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Moment of interia question

  1. Apr 13, 2010 #1
    1. The problem statement, all variables and given/known data
    A metal can containing condnsed mushroom soup has a mass of 215g, a height of 10.8cm and a diameter of 6.38cm. It is placed at rest on its side at the top a 3.00m long incline that is at 25.0 degrees to the horizontal and is then released to roll straight down. Assume energy conservation, calculate the moment of interia of the can if it takes 1.50s to reach the bottom of the incline. Which pieces of data if any are unnecessary for calculating the solution


    2. Relevant equations
    [tex]I = \sum mr^2[/tex]


    3. The attempt at a solution
    This is what i have done, however it is incorrect.
    [tex]I = \sum 0.215kg * 0.0319^2[/tex]
    [tex]2.19 * 10^{-4}[/tex]

    P.S
     
  2. jcsd
  3. Apr 14, 2010 #2
    can anyone help me??
     
  4. Apr 17, 2010 #3
    ok can someone tell me if this is the correct equation to find the moment of interia??

    [tex]mgh = \frac{1}{2}m(\frac{2x}{t})^2 + \frac{1}{2}I(\frac{v}{r})^2[/tex]

    [tex]mgh = \frac{1}{2}m(\frac{2x}{t})^2 + \frac{1}{2}I(\frac{x}{tr})^2[/tex]
     
  5. Apr 17, 2010 #4
    can you tell me what the correct answer for I is? I worked out a solution but I don't know if it's right
     
  6. Apr 17, 2010 #5
    1.21 * 10^-4 kg.m^2
     
  7. Apr 17, 2010 #6
    argh i got this right the first time but i crossed out a variable a little too darkly and i ended up wrong by eight times lol..


    your solution is correct, except you made a little mistake in it

    because v = 2x/t,

    v/r should be 2x/tr NOT x/tr


    if you want me to explain how to arrive at the equations you posted then ask
     
  8. Apr 17, 2010 #7
    I cannot seem to get the right answer

    [tex]\frac{1}{2}(\frac{2*3}{1.50})^2 + \frac{1}{2}(\frac{2*3}{1.50*0.0319})^{2}I = 9.8 * 3sin(25)[/tex]

    [tex]8+ 7861.557964I = 12.4249769[/tex]

    [tex]I = 5.53 * 10^{-4}[/tex]
     
    Last edited: Apr 17, 2010
  9. Apr 17, 2010 #8
    you tried to cancel out the mass terms, but you actually can't because there is no mass on the rotational energy term. if you tried to divide both sides by mass then you would have to divide the middle term by 0.215



    warning: do not get confused if you read this
    I think you may be confused because the rotational inertia does in fact depend on mass.

    You should talk to your teacher about this or consult a textbook about this, but the inertia is actually an integral, and after evaluating the integral you would find that
    I = some constant x MR^2.

    For a perfect cylinder, I = 0.5 MR^2 (although a soup can is not a perfect cylinder)

    The lesson is that I is its own term, don't get confused by its dependence on mass - you can't divide out mass from it unless you know what the constant is
     
  10. Apr 17, 2010 #9
    thanks i understand now.
     
  11. Apr 17, 2010 #10
    Just a quick question, i don't quite understand why would i use the 3sin(25) instead of the 10.8cm as the height when finding the PE?


    *Edit don't worry i know, because 10.8 is the height of the can soup.
     
  12. Apr 17, 2010 #11
    another question, why wouldn't you use I=MR^2 to find moment of inertia?
     
  13. Apr 18, 2010 #12
    well i touched on it in my spoiler, but because a soup can is a continuous distribution of particles, you would actually have to use an integral (this is all calculus concepts)

    I = ∑ M R²

    if you consider a small section of mass ∆M then all the particles in this section are close together and have approximately equal radius Ri so

    I = ∑ Ri² ∆M

    Then you take this limit as ∆M → 0, which is a Riemann sum, so

    I = ∫ R² dM

    What all this means is that if all the mass were located at the same distance from the axis of rotation, then the integral would become I = M R²

    But for a soup can a lot of the mass lies inside the outer shell, so you would have to add up all of the mass-radius inside, which has smaller Ri. If you know integral calculus then you can compute the integral for a cylinder to be I = .5 MR²
     
  14. Apr 18, 2010 #13
    so basically in order for you to find the moment of inertia using [tex]I = \ssum MR^2[/tex] you would have to add up all the mass and radius from the center. Hence a faster way is using the approach in my above posts.

    For any object that doesn't have a defined equation, would you use the approach like what i've done in this question? (assuming it is exactly the same question but different object)
     
  15. Apr 18, 2010 #14
    well evaluating the integral of I = ∫ R² dM is used for general circumstances when you want to find the moment of inertia for ALL cylinders or ALL spheres or ALL rectangular prisms etc.

    The method used to solve this problem is good for evaluating the moment of inertia of a particular object you're experimenting on. But sometimes you just want to skip the experiments and go to a general expression.

    As for whether one method is faster or not, I like to think of physics concepts as tools instead of being on a ladder of absolute usefulness. In some situations the method used for this problem is better, in others computing the integral may be faster, but it isn't always faster for every case.

    In my physics classe our teacher just gives us a list of moment of inertia's for common objects but we still have to do the problem for weirder distributions.
     
    Last edited: Apr 18, 2010
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