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Cone-shaped drain speed based on R and time

  1. Oct 13, 2008 #1
    1. The problem statement, all variables and given/known data

    A basin surrounding a drain has a shape of a circular cone opening upward, having everywhere an angle of 35 with the horizontal. A 25 g ice cube is set sliding around the cone without friction in a horizontal circle of radius R.
    (a) find the speed the ice cube must have based on r
    (b) is any piece of data unnecessary for the solution? suppose R is two times larger
    (c) will the required speed inc, dec, or stay constant? If it changes by what factor?
    (d) will the time required for each revolution inc, dec, or constant? by what factor if it changes?
    (e) do answers to part c and d seem contradictory? Explain how they are consistent

    2. Relevant equations


    3. The attempt at a solution
    I don't know where to start. I wasn't even able to come up with a FBD
    guessing if I use (SIGMA)Fr=mv^2/r
    you can solve for speed for part a but what would be on the other side.
    i really have no idea
  2. jcsd
  3. Oct 13, 2008 #2


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    Homework Helper

    Think of it as a banked curve. What speed needs to be maintained so the centripetal acceleration will balance the downward force of gravity along the incline.
    (Hint: The angle affects each. Which functions of the angle need to be applied to each?)
  4. Oct 13, 2008 #3
    SIGMA Fr=(mv^2)r

    then from vertical component we find
    so N=(mg)/cos@
    mg/cos@ * sin@ =(mv^2)r
    do some work~~~~
    did i do this right?
  5. Oct 13, 2008 #4
    mass is unnecessary because it is crossed out
  6. Oct 13, 2008 #5
    (c) inc by a factor of sqrt(2)

  7. Oct 13, 2008 #6
    then part d

    and then period will decrease by a factor of 1/sqrt(2)
  8. Oct 13, 2008 #7
    okay i think i got it.
    haha THANKSS for the clue!!!
  9. Oct 13, 2008 #8


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    Homework Helper

    Good enough.
    Good again.
    You're on a roll.
    Almost. V increases by Sqrt(2) bur R doubles in that equation.
  10. Oct 13, 2008 #9
    thanks again !
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