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Have a problem with this Rotational Motion question

  1. Jun 19, 2017 #1
    1. The problem statement, all variables and given/known data

    A wheel, of radius 200mm, rolls over the top of a hill with a speed of 20m/s and negligible friction losses. (I = 1/2mr^2)

    2. Relevant equations

    Find the speed of the wheel when it is 10m below the top.

    3. The attempt at a solution

    mgh = 1/2mv^2 + 1/2IW^2

    W= v/r

    mgh = 1/2mv^2 + 1/2(1/2mr^2)(v/r)^2
    mgh = 1/2mv^2 + 1/4mv^2
    gh = 3/4mv^2
    v^2 = 4gh/3
    v^2 = 4(9.81)(10)/3
    v = 11.4m/s

    I got v = 11.4 m/s
    but my answer is incorrect as it is different from the answer given which is 23m/s.
    I want to know the correct solution.

    I've also tried searching for the height first.

    h = (3/4v^2)/g
    h = 30.58m

    and then 10m below from the top

    h = 20.58m

    v^2 = 4gh/3
    v^2 = 4(9.81)(20.58)/3
    v = 16.41m/s

    Which is still far from the answer also.
     
  2. jcsd
  3. Jun 19, 2017 #2
    h = 30.58m

    I tried adding 10m to h instead of subtracting 10m

    so, h = 40.58

    v^2 = 4gh/3
    v^2 = 4(9.81)(40.58)/3
    v = 23m/s

    Doing it this way, I got the correct answer.
     
  4. Jun 19, 2017 #3

    haruspex

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    Do you now understand why that is correct?
     
  5. Jun 19, 2017 #4
    The wheel is already rolling on the top of a hill
    my mistake is I assume the initial kinetic energy is 0 but it is actually not.
    So initial energy is mgh + 1/2mv^2 + 1/2IW^2


    mgh + 1/2mv^2 + 1/4mv^2 = 1/2mv^2 + 1/4mv^2

    gh + 1/2v^2 + 1/4v^2 = 1/2v^2 + 1/4v^2

    gh + 3/4v^2 = 3/4v^2

    (9.81)(10) + (3/4)(20)^2 = 3/4v^2

    398.1 = 3/4v^2

    v^2 = 4(398.1)/3
    v^2 = 530.8
    v = 23m/s

    Here is another way to answer the question.
    This solution is more understandable for me than the earlier solution.
     
  6. Jun 19, 2017 #5
    Although I got the correct answer for that, I still don't understand why do I have to add 10m to the height that I got instead of subtracting 10m to it.
     
  7. Jun 19, 2017 #6
    I also cannot assume that the final potential energy is 0 so

    I think this is the most understandable solution.

    mgh + 1/2mv^2 + 1/2IW^2 = mgh + 1/2mv^2 + 1/2IW^2

    mgh + 1/2mv^2 + 1/2mv^2 = mgh + 1/2mv^2 + 1/2mv^2

    gh + 3/4v^2 = gh + 3/4v^2

    (9.81)(30.58) + (3/4)(20^2) = (9.81)(20.58) + 3/4v^2

    (9.81)(30.58) - (9.81)(20.58) + 300 = 3/4v^2

    3/4v^2 = (9.81)(30.58-20.58) + 300

    3/4v^2 = (9.81)(10) + 300

    3/4v^2 = 398.1
    v^2 = 530.8
    v = 23m/s
     
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