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Given Mu find the radius.

  1. Apr 11, 2006 #1
    I am not sure where this problem belongs, but it was proposed by my math teacher as a challenge.

    Problem:
    An ant is located on a bowl that has the shape of a semi sphere and he is just able to get out of the bowl. The coefficient of static friction between the ant and the bowl is 1/3. Find the radius of the bowl.

    My attempt:

    Mu is the tangent ratio between Friction and Normal forces. The angle that is found by taking the arctangent(1/3) is also the angle that the radius is currently making with from the 0 position(bottom of the bowl).

    Since this is a bowl, the line tangent to the circle would be the slope used to calculate friction. Therefore as the line changes so does Mu, as of course is the angle. There is no time interval give.

    Since we are given a specific angle and nothing else relating to the bowl itself (my assumption) I decided to put this problem in polar coordinates. I represented the bowl as r=sin(?) and plugged in the angle. This produces a radius of (1/10)^(.5)meters. However, this seems like a forced solution to me.

    I thought about trying to find the position of the ant, but because the bowl could be located anywhere I couldn't treat this as a freefall problem.

    Thanks for your help.
     
  2. jcsd
  3. Apr 11, 2006 #2

    Kurdt

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    Consider coulomb friction F = mu*N where N is the normal force and then for the ant to just make it out of the bowl the force due to gravity must be equal to the force holding the ant on the surface. but remember the normal force will change because the bowl is curved.
     
  4. Apr 11, 2006 #3

    Doc Al

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    I guess I don't understand the problem (or I'm missing something). While it makes sense to ask for the maximum height (or angle) that the ant could climb up the side of the bowl (assuming the usual model for static friction--which doesn't quite apply to real ants, of course), I don't see how the ant could climb out of the bowl. (At the lip of a hemispherical bowl, the friction force would be zero.)

    That's the way to solve for the maximum angle (from vertical) that the ant could climb.
     
  5. Apr 11, 2006 #4
    There is either something wrong with the question or you have made a mistake in taking down the question.
    The ant can never get out of the bowl if it is a perfect hemisphere.
    This is because as the ant moves up, the hemisphere also curves upward until near the top where a tangent to the ant's motion starts becoming perp. to the floor (the ant moves in almost vertical direction). In this case the ant cannot make the bowl exert a normal force to bring friction into play and it falls off(no matter how large frictional forces are).
    Just like you can't scale a wall using friction between the shoes and the wall.

    Perhaps the question should be concerned with a bowl(non hemispherical) that is part of a sphere of given radius and the radius of its top circular cross section.

    Or perhaps if the ant could make an adrenaline burst and jump over the bowl before friction becomes 0.:biggrin:

    Edit:Doc Al is faster than me :biggrin:
     
  6. Apr 11, 2006 #5

    Kurdt

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    I too was confused about the shape and then thought he says it has the shape of a semi-sphere which does not necessarily say it is one. Just like saying a mirror is a parabolic shape does not mean it carries on to infinity. I have to agree the ambiguity is rather disturbing though.
     
  7. Apr 11, 2006 #6

    Doc Al

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    Nonetheless, without more data--such as the height of the bowl--one cannot solve for the radius.
     
  8. Apr 11, 2006 #7
    Perhaps this is the reason I have been having so much trouble with the problem. I too noticed that the ratio becomes Ff(max)/0 at the very edge, which is impossible.

    Would the problem change if the fact that it is a semi-sphere was ignored? If one is to assume that this is a bowl of any possible shape I am at a complete loss as to how to start this problem aside from the fact that an equation for the curve that the bowl makes would probably be the goal.
     
  9. Apr 11, 2006 #8
    Height is what is asked to find by the problem. The height of the bowl would be it's radius if I am not mistaken?
     
  10. Apr 11, 2006 #9

    Doc Al

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    If the bowl were a complete hemisphere, then the height would be the radius. But, as we've pointed out, the ant would not be able to climb out of such a bowl. I would check if your teacher misworded the problem.

    If the bowl is only part of a hemisphere, then you can find the angle the edge must make with the vertical.
     
  11. Apr 11, 2006 #10
    If the bowl is not a complete hemisphere and therefore the edge doesn't force an infinite value for Mu; then would it be correct to state the following.

    If the ant is just able to get out of the bowl then the Frictional force is maximum at that point. The problem states that the ant is "just able to get out" and then gives the mu of static friction. Therefore the angle we are given could be the maximum angle that the ant reaches on this portion of the sphere before escaping. If this reasoning is correct, then my answer of (1/10)^(.5) would be correct, or am I still making some kind of mistake?

    Thanks for all the replies.
     
  12. Apr 11, 2006 #11

    Doc Al

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    At that point, the frictional force will just be enough to balance the ant's weight.


    You can find the angle, but I don't see how you can find the radius. (Is that answer supposed to be a radius? What units?)
     
  13. Apr 11, 2006 #12
    Like i mentioned previously, I got the answer by solving the equation r=sin(theta) in polar coordinates, since the angle in this case determines the position of the ant.

    The answer would be in meters since there were no conversions to lesser or greter magnitudes along the way, but I realize that there is no mention of the scale in the problem as well, which would also lead me to say that its just "units" whatever they happen to be.
     
  14. Apr 11, 2006 #13

    Doc Al

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    I don't know where you got r = sin(theta), but it seems dimensionally incorrect: r has units of length while sin(theta) is a pure number.
     
  15. Apr 11, 2006 #14

    Kurdt

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    Well i considered total energy was converted to potential when the ant eventually makes it out of the bowl. its easy to find the angle from balancing friction with weight and i came up with this that depends on velocity which is unknown.

    r=(v^2)/(2*g*sin(theta))-1

    where g is the gravitational constant and v the velocity.

    Dimensionally correct anyway.
     
  16. Apr 11, 2006 #15

    Doc Al

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    Almost dimensionally correct. (You have a 1 floating around.)

    I suspect you meant to write something like this:
    r=(v^2)/(2*g*(1 - cos(theta)))

    (That's just applying conservation of energy to a frictionless bowl. Start at the bottom with a given speed; express the radius of the bowl in terms of the angle that the "ant" slides.)

    But it seems like you are mixing and matching ideas at random. You use static friction to find the angle, then add KE and use conservation of energy. What happened to friction? Is the ant now a projectile?

    I think that kd092 should get the corrected version of the problem before spending anymore time on this. :smile:
     
  17. Apr 11, 2006 #16

    Kurdt

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    That 1 is actually separate (I'll have to sort my latex) not part of the denominator. I thought it was kinetic friction, silly me. Anyway th friction drops out because from balancing you find that its equal to tan (theta), so its in the equation somewhere just rather obscurely. Anyway assuming that as kinetic friction will give a rough estimate as kinetic is usually not too much less than static.

    It will be interesting if he does post an answer but fortunately i have no more time left to attempt this as interesting as it is.
     
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