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Given

$\alpha = 2.44 $ rad/$s^2$

$\omega = 2.44t+8.35$ rad/s

$\theta = 1.22t^2+8.35t$ rads

Find an expression for the magnitude of translational acceleration at $t=1.82s$, given that the radius of the circle is r.

What I did was use the fact that $a = r\omega^2 = r[(2.44)(1.82)+8.35]^2 = 164r$. This answer is correct, however, I was just wondering about something.

Isn't the translational acceleration also defined as $a=r\alpha$, which follows from $v=r\omega$ by taking the time derivative of both sides. In this case, $a = 2.44r$, as $\alpha$ is a constant angular acceleration. So then is the answer $a = 164r$ or $a=2.44r$? Am I seriously overlooking something here?

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Neither result is correct, but both are part of the solution!

The angular motion is given by ($r$ is the radius and $\theta$ is a function of time): \begin{align*} \vec r &= r \begin{pmatrix} \cos(\theta) \\ \sin(\theta) \end{pmatrix} \end{align*} From this we define $\omega = \partial_t \theta$, $\alpha = \partial_t \omega$. The translational acceleration on the other hand is $\vec a = \partial_t^2 \vec r$. By doing this derivative and inserting the definition of $\omega$ and $\alpha$ you will arrive at the general result (and see how it relates to the formulas you used).

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  • $\begingroup$ The answer was 164r though, the magnitude of translational acceleration. From what I understand, perhaps translational acceleration is different from tangential acceleration. $\endgroup$
    – Programmer
    Apr 16 '18 at 13:45
  • $\begingroup$ Yes, that is the key point: translational acceleration is a vector which decomposes into a normal and a tangential component (from which the magnitude can easily be computed if it is needed – since the decomposition is orthogonal), this decomposition can directly be seen when computing the derivatives. $\endgroup$ Apr 16 '18 at 16:19

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