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Humidifier disk and wetted area if partially submerged

  1. Oct 23, 2007 #1
    I seem to have down 1-4 on this question but Im struggling on the rest... any help??

    A circular disk of fixed radius r is used in a humidifier. It is partially submerged in water and rotates on its center. The goal is to determine the height above the water, h, at which to fix the center so that the exposed wet surface area is maximized.

    1. Find the surface area that is dry, i.e. never is submerged as the disk rotates, as a function of r and/or h.
    2. Find the total surface area that is wet as a function of r and/or h. (i.e. either submerged or exposed)
    3. Find the area of the sector of the circle that is created by radii that intersect the water as function of r and/or h. (HINT: Use theta as half of the angle from radius to radius of your sector.)
    4. Find the area of the triangle that is in this sector, but above the water, as a function of r and/or h.
    5. Find the surface area that is submerged at any time as a function of r and/or h.
    6. Subtract to find the desired area, area that is exposed and wet, as a function of r and/or h.
    7. Define this function in Derive and follow our standard method for finding maxima and minima. Remember that r is the fixed radius and h is the height, which we can change. (Hint: Recall 0 < h < r so you know the endpoints and are trying to find the absolute max on this interval. )
    8. Note your final answer should be a way to find h as a function of r. If Derive will not give an expression of the form h =, with only a function of r on the right side, you may need to do some algebra yourself to find a nice form.
    9. For disks of radii 1, 2, 3, 4, and 5 give the value for h that should be used.
  2. jcsd
  3. Oct 23, 2007 #2


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    Any thoughts? Is this homework?
  4. Oct 24, 2007 #3


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    First of all, state the problem correctly! Is the disk horizontal or vertical? If horizontal, the problem is trivial. If vertical (which I think is the correct assumption) it's only a little harder.
  5. Oct 24, 2007 #4
    the disk is vertical..... and so far i have the following...

    1. Dry Area = pi*h^2
    2. Wet area = pi*r^2 - pi*h^2
    3. (acos(h/r))/2 * r^2 = sector area
    4. sqrt(r^2-h^2)*h = area of triangle

    im working on 5 being the wet area - (sector area - area of triangle)...

    and this is a challenge problem from my professor for an extra point...
    Last edited: Oct 24, 2007
  6. Oct 25, 2007 #5
    ANY MORE HELP??? with parts 6-8
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