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Homework Help: Force on a diving board, work to stop a flywheel

  1. Oct 22, 2006 #1
    Hi. I'm having difficulty answering three questions of my homework and hope someone will lend a hand. The questions seem beyond what we've had in class but I need to do them regardless.

    Diving Board Problem
    Consider a diving board:
    Code (Text):

    [- 1m -][-  3.4m -]|_   <-- person at end of board
    |       |
    A       B
    I'm to find the force exerted on support A and support B. Givens: m_(board) and m_(person)

    I know that the sum of the torques is 0, and looking at someone else's work on another webpage, I came up with:

    F_b * 1m - m_(person)*g*4.4 - m_(board)*g*2.2 = 0

    Anyway, solving for F_b using this equation came out wrong plus I still need to solve for F_a.

    Work to Stop a Flywheel
    Given a flywheel with radius r, moment of inertia I, and it's rotation rate (revolutions/sec), if a force of 4.3 N is applied tangentially to the flywheel to slow it down, how much work will be done by this force in brining the flywheel to a stop?

    I know that W = delta KE = KE_f - KE_i

    and that KE_r = (1/2) * I * (omega)^2,

    now, rotation rate = 7.6 revolutions/sec = 7.6 * 2 * pi/sec = 47.7522 rads/sec

    Thus KE_r = (1/2) * I * (47.7522)^2

    Which looks good; but how do I fit the force applied to the flywheel in the W = KE_f - KE_i??

    Thank you, Elizabeth
  2. jcsd
  3. Oct 22, 2006 #2


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    For your first question you have two unknowns (i.e. the reaction forces at A and B); now what would happen if you were to take moments about point A?

    As for your second question it is not necessary to include the torque in your calculations, you have all the information you need already.
  4. Oct 22, 2006 #3
    Divide by zero? F_a * 0 - m_(board)*g*2.2 - m_(person) * g * 4.4 = 0

    How do I determine the work? I mean, if newtons are applied tangentially, it is not in the correct units (it should be joules).
  5. Oct 22, 2006 #4


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    Note quite, I said take moments about A, which means set your pivot at point A. And I apologise I didn't notice that on your post you said;
    Which is correct, this is what I meant by taking moments about A. Why do you think the above is incorrect?

    Think conservation of energy.
  6. Oct 22, 2006 #5
    ok, with a pivot point at A, with have:

    sigma tau = F_a * 0 + F_b * 1 - m_(board)*g*(L/2) - m_(person)*g*(L) = 0 (1)

    But we have this, too:

    sigma F_y = F_a + F_b - m_(board)g - mg = 0 (2)

    so using (2) I should be able to find F_a from (1). Look good?
    EDIT: I correctly solved this {the first part of the larger set of questions}. Thank you.

    As for the other question regarding the flywheel, I set up this:

    k_f + u_f = k_i + u_i

    Where k_i is the kinetic energy of the flywheel, u_i = 0, and somehow work in the newtons onto the LHS?

    Thank you.
    Last edited: Oct 22, 2006
  7. Oct 22, 2006 #6


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    Good stuff, my pleasure.:smile:
    Hmm, ok. Let's forget about this question for a moment. Say you have a box sliding across a frictionless surface with a kinetic energy of one joule. How much work would you have to do stop the box?
  8. Oct 22, 2006 #7
    work = delta KE = KE_f - KI_i

    So, KE_f = 0 (because it is stopped), so:

    work = 0 - 1 J

    KE_r = (1/2) * I * (47.7522)^2

    W = F * d = KE_f - KE_i = 0 - KE_r

    Substitute the value given (the one in Newtons) as F, solve for d, and the answer is F d?
    Last edited: Oct 22, 2006
  9. Oct 22, 2006 #8


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    That is correct.
    You could do that but since Fd is equal to the change in kinetic energy (which you already know) why would you want to do that? You do not have to use the toque in order to solve this problem.
  10. Oct 22, 2006 #9
    Yes, there was no reason to even use the force applied tangentially. But, I got the answer!! and I accomplished my goal of completing my physics homework today and I owe you a big Thank You for your help!!

    Best wishes, Elizabeth
  11. Oct 22, 2006 #10


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    It was my pleasure. :smile:
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