1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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! :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Angular acceleration of a discus
  1. Angular acceleration (Replies: 1)

  2. Angular acceleration! (Replies: 2)

Loading...