# Finding translational acceleration

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?

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).