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Angular Acceleration

  1. Sep 16, 2016 #1
    1. The problem statement, all variables and given/known data
    A sprinter runs the curve of this 200 m in 12.96 s. Assume he ran in a lane which makes a semicircle (r = 36.8 m) for the first part of the race. At 3.36 s into the race his speed is 5.1 m/s. At 7.6 s into the race his speed is 9.6 m/s. His speed after the curve was 11.1 m/s.

    What was the sprinter's average angular acceleration after 12.96 s? Answer in deg/s2. I.e., what was his average acceleration while running the curve?

    2. Relevant equations
    This is for a biomechanics class. It's just a practice question to study but I am not in a physics program and this is my only math related course and i literally know very little about this. I have tried all of the formulas I know and I still can't get the right answer. Please help lol.

    3. The attempt at a solution
     
  2. jcsd
  3. Sep 17, 2016 #2

    haruspex

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    To find the average acceleration, or average angular acceleration, we would need to know the velocity at the end of the curve, but we are not given that or any way to find it.
    E.g., suppose he stopped running after 12.95s and walked at 1.296m/s for 0.02s. His average acceleration over the first 12.96s would be (1.296m/s)/(12.96s)=0.1m/s2.
     
  4. Sep 17, 2016 #3

    Simon Bridge

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    Welcome to PF;
    I notice the question also explicitly asks for the acceleration after 12.96s ... but that is after the curve has been run.
    Is this exactly how the question is written in front of you?

    Niggle: the units go hard against the quantity ... so 12.96s and 200m not 12.96 s and 200 m.
     
  5. Sep 17, 2016 #4
    I don't understand the question.

    "Average angular acceleration after 12.96s", ie after the curve, is 0. No curve, no angle. No angle, no angular acceleration.

    "Average angular acceleration while running the curve", ie during the curve, is easily calculated by time and angle. (12.96s, semicircle).

    "During" isn't "after".

    And you've still got all that extra data laying around doing nothing.
     
    Last edited: Sep 17, 2016
  6. Sep 17, 2016 #5

    CWatters

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    I agree the question is confusing. Can you check its posted word for word. Is the "ie" something you added?
     
  7. Sep 17, 2016 #6

    haruspex

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    No, you can calculate average angular velocity that way, but not average acceleration, angular or otherwise. See my post #2.
    It is poorly worded, but I think it is clear they are using "after" in the sense of "average so far after 12.96s have elapsed."
    Also, whether there is angular acceleration on the straight depends on the reference axis. If we take it to be fixed at the centre of curvature of the curved portion of track, and the runner continues in a straight line at constant speed thereafter, the angular veocity diminishes, so the angular acceleration goes negative.
     
  8. Sep 17, 2016 #7
    Given a redefinition of "after" as "at", then
    becomes relevant.
     
  9. Sep 17, 2016 #8

    PeroK

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    Perhaps British English is different, but this is a perfectly value use of the word "after":

    "After half an hour, I was still no further forward with the problem."

    Means, precisely, "during the first half hour I made no progress."

    And, in this context:

    After 12s his average speed was 6.6m/s makes perfect sense to me. That's his average speed during the 12s.
     
  10. Sep 17, 2016 #9
     
  11. Sep 17, 2016 #10

    PeroK

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    That's the thing about English. It isn't as simple as that! I would never have doubted what "after 12.96s" meant in this context. For example:

    After two hours our average speed was 50mph.

    After we stopped, we had a coffee.

    That's the way English is.
     
  12. Sep 17, 2016 #11

    haruspex

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    That is certainly a reasonable interpretation, and of course the right one here, but it is at the least ambiguous.
    "A runner ran two laps. Her average speed for the first lap was 15kmh. Her average after the first lap was 14kmh."
     
  13. Sep 17, 2016 #12
    Sorry, I know it is not a very well worded question. I haven't changed anything and that is word-for-word what the question is. I'm pretty sure it is asking the average angular acceleration for the whole curve.
    I've managed to find the average velocity which is 13.88 degrees, but I can't find the forumla anywhere for average acceleration. I've tried taking the acceleration at each time interval and averaging that. I've tried taking the average velocity and using that divided by the time to find it, and i've tried a bunch of random formulas and I still can't get the right answer.
     
  14. Sep 17, 2016 #13
    13.88 degrees/second***
     
  15. Sep 17, 2016 #14

    haruspex

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    As I demonstrated in post #2, you do not have enough information to find the average acceleration.
     
  16. Sep 17, 2016 #15
    My notes say that Average Acceleration = the change in angular velocity divided by time?
    So would the change in angular velocity be 13.88 deg/s because thats what the velocity is at the end of the curve and he would have started at zero?
     
  17. Sep 17, 2016 #16

    haruspex

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    How do you know that is the angular velocity as the runner leaves the curve? As I wrote in post #4, his velocity at that particular point can be anything at all. We have no information on it. If he reached the end of the curve and stopped, his average acceleration would be zero.
    Do not confuse the instantaneous velocity at a given point with the average velocity up to that point.
     
  18. Sep 17, 2016 #17
    Sorry, the question is asking about his acceleration during the curve. Like throughout the whole 12.96 seconds. The previous question asked what the average velocity of the curve was so i took the time, displacement (180 degrees) and radius and found it was 13.88 degrees/s (which was the right answer). Im sorry this question is silly and not worded well. I feel like there is a simple way to find the answer but i just seem to be overlooking it
     
  19. Sep 17, 2016 #18

    haruspex

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    I understand that.
    I may have been a bit unfair to the question earlier. Look at the last piece of information, that the runner runs at 11.1m/s after exiting the curve. Let's assume that means also that the runner is running at that speed as he exits the curve. What does that give you for the angular velocity at that point? What then is the net change in angular velocity from to start to finish of the curve?
     
  20. Sep 17, 2016 #19
    Got it. Thanks!
     
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