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Speed boat question: calculating centripetal acceleration

  1. Jun 15, 2008 #1
    1.Speedboat
    A speedboat with a length of 10.2m and a mass of 247 kg is negotiaging a circular turn (radius = 33 m) around a buoy. During the turn, the engine causes a net tangential force of magnitude 521 N to be applied to the boat. The intial speed of the boat going into the turn is 4 m/sec.

    a) What is the tangential acceleration of the boat, 2 sec into the turn?
    aT = m/sec2 *

    b) What is the tangential velocity of the boat, 2 sec into the turn?
    vT = m/sec *
    8.2 OK

    c) What is the centripetal acceleration, 2 sec into the turn?
    acen = m/sec2
    528 NO

    d) What is the total acceleration, 2 sec into the turn?
    a = m/sec2






    2. I got parts A and B, but for part C I tried using the formula: a(cp)= rw^2, i tried using the intitial velocity, 4 m/s and also tried using the calculated velocity from part b, 8.2 and neither are being accepted, am i making a conceptual mistake here in trying to apply this formula here?



    3. Solution attempt: a=(33)(4)^2 / 33 and also, a=(33)(8.2)^2 / 33
    I think I understand that the total will just be (At^2 + Acp^2) ^ .5 This is my first post so I hope this was in a clear enough format and not misplaced, thanks
     
  2. jcsd
  3. Jun 15, 2008 #2
    sorry, the answer from part A was 2.1m/s^2
     
  4. Jun 15, 2008 #3
    hmm i've tried using different equations and not seeming to get them to work either, could it just be a problem with the computer system taking my answers?
     
  5. Jun 16, 2008 #4

    alphysicist

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    Homework Helper

    Hi zooboodoo11,

    For part c, in your formula for centripetal acceleration:

    [itex]
    a_c [/itex]= r w^2


    w is the angular velocity, but it looks to me like you are plugging in the linear velocities. There is another formula for centripetal acceleration that uses the linear velocity (which is what you found in part b). What is that formula? Once you find that, I think you'll get the correct answer.

    (Or equivalently you could convert the linear velocity from part b into an angular velocity and use your original formula.)
     
  6. Jun 16, 2008 #5
    A(t)= linear velocity/time
    A(centripetal)= velocity(linear)^2/radius

    A(total)= [A(t)^2+A(c)^2]^1/2
     
  7. Jun 16, 2008 #6

    alphysicist

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    Hi physixguru,

    I don't believe your first equation is quite right. In this problem the tangential acceleration magnitude would be

    [tex]
    a_T = \frac{\mbox{change in linear speed}}{\mbox{time interval}}
    [/tex]

    (because we know the tangential part is constant). More generally it is the derivative of the linear speed with respect to time.
     
  8. Jun 17, 2008 #7
    That was not said with respect to the problem in any case.It was a generalization from my side.It was not question specific at all.
     
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