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Two circles and revolutions

  1. Feb 28, 2008 #1
    There are two disks of uniform density that touch at one point. their masses are in a ratio of 1:9. how many revolutions does the smaller disk make as it makes one rotation around the big circle? (assume that the disks do not slip)

    This is my try:

    a is small circle
    b is big circle

    M(a)/R^2(a)=M(b)/R^2(b)

    M(a)R^2(b)=M(b)R^2(a)

    M(a)R^2(b)=9M(a)R^2(a)

    R^2(b)=9R^2(a)

    assume radius of a=1

    R^2(b)=9

    R(b)=3

    Circumference of B

    2(pi)(3)=6pi

    Circumference of A

    2(pi)(1)=2pi

    number of rotations a makes around b

    6pi/2pi=3

    so three rotations.

    the answer is four.

    help.
     
  2. jcsd
  3. Feb 28, 2008 #2

    tiny-tim

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    Looks right to me.

    When the little disk has gone a quarter round, from say 12 o'clock to 3 o'clock on the big disc, the bit that touches changes from 6 o'clock to 9 o'clock. So 9 o'clock is touching 3 o'clock, which means that the little disc has gone round exactly once!

    Either you're right or I'm as dumb as you are!

    help.
     
  4. Feb 28, 2008 #3
    that's the logic. in one quarter of the big disk, the little disk has made one revolution. so in another quarter, it will go again only once. catch my drift?

    four quarters

    four revolutions....

    now to the math part of that....

    by the way, this shows that you're smarter than i am, so none of your options :wink:
     
  5. Feb 29, 2008 #4

    tiny-tim

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    … perhaps I'm just sleepier than you are! …

    oops! I wrote my last post just before going to bed, and I think I mis-read your original post.:redface:

    When you wrote "the answer is four", I thought you meant your answer was 4, and that you'd added the 1 (see below) to the 3 to get 4.

    But after a good night's sleep I see you meant your answer was 3 and the official answer was 4.

    mmm … :redface:

    It would have been better if I'd said that the math is that you'd correctly worked out how many revolutions it made relative to the point of contact, but you had to add on the number of revolutions of the point of contact itself (which is 1)!

    It's like the question: how many times does the earth revolve in one year? The answer is 366ish, not 365ish, because you have to add on one extra turn for its orbit round the sun. :smile:
     
  6. Mar 1, 2008 #5
    can you explain that more?

    are you saying that the point that is touching both circles when we start is the "point of contact"?

    so i calculated how many times that point of contact on the small circle touches the big circle?
     
  7. Mar 2, 2008 #6

    tiny-tim

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    Point of contact

    No

    The "point of contact" is moving round both circles. At any particular time, it's the point on each circle that just happens to be touching the other circle at that particular time.

    Imagine that the centres of both circles are fixed, and that both circles are free to rotate (like a pair of cogwheels).

    Then obviously the little circle goes round three time while the big circle goes round once.

    Now glue the circles together, and rotate both of them once around the centre of the big circle.

    You've now achieved the same as in the original exercise, except that instead of the little circle gradually going round the big one, it did all its "own" turning first, and then whizzed round the big circle all in one go!

    And that's why you always have to add one! :smile:

    (btw, the Earth's year is 365ish "solar days", but 366ish "sidereal days" - a sidereal day is the time it takes a star to "go round the Earth".)
     
  8. Mar 4, 2008 #7
    thanks!
     
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