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Differentiation problem

  1. Nov 1, 2009 #1
    1. The problem statement, all variables and given/known data

    i have been told to calculate DI/Dx i calculated I as being = m(L^2)/16

    how do i differentiate this? im confussed with what happens to the m is it a constant an just removed? same with the 1/16 i assume that is removed as a constant?

    just need some clarification.

    thanks

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 1, 2009 #2

    danago

    User Avatar
    Gold Member

    Is that all you are told? If m and L are functions of x, then you will need to take that into consideration.

    If m and L are just constants, then treat them like you would any other number.

    EDIT: Hmm just re-read your question; did you mean you want to calculate dI/dL and not dI/dx?
     
  4. Nov 1, 2009 #3
    Well the question says that it passes a distance of x from the perpendicular axis ? does that mean i have to substitute a value for dx in?
     
  5. Nov 1, 2009 #4

    Mark44

    Staff: Mentor

    How about posting the problem exactly as it was given to you? That would save us a lot of time.
     
  6. Nov 1, 2009 #5
    OK.

    Two point masses m1 and m2 are seperated by a massless rod of length L.
    a) write an expression for the moment of inertia I about an axis perpendicular to the rod and passing through it a distance x from mass m1
    b)Calculate dI/dx and show that I is a minimum when the axis passes through the centre of mass of the system
     
  7. Nov 1, 2009 #6

    Mark44

    Staff: Mentor

    So m and L are not functions of x.
    Now show us how you calculated I. If you haven't drawn a picture, now would be a good time to do that.
     
  8. Nov 1, 2009 #7
    i think i calculated I i drew a picture an since its two point masses i dont think i need to intergrate so therefore the Inertia is equal to the sum of the product individual mass and the distance from the axis squared.

    so i then got [m1L^2]/32 +[m2L^2]/32 so simplify this to get
    mL^2/16
     
  9. Nov 1, 2009 #8

    Mark44

    Staff: Mentor

    You are assuming that the two masses are equal, which is not given. Also, you have not used the information that the axis passes through a point a distance x from m1.
     
  10. Nov 1, 2009 #9
    should i replace L with X then? i dont really kno where to go
     
  11. Nov 1, 2009 #10

    Mark44

    Staff: Mentor

    To calculate the moment of inertia you need to know the mass and the length of the lever arm (i.e., the distance from the mass to the point for which you're calculating the moment of inertia).
    What is the M.I. for mass m1?
    What is M.I. for mass m2?
     
  12. Nov 1, 2009 #11
    so the M.I for mass one =m1x^2 ? that was using the equation I=mr^2 where r is the radial distance

    so the M.I for mass two is = m2(L-x)^2

    ?
     
  13. Nov 1, 2009 #12

    Mark44

    Staff: Mentor

    Yes for both. The combined M.I. is I = m1x2 - m2(L - x)2, right? The two masses are on opposite sides of the axis, so their lever arms are acting in opposite directions.

    Now, can you find dI/dx and finish off part b?
     
  14. Nov 1, 2009 #13
    ahh cheers thanks for the help ill give b a shot now. =]
     
  15. Nov 1, 2009 #14
    so b)

    dI/dx = m.2x+x^2-[-2(L-x)]+(L-x)^2

    not so sure if iv done that right :/
     
  16. Nov 1, 2009 #15

    Mark44

    Staff: Mentor

    No, not right. dI/dx will have both m1 and m2 in it.

    To make your work more readable here, use the X2 button for exponents and the X2 button for subscripts. These buttons are in the advanced menu bar. If you don't see this menu bar, click the Go Advanced button just below the text entry area.
     
  17. Nov 2, 2009 #16
    sorry was supposed to put

    m12x+x2-m2(-2[L-x])+(L-x)2
     
  18. Nov 2, 2009 #17

    Mark44

    Staff: Mentor

    The first term is right (2m1x) but the rest isn't, and I have no idea how you got what you show.

    Start with I = m1x2 - m2(L - x)2, and show me your steps in getting dI/dx.
     
  19. Nov 2, 2009 #18
    Im not really sure i just used the product rule. im not sure what happens to the mass or the length or if they are taken out as constants so need some help really
     
  20. Nov 2, 2009 #19

    Mark44

    Staff: Mentor

    The masses and length L are constants. You don't just "take them out."

    I = m1x2 - m2(L - x)2
    dI/dx = d/dx(m1x2) - d/dx(m2(L - x)2)
    = 2m1x - m2 d/dx ((L - x)2)

    Use the chain rule to get the remaining derivative. If you don't know the chain rule, expand (L - x)2 and differentiate that instead.
     
  21. Nov 2, 2009 #20
    so m2d/dx(L-x)2
    will that go to

    m22(L-x)L ?
     
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