1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Moment Of Inertia Problem using Algebra

  1. Mar 7, 2008 #1
    1. The problem statement, all variables and given/known data

    1a . By dividing a uniform narrow bar of length l into a number of small masses, find the moment of inertia of the bar rotating about an axis perpendicular to the bar and passing through its end

    1b. use the parallel axis theorem and the result above to find the moment of inertia about an axis parallel to that above passing through the centre of mass of the bar

    1c. calculate the moment of inertia of the turbine blade. Each blade has mass m and length l

    2. a car with its door open moves off from rest with an acceleration of 10ms-2. how long will it take the door to close


    2. Relevant equations



    3. The attempt at a solution

    I think i got 1 a) right

    I=m(l/10)^2+m(2l/10)^2...+ml^2
    =m/100(l^2+(2l)^2...(10l)^2)
    =385l^2/100

    For the rest I have not really any idea, so any tips would be great

    thx thx =D
     
  2. jcsd
  3. Mar 8, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That's a start on a). But shouldn't m for each part be m/10? And what happened to m in your final formula? And why just divide it into 10 parts? Why not into 100 or 1000 etc? I think you are actually supposed to set up an integral.
     
  4. Mar 8, 2008 #3
    I dont think I can use calculus on this problem because we haven't learnt it yet.

    I only know difrrentiation, not much integration =/.

    On the m, forgot that XD

    Is it possible to use mathematical induction to solve the infinite series if i use n lengths?
     
  5. Mar 8, 2008 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You can write out the formula for a general value of n. Then find or derive a formula for the sum of the first n squares and you can write out an explicit formula. Then take the limit as n goes to infinity.
     
  6. Mar 8, 2008 #5
    yup.. if you divide it into 'n' small parts.. you get the lengths as: [itex]\frac{l}{n}[/itex], [itex]\frac{2l}{n}[/itex] etc. Since you need to square that.. take [itex]n^2[/itex] and [itex]l^2[/itex] common. Then you'll have to solve for the series:

    [tex]
    \frac{l^2}{n^2}\left(1 + 2^2 + 3^2 + 4^2 + ...\right)
    [/tex]

    use the formula for sum of squares and use [itex]n \rightarrow \infty[/itex] and you shall get your answer. This however, will be much easier if worked with an integral..
     
  7. Mar 8, 2008 #6
    So I get

    lim n-->infinity = 3l^2
     
    Last edited: Mar 8, 2008
  8. Mar 8, 2008 #7
    well.. this is your radius of gyration.. Multiply with the mass and get moment of inertia:

    [tex]
    I = \frac{3Ml^2}{2}
    [/tex]

    For the second question.. you know the moment of inertia for an axis [from the first question] and there is another axis at a distance l/2 from the first one and parallel to the first one. Can you use the parallel axis theorem now?
     
  9. Mar 8, 2008 #8
    ok.. we seem to have made a huge mistake here...

    the sum:

    [tex]
    \frac{l^2}{n^2}\left(1 + 2^2 + 3^2 + 4^2 + ...\right)
    [/tex]

    tends to infinity as [itex]n \rightarrow \infty[/itex]. As Dick said, the mass of each such small element will be M/n and not 'M', hence the sum we need to evaluate is:

    [tex]
    \frac{M}{n}\frac{l^2}{n^2}\left(1 + 2^2 + 3^2 + 4^2 + ...\right)
    [/tex]

    which would give you:

    [tex]
    \frac{Ml^2}{3}
    [/tex]

    this is the required moment of inertia...
     
  10. Mar 8, 2008 #9
    Oh thank you

    i think its cos i forgot the 6 as the denominator =/

    parallel axis theorem is I=I(centre of mass)+Mr^2 I think

    so taking that I you get

    Ml^2/3=I(cm)+Mr^2

    I(cm)=Ml^2/3-Mr^2
     
  11. Mar 8, 2008 #10
    For most of the part, you have done it right.. but what is 'r' here?
     
  12. Mar 8, 2008 #11

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Ok so far. What the relation between l and r?
     
  13. Mar 9, 2008 #12
    i think r is the distance from the original point to the centre so l/2
     
  14. Mar 9, 2008 #13
    well.. you got it.. just put it in the equation and solve.. you should have ur solution.. :D
     
  15. Mar 10, 2008 #14
    cheers everyone =]

    especially rohanprbhu :D
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Moment Of Inertia Problem using Algebra
Loading...