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Kinematics of Rotational Motion Part II

  1. May 15, 2012 #1
    1. The problem statement, all variables and given/known data
    I cannot get the answer for part c, -13.02rad/s^2


    3. The attempt at a solution

    52.08=260.42+α(10)
    -208.34=10α
    α= -20.83rad/s^2 (Correct)


    0=52.08+α(10)
    -52.08=10α
    α=-5.208rad/s^2 (Wrong)


    Please help, thanks..
     

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  3. May 15, 2012 #2

    Doc Al

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    How long did it take to come to rest? (How long was it moving at constant speed?)
     
  4. May 15, 2012 #3

    tiny-tim

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    hifreshbox! :smile:

    (try using the X2 button just above the Reply box :wink:)
    the 10 seconds for 300 to 60 mph is correct

    the 10 seconds for 60 to 0 mph is not correct …

    you need to allow for the time spent cruising :wink:
     
  5. May 15, 2012 #4
    How long did it take to come to rest? (How long was it moving at constant speed?)

    Since it states that total is 20sec, 300km/h-60km/h is 10sec, so 60km/h-0km/h is 10sec since 20-10=10
     
  6. May 15, 2012 #5

    tiny-tim

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    but in between, "it travels for another 100 m at the constant speed of 60 km/h" !!
     
  7. May 15, 2012 #6
    but it says
    "and it takes a total of 20 seconds to stop at the pit from its top speed of 300km/h"

    total =20 secs?
     
  8. May 15, 2012 #7

    Doc Al

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    Yes, the total time is 20 seconds. But that includes three segments:
    - initial slowdown to intermediate speed
    - constant speed motion at intermediate speed
    - final slowdown to zero speed

    Do not ignore the 'constant speed' segment. How much time is spent there?
     
  9. May 15, 2012 #8
    Is 2nd segment vi=16.67m/s v=16.67m/s a=0m/s^2 ?
     
  10. May 15, 2012 #9
    How do i calculate the time when my acceleration is 0 for segment b?

    ω=ωi+αt,v=vi+at
    a times t = 0

    No t in the equation anymore..:redface::cry:
     
  11. May 15, 2012 #10

    Doc Al

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    It's constant velocity motion: distance = speed * time.
     
  12. May 15, 2012 #11
    my book only has this formula.. no wonder i keep getting stucked..

    So S=d/t is the only way to solve?

    Thanks btw Doc Al
     

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  13. May 15, 2012 #12
  14. May 15, 2012 #13

    Doc Al

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    What formula? The attachment was blank.
    That's the basic definition of speed, so sure.
     
  15. May 15, 2012 #14

    Doc Al

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    The equation you want from your list is:
    s = vit + 1/2at2

    (s is distance, of course.)

    Just plug in a = 0.
     
  16. May 15, 2012 #15
    Ah... yea why didn't i see that hehe.. i understand already, thanks for the help..

    Btw if you don't mind can you take a look at https://www.physicsforums.com/showthread.php?t=605830..I just want to clarify some doubts..

    Since the question says that "Gear A is accelerated uniformly from rest making 20 revolutions in 12secs"

    Gear A: Vi=0, t=12s

    Can I say that Gear B/Drum C is also Vi=0, t=12s since they are all connected together.
     
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