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Angular Motion Question Understanding help ?

  1. Nov 27, 2007 #1
    1. The problem statement, all variables and given/known data

    A figure skater completes a double axle (2 complete rotations) in 0.5 seconds. Calculate
    the skater’s angular velocity and average angular velocity in a) deg/sec, and b) rad/sec.

    If the skater manages to stop spinning in a time of 1.5 seconds,
    what was the angular acceleration and average angular acceleration during this period (in deg/s)?

    2. Relevant equations

    Average Angular Velocity => w=(Theta2-Theta1)/(t2-t1)
    Angular Velocity = theta/time

    Average Angular Acceleration => Alfa = (w2-w1)/(t2-t1)
    Angular Acceleration => Alfa = (w)/(t)
    wf=wi + alfa*(t)

    3. The attempt at a solution


    Angular Velocity => w=(360*2)/0.5 = 1440 deg/s
    1440 * (pi/180) = 23.132 rad/s

    Average Angular Velocity => w=(Theta2-Theta1)/(t2-t1)
    Average Angular Velocity => w=(360-360)/0.5 = 0 rad/s


    Angular Acceleration => Alfa = (w)/(t)
    Angular Acceleration => Alfa = 1440/1.5 = 960 rad/s^2

    Average Angular Acceleration => Alfa = (w2-w1)/(t2-t1)= (0-1440)/1.5 = -960rad/s^2

    wf=wi + alfa*(t) = > alfa = (wf-wi)/t ==> alfa = (0-1440)/1.5 = -960 rad/s^2


    im almost sure about my results in the (Angular Velocity and Average Angular Velocity ) but for the (Angular Acceleration and Average Angular Acceleration i am not )

    could someone correct my answers and explain the changes that has been made.

    Thanx in Advance.
     
  2. jcsd
  3. Nov 27, 2007 #2

    Astronuc

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    Staff: Mentor

    One is correct. One revolution is 360° or 2[itex]\pi[/itex] radians, and one correctly used the relationship [itex]\pi[/itex]/180 rad/deg.
     
  4. Nov 27, 2007 #3
    Thanx but can anyone explain what is the correct angular acceleration (+960 or -960)and why ? and what is the difference in the pronouncings Average angular acceleration and angular acceleration hence

    average speed = total distance / time.
    average velocity = displacement / time.
     
    Last edited: Nov 27, 2007
  5. Nov 28, 2007 #4
  6. Nov 28, 2007 #5

    rl.bhat

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    Homework Helper

    Average velocity or acceleration is calculated if they are not uniform. In your problem there is no indication of that.
     
  7. Nov 28, 2007 #6
    What do u mean not uniform ? and what is the indication or the thing that will make me know wheather its uniform or not ?
    [​IMG]

    i still dont know how the second part is solved. Can someone clarify things to me ?
     
  8. Nov 28, 2007 #7

    rl.bhat

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    Homework Helper

    Average Angular Velocity => w=(Theta2-Theta1)/(t2-t1)This is not true.
    average angular velocity = total angular displacement / total time.
     
  9. Nov 28, 2007 #8
    ok
    applying the law
    average angular velocity = total angular displacement / total time.

    average angular velocity = 1440/ 1.5 = 960


    how is it -960 ?
     
  10. Nov 29, 2007 #9
    and what would that


    angular velocity be equal to in this case ?!
     
  11. Nov 29, 2007 #10

    rl.bhat

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    Homework Helper

    Angular Velocity => w=(360*2)/0.5 = 1440 deg/s
    1440 * (pi/180) = 23.132 rad/s. = Average angular velocity
    Angular Acceleration => Alfa = (w)/(t)
    Angular Acceleration => Alfa = 1440/1.5 = 960 rad/s^2

    When the body comes to rest the acceleration cannot be positive.
     
  12. Nov 29, 2007 #11

    Astronuc

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    Staff: Mentor

    As one worked out initially - Average Angular Acceleration => Alfa = (w2-w1)/(t2-t1)= (0-1440)/1.5 = -960rad/s^2

    wf=wi + alfa*(t) = > alfa = (wf-wi)/t ==> alfa = (0-1440)/1.5 = -960 rad/s^2

    The skater starts with an initial angular velocity wi at ti, and then decelerates to wf at tf.

    The change in angular velocity is wf-wi and the change in time is tf-ti, and the angular acceleration is alfa = (wf-wi)/(tf-ti). If the body comes to rest, wf=0, to alfa = -wi/(tf-ti), and since tf > ti, the difference is positive, to the angular acceleration is negative.


    w=(Theta2-Theta1)/(t2-t1) is correct, but one must be careful that Theta2 and Theta1 represent cumulative angular displacements from the same reference angle, and not just the angular displacement on a circle, i.e. Theta2 and Theta1 could > 360° or 2pi rad. In the given expression Theta2-Theta1 is the total angular displacement occuring between t2 and t1.
     
  13. Nov 29, 2007 #12
    thanx for your help but sorry for insisting ...

    Understood
    Thats just the same its converting from degree to radian( so converting from degree to radian is average angular velocity

    Not sure about that.

    Understood.


    Once again why was the average angular velocity = 0 but there was a valus for the angular velocity. Not only that but what law should be used when calculating
    Angular Acceleration & average Angular Acceleration.
     
    Last edited: Nov 29, 2007
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