Torque is the cross product $\vec \tau = \vec r \times \vec F$, which means it is perpendicular to both $\vec r$ and $\vec F$.
Consider some essentially two-dimension problem, like a horizontal iron bar with one end fixed, affected by gravity. The direction of the torque is perpendicular to the bar and gravity.
I also see a vector formula like $\vec \tau = I \vec \alpha $. Since the moment of inertia $I$ is a positive scalar, it does not change the direction of vectors. Hence, this kind of formula implies that the angular acceleration is perpendicular to the force causing it.
In our example, the non-fixed end of the iron bar would start moving down, but this acceleration is perpendicular to torque. This implies that it is perpendicular to $\vec \alpha$, above.
This leaves me quite confused; given torque, how can I determine how an object starts moving? There should be a cross product involved, somewhere; otherwise, the perpendicularity do not work out correctly, I think.
Best Answer
Let's start with determining velocity $\newcommand{\v}{\mathbf{v}}\v$ from angular velocity $\newcommand{\w}{\boldsymbol{\omega}}{\w}$. If an object is currently at position $\newcommand{\r}{\mathbf{r}}\mathbf{r}$, and is rotating about a fixed point, which we will take to be the origin, with angular velocity $\w$, then the object's velocity is given by $\v =\w \times \r$.
Now to find the object's linear acceleration $\newcommand{\a}{\mathbf{a}}\a$, simply differentiate the above equation:
$\begin{equation} \begin{aligned} \a = \dot{\v} &= \dot{\w}\times \r + \w \times \dot{\r} \\ &= \newcommand{\al}{\boldsymbol{\alpha}}\al \times \r + \w \times \left( \w \times \r \right) \\ &= \al \times \r + \w \left(\w \cdot \r\right) - \omega^2 \r \\ &=\al \times \r - \omega^2\left(\mathbb{I} - \hat{\omega}\otimes\hat{\omega} \right)\r. \end{aligned} \end{equation}$
Above, the second line introduces the angular acceleration $\al$, defined as the time derivative of $\w$. Also in the second line, but in the second term, we used the result for velocity that $\dot{\r} = \v = \w \times \r$. In the end, we got that the linear acceleration $\a$ consists of two terms. The second term is the usual centripetal acceleration term, which looks like $-\omega^2r$, but there is a projection which makes sure you are using the separation from the closest point on the axis of rotation (that is, you subtract of the component of $\r$ along $\w$).
The tangential acceleration, then, must be contained in the first term. Since it is given by a cross-product with $\r$, we see it is perpendicular to $\r$ and therefore is "tangential" in the sense that is tangent to the sphere of radius $r$. Notice in the special case where the axis of rotation is fixed, so that $\a$ and $\w$ are colinear, $\a= \al \times \r$ is colinear with $\v = \w \times \r$, so the tangential acceleration is colinear with the velocity as expected.