Problem: A woman is running at 8m/s in a circular motion, the radius of the circle is 100m. What is her acceleration in m/s/s?
I don't have a clue how to start! I wanted to use this:
\begin{align}\omega^2 - \omega_0^2 = 2\alpha(\theta-\theta_0)\end{align}
but I don't have the angle?
This is a question pertaining to centripetal acceleration. Keep in mind that:
\begin{align}a_c = \dfrac{v^2}{r}\end{align}
Does this make sense now? You are not dealing with rotational kinematics in this problem.
The question may seem odd at first because she is running at a constant speed and it is asking for her acceleration. But because she is running in a circle her direction is constantly changing, which means she is indeed accelerating. This is called her centripetal acceleration (Which is what the question asks for by saying "her acceleration"). The formula for centripetal acceleration is her velocity squared divided by the radius of the circle she is running in.
$$a_c = \frac{v^2}{r} $$
Hopefully this answers your question!
Use 2-dimensional vectors. Let the woman's position be the vector $\mathbf{r}$. Circular motion means that:
$$\mathbf{r} = r \begin{pmatrix}\cos\, \omega t \\\ \sin\, \omega t \end{pmatrix}$$
Assuming constant $r$ and $\omega$, it is easy to see by taking the time derivative of each vector component that the velocity vector $\mathbf{v}$ is:
$$\mathbf{v} = \frac{d\,\mathbf{r}}{dt} = \omega\, r \begin{pmatrix}-\sin\, \omega t \\\ \cos\, \omega t \end{pmatrix}$$
And the acceleration vector $\mathbf{a}$ is:
$$\mathbf{a} = \frac{d^2\,\mathbf{r}}{dt^2} = \omega^2\, r \begin{pmatrix}-\cos\, \omega t \\\ -\sin\, \omega t \end{pmatrix} = -\omega^2\, \mathbf{r}$$
The problem states that $r=|\mathbf{r}|$ is 100m and that $|\mathbf{v}|$ is 8m/s. That is all you need to calculate first $\omega$ and then $\mathbf{a}$.
