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Calculate linear acceleration of a pivoted stick

  1. Jul 16, 2016 #1
    1. The problem statement, all variables and given/known data
    A uniform thin stick of mass M and length L has a moment of inertia about its centre of mass of I = (1/12)ML^2. The stick, which is pivoted at one end, is held horizontal and then released. Assuming the pivot is frictionless, calculate:

    i) the linear acceleration of the free end of the stick (due to gravity) immediately after its released, and
    ii) the speed of the free end of the stick as it passes thru the vertical position.

    2. Relevant equations
    $$ \tau = rF $$
    $$ \tau = I \omega $$
    $$ a_r = \omega^2 r $$
    $$ a_{tan} = \alpha r $$

    3. The attempt at a solution
    Using the first two equations: ## \tau = mg(L/2) ## and ## \tau = (1/12)ML^2 \omega ##
    I got ## \omega = 6g/L ##

    I presume the linear acceleration is the tangential plus radial acceleration, but I don't see how to work out the tangential acceleration. Also, I presume you use conservation of energy to calculate part 2, but I am not sure if it is just gravitational potential energy (at horizontal) equalling translational kinetic energy or translational plus rotational kinetic energy.
     
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  3. Jul 16, 2016 #2

    Orodruin

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    How would you find the velocity of the free end? How is velocity and acceleration related?

    When you have a fixed point, the entire kinetic energy is given by the rotational energy given by ##I\omega^2/2## where I is the moment of inertia relative to the fixed point. You could also write it as the rotational energy relative to the center of mass plus the transational energy relative to the center of mass, but in this case it would just complicate things. For example, you would have to compute the moment of inertia relative to the center of mass.
     
  4. Jul 16, 2016 #3

    haruspex

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    You don't know how to get the tangential acceleration from the angular acceleration? How would you get tangential velocity from angular velocity? Or did you mean you don't know how to get the radial acceleration?
    You need to be consistent about the reference axis. What axis have you used in each of those two equations?
    Also, it is usual to use α, or ##\dot \omega##, for angular acceleration and ω for angular velocity.
     
  5. Jul 16, 2016 #4
    So, ## v = \omega r = 6g ## . I'm not sure how to get the tangential acceleration from that. I know you can differentiate angular velocity (to get angular acceleration etc) (and apply the formula), but there is nothing to differentiate (it would be 0). Sorry, I'm just not seeing it yet.

    The axis was at the pivot. Am I going about this wrong?
     
    Last edited: Jul 16, 2016
  6. Jul 16, 2016 #5

    haruspex

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    But the expression you used for the moment of inertia was 1/12 ML2, which is:
    I don't understand that statement. Differentiate ##v=\omega r## with respect to time, taking r as constant.
     
  7. Jul 16, 2016 #6

    Orodruin

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    No, the angular velocity (and the velocity) are zero! Your equation also does not match dimensionally. The first step is true. What do you get if you differentiate it with respect to time?
     
  8. Jul 16, 2016 #7

    Orodruin

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    This is not correct. The torque is the time derivative of the angular momentum ##L = I\omega##, not the angular momentum itself.
     
  9. Jul 16, 2016 #8
    Sorry I am a bit slow with this. ## \tau = I \alpha = (1/12)ML^2 \alpha ## , so ## \alpha = 6g/L ##
    Differentiating ## v = \omega r ## wrt t gives ## a = \alpha r ##
    So, if my moment of inertia is about the CM, then my axis should also be there? Which would mean r = L/2, so ## a = 3g ##
    Is that better?
    Do I then need to integrate ## \alpha ## to get ## \omega ## ? Which would be 6gt/L + C, which doesn't seem right
     
    Last edited: Jul 16, 2016
  10. Jul 16, 2016 #9

    Orodruin

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    Yes, it is better, but not correct. You also need to consider the comments of haruspex about which point your momentum of inertia is computed relative to.

    Edit: Also, at the free end r=L.
     
  11. Jul 16, 2016 #10
    I'm not sure what to do. Is this wrong - ## \omega = mg(L/2) ## . I don't know what else it would be - mg(L/4) ?
     
  12. Jul 16, 2016 #11

    Orodruin

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    This even has the wrong units. Start by recomputing the moment of inertia about the pivot point. You can do this using the parallel axis theorem or compute it from first principles.
     
  13. Jul 16, 2016 #12

    haruspex

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    They need to be about the same point, whatever. If you take the mass centre as axis then you have to go back to calculating the torque. The weight of the rod, mg, has no moment about that axis. Instead, the reaction from the pivot will have a moment about it. But this way makes things more complicated, so better to use the pivot as the axis for both torque and moment of inertia.
     
  14. Jul 16, 2016 #13
    so ## I = I_{cm} + ML^2 = (1/12)ML^2 + ML^2 ##
    ## mg(L/2) = ((1/12)ML^2 + ML^2) \alpha ##
    ## \alpha = 6g/13L ##

    Is that better?

    If that is correct, then a = 6g/13. And this would be the linear acceleration.
    Thank you so much for your help, I really appreciate it. I'm studying past exam papers, and was really stuck on this one.
     
    Last edited: Jul 16, 2016
  15. Jul 16, 2016 #14

    Orodruin

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    What is the distance from the CoM to the pivot?
     
  16. Jul 16, 2016 #15
    It's L/2. Does that mean ## (1/12)ML^2 + M(L/2)^2 ## for I?
    So, ## \alpha = (3g)/(2L) ##
    and ## a = 3g/2 ##
     
    Last edited: Jul 16, 2016
  17. Jul 16, 2016 #16

    haruspex

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    Yes.
     
  18. Jul 17, 2016 #17
    Thank you both so much, I really appreciate it.
     
  19. Jul 17, 2016 #18

    TSny

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    It's interesting that the free end of the stick has greater acceleration than "free fall" (g).

    As you can show with your equations and as the coins show in the video, the particles in the far 1/3 of the stick initially have tangential accelerations greater than g.
     
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