Angular acceleration of a discus

[sona1177]

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.

 

[nickjer]

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.

 

[sona1177]

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 doesn't that mean that at that point the final w is 2 pi rad/second therefore the change is 2 pi radians/Sec?

 

[nickjer]

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.

 

[sona1177]

Thank you for taking the time to clarify that for me! :)