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Rotation in Vertical Plane

  1. Feb 17, 2004 #1
    I have a question about rotational movement in the vertical plane. I know I'm supposed to use polar coordinates to solve it, but since our professor only taught us how to derive the formulas and gave us no examples, I have no idea how to approach this problem.

    A 227-g block B fits inside a small cavity cut in arm OA, which rotates in the vertical plane at a constant rate. When θ=180 degrees, the spring is stretched to its maximum length and the block exerts a force of 3.56 N on the face of the cavity closest to A. Neglecting friction, determine the values of θ for which the block is not in contact with that face of the cavity.

    Diagram

    What I've done so far:

    At θ=180 degrees, the arm OA is vertical, with A pointing downwards. And I figure that:

    ΣF=kx-mg in the upward direction. And this should be equal to ma(r), where a(r) = [tex]\ddot{r} - r\ddot\theta[/tex].

    And this is equal to 3.65 N [?? I'm not sure on this one. ]

    And this is where I'm lost, because the problem doesn't give any distances, it doesn't tell me the value of [tex]\dot\theta[/tex] or the value k for the spring. And we're expected to come up with a numerical answer of -53.2 degrees to 53.2 degrees.

    Can anyone help me and point me to the right track?
     
  2. jcsd
  3. Feb 19, 2004 #2

    jamesrc

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    The sum of forces in the r-direction at θ=180 degrees is actually:

    [tex] \Sigma F_r = ma_r = m(\ddot r - r\dot \theta^2) = mg - N - k(r-r_o) [/tex]

    The good news, however, is that the spring is little more than a red herring here. Imagine that instead we have a mass in a slot with no spring. The only forces acting on it are the normal force from the contact point (between the slot and mass) and the weight of the object. For this problem, we are only concerned with the situations where the block is up against the end of the slot. The spring force only reduces the normal force.

    Let's write the general expression that will allow us to calculate the normal force for this made up problem:

    [tex] mr\dot \theta^2 = N_\theta + mg\cos\theta [/tex]

    You should be able to derive that without much trouble (Nθ is just the normal force as a function of θ). Notice that there is no d2r/dt2 dependence because r is fixed for this problem.

    Now let's look specifically at the case where θ=180:

    [tex] mr\dot\theta^2 = N_{\rm max} - mg [/tex]

    Nmax is the maximum normal force we will encounter in this situtation (when the arm is pointing straight down).

    We can eliminate the rotational velocity term by putting the two equations together:

    [tex] N_{\rm max} - mg = N_\theta + mg\cos\theta [/tex]

    Now we can bring in what we know about the spring: if the "normal force" in the simplified situation 3.56 N less than what it is when it is pointing down, then the tension in the spring will be great enough to keep it from contacting the wall:

    Condition for no cavity-wall contact:

    [tex] N_{\rm max} - N_\theta < 3.56 [/tex]

    Solving for the difference in normal forces:

    [tex] mg(1+\cos\theta) < 3.56 [/tex]

    Plug in from here to find the angle range.

    I hope I explained that well enough. Try to see why the analogy between the two problems is valid. Good luck.
     
  4. Feb 19, 2004 #3
    Oh, thank you so much! :-) I was stuck on the problem because I treated r as a variable, and had trouble coming up with [tex]\dot\theta[/tex]. I guess for this situation, we're only concerned about the two ends of the range values for &theta; where r is fixed at the end of the arm. Is that correct?

    Thanks again!
     
  5. Feb 19, 2004 #4

    jamesrc

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    Yep, that's how I approached it.
     
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