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Homework Help: Angular acceleration of a discus

  1. Jul 28, 2010 #1
    To throw a discus, the thrower holds it with a fully outstretched arm. Starting from rest, he begins to turn with a constant anguLar acceleration. The diameter of the circle in which the discus moves is about 1.8 m. If a thrower takes 1.0 s to complete one revolution, starting from rest, what will The speed of the discus be at release?

    I'm just starting this question and am wondering why when I calculate the angular acceleration using formula:

    Angular acceleration= change in Angular velocity/change in time which for this problem is 2pi rad/1 sec/1sec I get 2 pi rad/s^2

    But when I use the formula:

    Change in theta=initial angular velocity * initial change in time time + .5 * angular acceleration * time^2

    I get 4pi rad/s^2 bc
    2pi=.5(angular acceleration) (1)^2 I get 4pi rad/s^2. So why am I getting two different angular accelerations with these two formulas? I know the problem Is asking for speed but first I want to know why I get two different answers for angular acceleration.
  2. jcsd
  3. Jul 28, 2010 #2
    For your first equation you are assuming the change in angular velocity is 2*pi rad/s. This isn't a correct assumption since you don't know the final angular velocity.
  4. Jul 28, 2010 #3
    But If the discus starts at rest, then Wi=zero rad/s and after 1 second it goes 1rev which is 2 pi radians so doesnt that mean that at that point the final w is 2 pi rad/second therefore the change is 2 pi radians/Sec?
  5. Jul 28, 2010 #4
    You can use:

    [tex]\omega = \frac{\Delta \theta}{\Delta t}[/tex]

    only if there is no angular acceleration. If there is angular acceleration then you need to use:

    [tex]\omega = \omega_0 + \alpha t[/tex]

    or many other kinematic equations you use for constant acceleration.

    One that might prove particularly useful to get [itex]\omega[/itex] right away is:

    [tex]\frac{\Delta \theta}{\Delta t} = \frac{\omega + \omega_0}{2}[/tex]

    taken from:

    [tex]\frac{\Delta x}{\Delta t} = \frac{v + v_0}{2}[/tex]

    if you remember your kinematic equations for constant acceleration.
  6. Jul 29, 2010 #5
    Thank you for taking the time to clarify that for me! :)
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