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Homework Help: Stick falling and slipping

  1. Oct 23, 2013 #1
    1. The problem statement, all variables and given/known data

    A stick of mass M and length L is held vertically on a frictionless surface. A gentle push at the top causes the rod to fall from this position . Find the speed of its center of mass when it makes an angle θ with vertical .

    a) Use force method
    b) Use energy method

    2. Relevant equations


    3. The attempt at a solution

    First I tried using the Newtons laws of translation and rotation .

    1) The forces acting in the vertical direction are Mg and Normal from the floor at the lower end (N) .

    $$ Mg - N = Ma_{cm} $$

    2) Writing torque equation about CM ,

    $$ N\frac{L}{2}sinθ = \frac{ML^2}{12}\ddotθ $$

    3) ycm = (L/2)cosθ

    So , $$ v_{cm} = -\frac{L}{2}(\dotθsinθ) $$

    and , $$ a_{cm} = -\frac{L}{2}[\ddotθsinθ + (\dotθ^2)cosθ] $$

    Using the above three equations,I end up with

    $$ a_{cm} = \frac{1}{1-3sin^2θ}(-3gsin^2θ - \frac{L}{2}\dotθ^2cosθ ) $$

    But I need to find relation between vcm and θ .

    I would be grateful if somebody could help me with the problem .
    Last edited: Oct 24, 2013
  2. jcsd
  3. Oct 23, 2013 #2
    There was a thread about this problem a couple of days ago, where the same (difficult) approach was used.

    I suggest using energy.
  4. Oct 24, 2013 #3


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    You can write up acm also in terms of theta and derivatives. Then apply the trick that d2θ/dt2= dω/dt = (dω/dθ) ω =0.5 d(ω2)/dθ. You can consider ω2 a new variable, say z, and you get a first order linear equation for it (hopefully) .

    Last edited: Oct 24, 2013
  5. Oct 24, 2013 #4
    Is it something different from $$ a_{cm} = -\frac{L}{2}[\ddotθsinθ + (\dotθ^2)cosθ] $$ ?

    Are the three equations I have formed correct and relevant ?
    Last edited: Oct 24, 2013
  6. Oct 24, 2013 #5


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    I meant to write a de for theta.

  7. Oct 24, 2013 #6
  8. Oct 24, 2013 #7


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    As I read the OP, the requirement is to solve it twice, once using energy and once using forces.
  9. Oct 24, 2013 #8
    Okay ...this is what i get

    $$\frac{L}{6sinθ}(3sin^2θ-1)\ddotθ + \frac{L}{2}cosθ\dotθ^2 + g = 0 $$

    Putting θ '' = ωdω/dθ and θ' = ω

    $$ \frac{L}{6sinθ}(3sin^2θ-1)ω\frac{dω}{dθ} + \frac{L}{2}cosθω^2 + g = 0 $$

    Is it correct ?
    Last edited: Oct 24, 2013
  10. Oct 24, 2013 #9
    are you considering the translational velocity of the center of mass?
    I am too trying to solve this problem.
  11. Oct 24, 2013 #10
    vcm is the velocity of center of mass .
  12. Oct 24, 2013 #11
    I mean that the rod as a whole is also moving in horizontal direction(as there is no friction). I don't see you have considered this velocity.
  13. Oct 24, 2013 #12
    Why would it move horizontally?
  14. Oct 24, 2013 #13
    Please wait for some time before you put forward any query or suggestions...
  15. Oct 24, 2013 #14


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    I miss a picture...

    If the origin is at the ground just below the CM, the y axis points upward, and then macm=d2ycm/dt2=N-mg.
    Using acm and vcm is confusing, use the derivatives of y, and check the signs.

    At the end, use ω2 as variable, function of theta.

  16. Oct 24, 2013 #15


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    Check the signs. Read my previous post.

    ωdω/dθ=0.5 d(ω2/dθ). You have a first order equation for ω2, easy to solve.

    Last edited: Oct 24, 2013
  17. Oct 24, 2013 #16
    Here is the picture

    Attached Files:

  18. Oct 24, 2013 #17


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    Ok, I have the same picture.

  19. Oct 24, 2013 #18
    If it is going in circular motion it will experience a radially outward force.Isn't it?So i think due to this force the center of mass will have a translational motion too as there is no friction.Is that right??
  20. Oct 24, 2013 #19
    $$ \frac{(3sin^2θ+1)}{(12sinθ)}\frac{dz}{dθ} + \frac{L}{2}(cosθ) z - g =0 $$

    Does this makes sense ?
  21. Oct 24, 2013 #20


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    You miss an L in the first term. Otherwise it is OK :thumbs:

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