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Maximum angular velocity so that there is no stick

  1. Sep 6, 2016 #1
    1. The problem statement, all variables and given/known data
    We have a concrete mixer, and if the drum spins too fast, the ingredients will stick to the wall of the drum rather than mix. Assume that the drum of the mixer has radius R = 0.5 m and that it is mounted with its axle horizontal. What is the fastest the drum can rotate without the ingredients sticking to the wall all of the time?

    2. Relevant equations
    Newton's 2nd law

    3. The attempt at a solution

    So I believe that I have solved this problem. We define that the ingredients stick to the wall when, at the very top of the motion, there is some normal force pushing the ingredients toward the center. Thus to find the maximum angular velocity, we just need to find the angular velocity at which there is only gravity acting at the top and no normal force.
    Also, we are working in polar coordinates, so we define our reference frame such that towards the center of rotation is positive, and outward is negative.

    Thus, (imagine that our analysis is on a single particle, at the very top of the rotation)
    ##F_g = m a_{radial}##
    ##mg = m (\ddot{r} - r \dot{\theta^2}_{max})##
    ##g = -r \dot{\theta^2}_{max}##
    ##\displaystyle \dot{\theta}_{max} = \sqrt{-\frac{g}{r}}##

    This is the correct equation (I think), except there is a negative. Why is there a negative? Did I define my coordinate system wrong?
     
  2. jcsd
  3. Sep 6, 2016 #2

    kuruman

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    You mention the normal force, but there is no normal in your equations. Yes, the normal force must be zero at some point, but your solution implies that ## m\ddot{r} ## is the normal force because that's what you set to zero. What is ## m \ddot{r} ## according to Newton?
     
  4. Sep 6, 2016 #3
    Well, ##r## is not changing, so I thought that ##\ddot{r} = 0##. Also, at the very top, gravity would be the only force acting on the particle, so I still don't see what I am doing wrong
     
  5. Sep 6, 2016 #4

    kuruman

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    Please answer my question first, what does ## m\ddot{r} ## represent? Answering this will clarify the way you think about this problem.
    Hint: What does Newton's Second law say? Also, consider this: does the fact that r is not changing necessarily mean that ## \vec{r} ## is not changing?
     
  6. Sep 6, 2016 #5
    Is ##m \ddot{r}## the normal force, since ##\ddot{r}## is the acceleration of the radius vector from the center, which is constant in magnitude but not in direction?
     
  7. Sep 6, 2016 #6

    kuruman

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    No. More correctly, ## m \ddot{ \vec{r}} ## (mass times the acceleration vector) is ... ? Newton's Second law?
     
  8. Sep 6, 2016 #7
    ##\ddot{r}## is the acceleration due to a change in radial speed. So ##m \ddot{r}## would be the force that causes a change in radial speed...
     
  9. Sep 6, 2016 #8
    Sorry to bump, but I really need some help with this. I'm so close to the answer, I just need some pointers.
     
  10. Sep 7, 2016 #9

    jbriggs444

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    You have the right equation. Karuman is on the wrong track. The stated concern is the sign error resulting from:

    ##mg = m (\ddot{r} - r \dot{\theta^2}_{max})##

    In that equation, you are equating ##mg##, the downward force of gravity on the object, with ##(\ddot{r} - r \dot{\theta^2}_{max})## the radial acceleration of an object described using polar coordinates and a convention that out = positive.

    At the top of the path, do your sign conventions (down=positive and out=positive) line up properly?
     
  11. Sep 7, 2016 #10
    Ah ha! That fixes everything. Thank you!
     
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