Something weird about Rotational Kinematic Equations

AI Thread Summary
The discussion revolves around calculating the average angular acceleration of a gramophone record, with initial and final angular velocities provided. The user initially uses the equation w² = W² + 2aθ, resulting in an acceleration of 24.5 rad s^-2, but later realizes that multiplying by 2 gives 49 rad s^-2. Participants clarify that the equation t = θ/w is incorrect since angular velocity is not constant, suggesting it should be t = 2θ/w. The correct approach involves using standard rotational kinematic equations, emphasizing the need to account for angular acceleration properly. Accurate application of these equations is essential for determining the correct average acceleration.
Stefenng
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Homework Statement


I need to find the average acceleration of a gramophone record.

Initial angular velocity, W = 0
Final angular velocity, w = 3.50rad s^-1
angular displacement, 8 = 0.25rad

Homework Equations


The equation given by the answer is

t= 8/w
a = ( w - W ) / t
the answer is 49.0 rad s-2

The Attempt at a Solution


This is the equation I use

w2 = W2 + 2a8

and I get 24.5rad s-2

But if I multiply it with 2, the angular acceleration will become 49rad s-2

That mean my equation will become
w2 = W2 + a8

Can anyone explain this to me?
Or I use a wrong equation?
 
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Stefenng said:

Homework Equations


The equation given by the answer is

t= 8/w
a = ( w - W ) / t
the answer is 49.0 rad s-2


The Attempt at a Solution


This is the equation I use

w2 = W2 + 2a8

and I get 24.5rad s-2

You are right:

t= 8/w is wrong because the angular velocity is not constant

It should be t= 28/w
 
ok...thx... ^^
 
Stefenng said:

Homework Equations


The equation given by the answer is

t= 8/w
a = ( w - W ) / t
the answer is 49.0 rad s-2

Hi Stefenng! :smile:

(have a theta: θ and an omega: ω :wink:)

You must learn the standard https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations …

it isn't θ = ωt (or t = θ/ω), it's θ = ωt + 1/2 αt2 :wink:
 
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