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Moment of Inertia lab

  1. Oct 29, 2007 #1
    1. A solid cylinder is being rolled down an incline, and we are looking for acceleration. I can find the observed value of acceleration fairly easily, but am not sure how to find the accepted one. Below are the instructions given on how we are supposed to find the accepted acceleration equation:
    Show me, without any numbers, how to determine the equation for the Accepted Value for the acceleration down the incline of the Solid Cylinder. Make sure that you include your derivation of the moment of inertia of the Solid Cylinder and all the necessary equations to determine the acceleration. (In other words, don't just copy derived formulas out of the book willy-nilly.)

    It is assumed that there is no slipping while it rolls.

    2. I=(integral)r^2dm, acceleration(inst)=dV/dt

    3. using conservation of energy and solving for V, then taking the derivative of that with respect to t...
    Last edited: Oct 29, 2007
  2. jcsd
  3. Oct 29, 2007 #2
    Ok, so I have looked at it a little bit more, and tried the conservation of energy thing again, I get to a=[tex]\sqrt{(2mgh_{i}-I\omega^2)/m}[/tex]dt

    is this even on the right track?
  4. Oct 30, 2007 #3

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    Show us how you got that.

    Describe the physics of the whole situation, as you understand it. We can always give you the answer.
  5. Oct 30, 2007 #4
    Yea, ok, I got close now I think.

    Because it is rolling without slipping conservation of Energy is true.

    I already went through and derived the moment of inertia for a solid cylinder and got the equation, 0.5MR[tex]^{2}[/tex] (I know that is right :p)

    Ok so Me[tex]_{f}[/tex]=Me[tex]_{i}[/tex]

    Setting the Zero Line at the bottom of the incline:





    Now I know it is just algerbra... but how would I solve for V there... because I would get V in the anser if I distributed...

    But after I find V it is just Uniformly accelerated Motion so I can get it right after that one problem.

    EDIT: Don't know why the subscripts are messed up but they are a few times where they shouldn't be... Which is just strange...
  6. Oct 31, 2007 #5

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    1. Look at the 3rd last and 2nd last equarions. You have written, v/r in place of omega^2. Correct that. Then differentiate wrt t.

    2. Also, this vf is nothing but v, since you are considering any point on the path.

    3. If the angle the plane makes with the horizontal is theta, then h=x*sin theta, where x is the dist actually travelled down the plane. Draw a diagram.

    4. accn a=dv/dt and v=dx/dt.

    5. Tell us how much accn 'a' is in terms of g and theta.
  7. Oct 31, 2007 #6
    I did mess up writing the V/r it should have been (V/R)^2 because V=R(omega) so I can just get rid of omega in the equation.
  8. Oct 31, 2007 #7

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    Right. Now wrap it up, as I have said.
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