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Need help In Rotation Questions Please

  1. Aug 30, 2008 #1
    1. The problem statement, all variables and given/known data

    A system consists of two particles, of masses m1 and m2, is connected by a light rigid rod of length L.
    a) Find the rotational inertia I of system for rotations of tis object about an axis perpendicular to the rod and distance x from m1
    b) Show that I is a minimum when x=xcm

    2. Relevant equations
    I= Icom + mh2
    I= 1/12 mr2

    3. The attempt at a solution
    The answer provided was given as: I2=(m1+m2)x2+m2L2-2m2xL

    I understand that it is juz applyin the formula I= Icom + mh2 but wat i do not understand is that

    The thing that I dont understand is the mh^2 part..why is only m2 used in the equation?
    And why when h= (L-x)^2 is expanded..what happens to the m2x^2?

    And wat about the part b? By juz replacing the value it wif x...it juz doesnt make any sense...

    Thanks ya..any help will be very much appreciated..
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 31, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi brainracked! Welcome to PF! smile:
    No, it isn't.

    It's using the far simpler method of just using the basic formula for a point mass, I = mr2.

    Try that! :smile:
     
  4. Aug 31, 2008 #3
    Oh gosh.. -blush-
    Dat simple?
    Aiyo..I always complicate stuff....but..I still dont quite get it..haha..sorry ya..
    So..okay..tryin out d simpler way..i get d 1st part which is I=(m1+m2)x^2 but wat about the rest of the answer where there is the +m2L^2 - 2m2xL?

    And besides..i thought if it is not rotated about a central axis but at sum distance from it, the formula of I = Icom +mh^2 should b used as h comes into play as there is the mention of distance x from m1..

    I can solve other problems similar to tis one but wif the numbers & figures given such as mass and distance and rev...but i juz cant seem to get my head round tis..

    So sorry but I think I still need help in gettin tis rite.. =)
     
  5. Aug 31, 2008 #4

    tiny-tim

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    ok, let's do the easy one …

    I for m1 is momentum x distance, = m1x2.

    So I for m2 is … ? :smile:
     
  6. Aug 31, 2008 #5
    Haha..ok..so m2 should b...m2L^2? Correct?
    So the -2m2xL?
     
  7. Aug 31, 2008 #6
    No that is still incorrect. Do you even understand how tiny-tim gave you the answer for the inertia of [tex]m_1[/tex]? If so, then do the same for [tex]m_2[/tex] and then add their inertia's together and you will indeed arrive at the answer provided.
     
  8. Sep 1, 2008 #7
    At first I dont think I fully understood..but now I get it..

    Im1=m1x2

    So..Im2 should b..

    Im2=m2 (L-x)2
    = m2 (L2+ x2 - 2xL)

    Then as u said..by adding the Inertia's together..tadaa..the answer..Haha..

    Big thank you to tiny-tim & chislam!!! :smile:
    Take care ya ppl!
     
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