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Question:

A wheel starts is spinning at $27\text{ rad/s}$ but is slowing with an angular acceleration that has a magnitude given by $\alpha(t) = (3.0\;\mathrm{rad/s^4})t^2$. It stops in a time of: $\qquad$ ?

I happened to guess $3.0\text{ s}$ and it turned out to be right. But I want to understand why.

I tried using a kinematics equation and plugging $a$ in for $\alpha$, but it gave me the wrong answer. Am I missing something simple here?

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1 Answer

I figured it out. I needed to take the integral of the acceleration function, then since at t=0 w=27, my velocity function would be w = t^3 + 27. From there, it's just a simple matter of plugging in 0 for the final velocity and solving for t.

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That sounds correct! –  queueoverflow Feb 18 '12 at 16:24
    
Can you accept this answer so that it stupid being auto-bumped? –  Colin K Feb 18 '12 at 18:53
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