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Vertical speed of a point attached by two rods to a rotating hinge.

  1. Apr 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Not exactly homework, but an interesting problem I found for which I have some questions about the answer.

    A rod of length r is rotating around a point O with constant angular velocity w. Distance r away to the right from the point O is a rail. The end of the rod with length r is attached at point P to a rod with length L. Rod L attaches to a point Q on the rail. Point Q is below point O and P at all times The point Q is free to move up and down as the rod r rotates. Calculate the maximum and minimum speed of point Q, maximum speed is max magnitude up, min is max magnitude down. L > 2r

    Here is the drawing:

    1znq6bk.jpg

    2. Relevant equations

    3. The attempt at a solution

    The solution says that because the point P has a constant acceleration that is radial towards point O and since the projection of this acceleration to the rod is equal to the projection of the acceleration of point Q to the rod, we can find the situations where the projection of acceleration of point P to the rod is zero and find maximum and minimum velocities from that. These happen when the two rods are at right angles. Minimum v = -wr. For maximum v it uses the fact that a cyclic quadrilateral forms and halving it you get two triangles QPO and QOR (R is on the rail where there is a right angle between the rail and line connecting point R to O) of which one side is mutual, one angle is 90 and one angle is r on both triangles so they are congruent and max v is the projection of the velocity of point P to the rail, [itex]_{max} v = \frac{wr}{cos(\angle PQR)} = \frac{wr}{cos(2arctan(r/L))}[/itex]

    My question is that I am not quite sure the situations where the rods are at right angles would represent maximum and minimum velocities. Yes the projection of the acceleration to the rod would have a component still accelerating it upwards when projected to the rod, but it also has velocity to the left and maybe if it moves to the left with more velocity it makes the point Q move faster up, since moving P to the left would make Q go up when trying to achieve v max.

    I put together an equation for point Q depending on time: [itex]Q(t)=\sqrt{L^2-(r+r*cos(w*t))^2}+r*sin(w*t)[/itex]
    I assumed that zero angle is when the rod r points away from the rail and it starts moving clockwise.

    I put it into GeoGebra, took the first and second derivatives of it, put the roots of the second derivative (time when velocity is at an extrema) into the first derivative to find speed and found out that the beforehand solution and this equation have only a small error when r is small compared to L, but if r is 1 and L is 2.1 and w is 1, then they differ by 0.3; if I make L=4 then they differ by ~0.002

    What do you think, is the solution accurate or not and why or why not?
     
  2. jcsd
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