Angular acceleration of a discus

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The discussion focuses on calculating the angular acceleration of a discus thrown by an athlete starting from rest. The thrower completes one revolution in 1 second, leading to confusion when applying different formulas for angular acceleration. One formula suggests an angular acceleration of 2π rad/s², while another yields 4π rad/s² due to the assumption of constant angular velocity. The key point is that the correct approach must account for angular acceleration, necessitating the use of the appropriate kinematic equations. Understanding these distinctions is crucial for accurately determining both angular acceleration and the discus's speed at release.
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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.
 
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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.
 
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?
 
You can use:

\omega = \frac{\Delta \theta}{\Delta t}

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

\omega = \omega_0 + \alpha t

or many other kinematic equations you use for constant acceleration.

One that might prove particularly useful to get \omega right away is:

\frac{\Delta \theta}{\Delta t} = \frac{\omega + \omega_0}{2}

taken from:

\frac{\Delta x}{\Delta t} = \frac{v + v_0}{2}

if you remember your kinematic equations for constant acceleration.
 
Thank you for taking the time to clarify that for me! :)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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