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Homework Help: Torque around an axis

  1. Feb 28, 2014 #1
    1. The problem statement, all variables and given/known data
    A 2.72kg ball is attached to a .65m rod that weighs 1.24kg. The rod rotates around it's axis 125 degrees. After rotating 125 degrees the ball is let go from the rod. How much torque is required to make the ball's exiting velocity 18m/s.

    2. Relevant equations
    vf=vi + at

    3. The attempt at a solution
    First I solved for how much distance the ball is traveling in an arc before being released=1.43m
    L = ((125)*2pi*.6558m)/360 = 1.43m.
    Then I found it's needed acceleration to reach 18m/s.
    18^2=0+a(1.43m), a=113.28m/s^2
    Then I solved for the amount of time it will take to reach 18m/s
    vf=vi+at, 18=113*t, t=.1558s
    Now this is where I got stuck.
    F=m*a, so i'm assuming (2.72kg)*113.28m/s^2=308.12N
    Is this correct? I think this may involve angular motion which I am inexperienced with. Ps, this is for a catapult idea/robot so any help is gratefully appreciated.
  2. jcsd
  3. Feb 28, 2014 #2


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    Doesn't the rod need some torque too to accelerate all through these 125 degrees ?
  4. Feb 28, 2014 #3
    The rod is accelerating around it's axis(another rod is spinning and providing torque). My question is how fast must the axis(rod) spin and/or how much torque must be produced. This question leads into gear ratios but I understand that stuff mostly.
  5. Feb 28, 2014 #4


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    Yes yes (the catapult idea is a clue for me which "it's" axis you mean (because a cylinder has a lot of axes -- in fact every body has).

    Point is, your rod and your ball don't differ that much in weight, so the rod cannot be ignored.

    And you might want to delve into angular velocity, acceleration etc. nevertheless!

    By the way, how did you measure the length of the rod so precisely ?
  6. Feb 28, 2014 #5
    So when finding the force, would mass =ball+rod?
    Okay so angular velocity is change in angular position/change in time. so it'd 2radian/.1558s=13.45 radians/s.
    a=v^2/r, 13.45^2/.65m=278.46
    I=mr^2, 2.72kg*(.65m)^2=1.14
    angular acceleration = a/r, 278.46/.65m=428.40
    torque=I * angular acceleration=1.14*428.40=492.31N*m
  7. Mar 1, 2014 #6


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    You still want to read up on the angular stuff if you want to do this quantitatively. It's not all that complicated.
    You have already used angle ##\phi## to get displacement ## Δ\vec s = Δ\phi\ \vec r##
    angular velocity ##\omega= {d \phi \over dt} ## along the rotation axis, with ##\vec \omega = {\vec r \times \vec v \over |\vec r|^2}##
    Angular acceleration ##\vec \alpha = {d \vec \omega \over dt} ##, also along the rotation axis, where the torque you want to find out comes in and also the moment of inertia ##I## that appears in the kinetics formulas instead of the mass.

    Also check out the list of moments of inertia, where you find I for the ball (yes, ##I = m\ R^2##) and the rod for an axis perpendicular to the rod at one end ((hey, ##I = {m\over 3} \ R^2## !)

    I take it you want to design something like in the picture to defend yourself or for an assignment. Start with drawing a few diagrams (##\phi = 0, t=0## then ## \phi = 125, t = .... \ ## See how gravity comes in, etc.

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