to do with the tangential force provided by the problem. I know F = ma, but I am not sure where that, or if it even does, fit in. Please help, this one has been stumping me for a while. Thanks for your time.
  2. jcsd
  3. Sep 14, 2007 #2


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    Gold Member

    The two rings are a blind. You can treat the construct as one assembly with a MoI given by the sum of the MoI of the two rings.

    So the problem is just to calculate the angular acceleration caused by a tangential force.
    Have a look here http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html and you'll see it's just like a = f/m.
  4. Sep 14, 2007 #3
    Okay, so I used the formula from the reference you gave me, alpha = F/I:

    alpha = (13 N)/(0.00346 kg*m^2)
    alpha = 3757.225434 rad/s^2

    also I know the equation for angular speed relative to angular acceleration:
    w = alpha * time

    w = (3757.225434 rad/s^2)(0.300 s)
    w = 1127.16763 rad/s

    Unfortunately I made a mistake because that is wrong. I don't think I need to use one of the constant acceleration equations because the angular acceleration isn't constant. I thought I was on the right track. Thanks for your time.
  5. Sep 14, 2007 #4
    Nevermind, I figured out where my error was. I don't know why I did this, but I set the torque to just the tangential force. Torque is defined as force * radius. Thus all I needed to do was to multiply the 13 N force in my problem by the outer radius since that is where the force is applied. My final answer came to 157.8034682 rad/s.

    Thanks, Mentz114 for leading me in the right direction.
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