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Please help before I go crazy

  1. Sep 15, 2009 #1
    1. The problem statement, all variables and given/known data
    A crankshaft with a diameter of 3.0 cm , rotating at 2300 rpm comes to a halt in 1.30 s . What is the tangential acceleration of a point on the surface of the crankshaft?


    2. Relevant equations
    W=v/r
    Wf=Wi+@t
    @=At/r

    3. The attempt at a solution
    -2300 rpm=138000r/s=13006.19m/s

    -v=13006.19m/s
    -r=.015m
    -t=1.3

    -W=867079

    -@=666983

    -@10004.7

    the answer is completely wrong.
     
    Last edited: Sep 15, 2009
  2. jcsd
  3. Sep 15, 2009 #2

    sylas

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    What are the units on those numbers?
     
  4. Sep 15, 2009 #3
    sorry, edited
     
  5. Sep 15, 2009 #4

    drizzle

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    Gold Member

    did you convert the units, cm to m and rpm to rps?

    edit: you did, but:


    2300 rpm=2300r/m=2300r/(60 sec)=2300/60 r/s
     
    Last edited: Sep 15, 2009
  6. Sep 15, 2009 #5
    yes, if you see i said the radius = 1.5cm=.015m and 2300rpm=13006.19m/s
     
  7. Sep 15, 2009 #6

    sylas

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    Giving names the variables so we know what is being discussed.

    W (angular velocity in radians per second)
    v (tangent velocity in meters per second)
    r (radius in meters)
    @ (angular acceleration in radians per sec per sec)
    A (tangential acceleration in meters per sec per sec)
    t (time in seconds)

    Wf is final and Wi is initial; but since it is going to zero, we can just use W = @t

    Drizzle is right. You've done the unit conversions in step 3 incorrectly.
     
  8. Sep 15, 2009 #7
    ahh finally!! im clumsy.. thank you
     
  9. Sep 15, 2009 #8

    drizzle

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    Gold Member

    no you’re not, just think logically next time after you did the conversion of units, i.e. could the number of rounds in a second be greater than the number of rounds in min of the same system or should it be less? just a bit of advice :wink:
     
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