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Rotating rod around a pivot

  1. Jan 12, 2008 #1
    1. The problem statement, all variables and given/known data
    A thin rod of mass 0.630kg and length 1.24m is at rest, hanging vertically from a strong fixed hinge at its top end. Suddenly a horizontal impulsive force (14.7î)N is applied to it. Suppose the force acts at the bottom end of the rod. Find the acceleration of its center of mass and the horizontal force the hinge exerts.

    2. Relevant equations
    Rotation of a rigid object.

    3. The attempt at a solution
    I found the acceleration of its center of mass without problems. First I computed the torque done by the external force.
    [tex]\sum{\tau}=FL[/tex]
    with L the length of the rod and F the magnitude of the external force.
    Because it's a rod that is rotated about one end, the moment of inertia is given by
    [tex]I=\frac{1}{3}ML^2[/tex]
    with M the mass of the rod.
    We can use that the net torque applied to a rigid object is proportional to the angular acceleration.
    [tex]\sum{\tau}=I\alpha \\
    \Leftrightarrow FL=\frac{1}{3}ML^2\alpha \\
    \Leftrightarrow \alpha = \frac{3F}{ML}[/tex]
    Furthemore, the acceleration of the center of mass is given by
    [tex]a_{CM}=\frac{L}{2}\alpha=\frac{3F}{2M}[/tex]
    I have problems with finding the horizontal forces that act on the hinge. I know that in the vertical direction it has a force exerted on it directed upwards and equal in magnitude to the gravitational force exerted by the rod. Otherwise, it wouldn't be in equilibrium. Is that correct? Now, how do I find the horizontal force it exerts?
     
  2. jcsd
  3. Jan 12, 2008 #2

    Doc Al

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    Staff: Mentor

    Hint: You found the acceleration of the center of mass. What does that tell you about the net horizontal force on the rod?
     
  4. Jan 12, 2008 #3
    Well, because
    [tex]\sum{F}=Ma_{cm}[/tex]
    then the force on the rod is
    [tex]Ma_{cm}=\frac{3}{2}F[/tex]
    Is that also the horizontal force on the hinge then?
     
  5. Jan 12, 2008 #4

    Doc Al

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    Staff: Mentor

    No, that's the net force at that instant. You know the applied force, so figure out what the hinge force must be.
     
  6. Jan 12, 2008 #5
    Hmmm, I think I get it. The net force on the center of mass is
    [tex]\sum{F}=F+H[/tex]
    with H the hinge force.
    Also according to Newton's second law,
    [tex]\sum{F}=Ma_{cm}[/tex]
    so
    [tex]F+H=Ma_{cm} \\
    \Leftrightarrow H=Ma_{cm}-F \\
    \Leftrightarrow H=\frac{1}{2}F[/tex]
    Is that the correct answer?
     
  7. Jan 12, 2008 #6

    Doc Al

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    Staff: Mentor

    Looks good to me!
     
  8. Jan 13, 2008 #7
    Ok thank you for your time. I find it a bit confusing that sometimes you look at the rod as a rigid object rotating about the hinge, and sometimes you look at it as a point-particle.
     
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