Anonymous
Anonymous asked in Science & MathematicsPhysics · 9 years ago

1 cm thick rings are hanging from a peg. top rings outside diameter is 20cm. outside diameter of each of outer?

rings is 1cm less then the one above it. bottom rings outside diameter is 3 cm. So how would I find the distance from the top of the top ring to the bottom of the bottom ring?

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  • 9 years ago
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    Answer ... 170 cm ...

    First, determine the total number of rings …

    Let … i = number of the ring … i = 1 is the bottom ring R (1)

    … i = 2 is the ring R (2) above the bottom ring … etc …

    Let … ODR (i) = outer diameter of ring R (i) … then

    … ODR (i + 1) = ODR (i) + 1 in cm … i = 1 , 2 , 3 , ⋯

    For i = 1 … ODR (1 + 1) = ODR (2) = ODR (1) + 1 …

    and since ODR (1) = 3 … ODR (2) = 3 + 1 = 4 …

    For i = 2 … ODR (2 + 1) = ODR (3) = ODR (2) + 1 …

    Substituting … ODR (2) = ODR (1) + 1 … we get …

    … ODR (3) = [ ODR (1) + 1 ] + 1 = ODR (1) + 2 = 5 …

    For i = 3 … ODR (3 + 1) = ODR (4) = ODR (3) + 1 …

    Substituting … ODR (3) = ODR (1) + 2 … we get …

    … ODR (4) = [ ODR (1) + 2 ] + 1 = ODR (1) + 3 = 6 …

    That is … for i = 1 , 2 , 3 , we have …

    … ODR (2) = ODR (1) + 1 = 4 …

    … ODR (3) = ODR (1) + 2 = 5 …

    … ODR (4) = ODR (1) + 3 = 6

    Generalizing … ODR (n) = ODR (1) + ( n – 1 ) = n + 2 …

    Then … ODR (n) = 20 for the topmost ring corresponds to

    … ODR (n) = n + 2 = 20 … --> … n = 20 – 2 = 18 …

    That means there are 18 rings in all …

    Consider next the outside diameter of the ring (ODR) expressed

    in terms of the inner diameter of the ring (IDR) … since each ring is

    1 cm thick … ODR (1) = IDR (1) + 2 = 3 … --> … IDR (1) = 3 – 2 = 1 ...

    … ODR (2) = IDR (2) + 2 = 4 … --> … IDR (2) = 4 – 2 = 2 …

    … ODR (3) = IDR (3) + 2 = 5 … --> … IDR (3) = 5 – 2 = 3 …

    Generalizing … IDR (n) = n in cm …

    Now … the rings hang in such a way that any ring R(i) rests on the ring

    R(i+1) above it, so that 1 cm thickness of R(i+1) extends below the

    upper inner edge of R(i) , while the ring R(i -1) below ring R(i) hangs also

    on R(i), so that 1 cm thickness of R(i -1) also extends above the lower inner

    edge of R(i) …that means … IDR(i) = 1 + Δ(i) + 1 = 2 + Δ(i) … Δ(i) = separation

    distance between the outer diameters of rings R(i+1) and R(i -1) inside the inner

    diameter of ring R(i) …i = 2 , 3 , 4 , etc...For i = 2 … Δ(2) = IDR(2) – 2 = 2 – 2 = 0 …

    so that inside ring R(2) , R(3) and R(1) touch each other …for the distance of the top

    of ring R(i) from the top of ring R(i -1) inside ring R(i) , we find that …

    … DTR(i) = 1 + 1 + Δ(i) = 2 + Δ(i) … i = 2 , 3 , 4 , ⋯ , 18 …for i = 2 , we have …

    … DTR(2) = 2 + Δ(2) = 2 … since … Δ(2) = 0 … for i = 3 , we have …

    … DTR(3) = 2 + Δ(3) … Δ(3) = IDR(3) – 2 = 3 -2 = 1 …

    … DTR(3) = 2 + 1 = 3 …for i = 4 , we have … DTR(4) = 2 + Δ(4) …

    … Δ(4) = IDR(4) – 2 = 4 -2 = 2 … DTR(4) = 2 + 2 = 4 … Notice that DTR(n) = n …

    Finally … distance of top ring from bottom ring is … 18 + 17 + 16 + ⋯ + 3 + 2 = 170

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