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Rotational Motion of a thin rod

  1. Nov 12, 2008 #1
    1. The problem statement, all variables and given/known data
    A 4.00 kg mass and a 3.00 kg mass are attached to opposite ends of a thin 44 cm long horizontal rod (see figure). The system is rotating at angular speed ω = 6 rad/s about a vertical axle at the center of the rod.

    (a) Determine the kinetic energy K of the system.


    (b) Determine the net force on each mass.


    (c) Repeat parts (a) and (b) assuming that the axle passes through the CM of the system.



    2. Relevant equations



    3. The attempt at a solution
    KE=1/2(0.3388m*N)(6rad/s)^2=6.10 J
    F1=(4kg)(6rad/s)^2(0.22m)=31.7 N
    F2=(3kg)(6rad/s)^2(0.22m)=23.8 N
    CM=(4kg)(0.22m)+(3kg)(0.22m)/4kg+3kg=0.22m

    How does the CM change the problem? I am having trouble figuring out the second set of answers.
     
  2. jcsd
  3. Nov 12, 2008 #2
    The moment of inertia of a system is given by the sum of the moments of intertia of each of the system's components, the spheres and the rod. Remember the parallel axis theorem when solving for the moments of inertia when the axis is moved to the center of mass.
     
  4. Nov 12, 2008 #3

    LowlyPion

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    What are you doing with the Center of mass equation? What lengths to the 2 masses do you get?
    And aren't they supposed to add to .44 m?
     
  5. Nov 12, 2008 #4

    djeitnstine

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    Center of mass = [tex]\frac{\sum{(distance)(mass)}}{\sum{mass}}[/tex]

    make sure to define your co-ordinate system

    remember the CM is not changing, the rod is only at [tex]\frac{d}{2}[/tex] of the masses not at the center of mass

    I just read your CM equation, you are using [tex]\frac{d}{2}[/tex] as your starting point, so wouldn't right be positive and left be negative?
     
    Last edited: Nov 12, 2008
  6. Nov 12, 2008 #5
    I am confused.
     
  7. Nov 12, 2008 #6

    LowlyPion

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    What is the distance of each mass from the center of mass?

    Can't you figure where the center of mass is and what those distances to each mass are?
     
  8. Nov 12, 2008 #7

    djeitnstine

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    Hey set your co-ordinates correctly, if your center point is 0, then to the right 0.22 m is positive to the left 0.22 is negative....---->THINK X AXIS <----- So you have [tex]\frac{(4)(-0.22)+(3)(+0.22)}{3+2}[/tex]

    Your displacements are vectors, they have direction!
    Its pretty straight forward

    [​IMG]
     
    Last edited: Nov 12, 2008
  9. Nov 12, 2008 #8
    So the center of mass is -.03143m

    Thanks!
     
    Last edited: Nov 12, 2008
  10. Nov 12, 2008 #9

    djeitnstine

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    Yes, or 0.0314 meters to the left.
     
  11. Nov 12, 2008 #10

    djeitnstine

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    More accurately CM is defined as [tex]\sum{\frac{m_{1}\vec{r_{1}}+m_{2}\vec{r_{2}}+...+m_{n}\vec{r_{n}}}{m_{1}+m_{2}+...+m_{n}}}[/tex]
     
  12. Nov 12, 2008 #11

    djeitnstine

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    It changes now because the axle is at the CM instead of [tex]\frac{d}{2}[/tex]
     
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