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How does the moment of inertia scale?

  1. Jul 26, 2013 #1
    1. The problem statement, all variables and given/known data

    If we multiply all of the design dimensions by a scaling factor f, it's volume and mass will be multiplied by f^3. (a) By what factor will its moment of inertia be multiplied? (b) And if a 1/48 scale model has a rotational kinetic energy of 2.5J, what will be the kinetic energy for the full-scale object of the same material rotating at the same angular velocity?

    2. Relevant equations

    3. The attempt at a solution

    I have absolutely no idea what they are talking about. I feel like I missed an entire chapter, but I am sure that I didn't. All the other questions in my book leading up to this one have been the usual ones that ask me to find a value when dealing with the moment of inertia. What is this asking.
     
  2. jcsd
  3. Jul 26, 2013 #2
    Scale Factors

    If you look at some of the moment of inertia equations you know, you'll see how we can address this question.

    For a point mass, it's [itex]I = \frac{2 m r^2}{5} [/itex]. For a solid sphere, [itex]I = \frac{2 m r^2}{5} [/itex]. For a hollow sphere, [itex] I = \frac{2 m r^2}{3} [/itex]. Rod about the end [itex]I = \frac{1 m r^2}{3} [/itex]. Rod about the middle: [itex]I = \frac{1 m r^2}{12} [/itex].

    The only thing that is changing is the constant in front.

    The moment of inertia is given by the equation [itex]I=k*m*r^2[/itex], where k is the constant that depends on the shape. So the moment increasesas the mass increases, and increases like the square of the radius.

    To find the scaling, you just multiply each quantity by a scale factor, f. We have to multiply by f3 for mass (since mass increases with increasing *volume*). We multiply each term of r by f.

    Hope that helps.

    Dr Peter Vaughan
    BASIS Peoria Physics
     
  4. Jul 27, 2013 #3
    Oh, thank-you so much!!!!
     
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