Given torque, what is the direction of motion? 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.
 A: Suppose the force $\vec F$ and displacement $\vec r$ are in the xy plane.

Then you would expect the the angular acceleration would be anticlockwise looking down from the top just from the direction of the force.
The torque $\vec \tau = \vec r \times \vec F$ is in the $\hat z$ direction but so is the the direction of the angular acceleration $\vec\alpha$ if you use the right hand grip rule.  
The linear acceleration is in the same direction as the five which is in the xy-plane and hence perpendicular to $\hat z$ and the angular acceleration $\vec \alpha = \alpha \hat z$.
A: Yes, $\alpha$ is perpendicular, which is a quite normal thing in rotational motion, But first things first. You shouldn't write it with vectors but $\tau = I \alpha$ as this form is true for a quasi 2D motion. And NO, $I$ not a scalar but a tensor and, in general, changes the direction of vectors. 
Now to the direction of $\alpha$. It is the same for angular velocity $\omega$ This is a vector as well, the length is the speed and the direction is the axis of rotation. As acceleration changes the speed you can imagine that $\alpha$ should also point in the direction of the axis of rotation, which is in the direction of $\tau$.
For the motion you have to consider
$$\dot {\vec L}=\vec\tau=\mathbf I \dot{ \vec \omega}$$
A: 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.
A: 

You can now work backward. Point your thumb along the direction of the torque, your fingers will give you the direction of the applied force (direction of the tangential acceleration).
If the torque is coming out of the plane, the tangential acceleration is in the anti-clockwise direction.
If the torque is going into the plane, the tangential acceleration is in the clockwise direction.
If you know the tangential acceleration, you can now determine how the motion of the rotating object changes. Suppose the object was at rest, then the direction of the tangential acceleration will give you the direction along which the body begins to rotate.
A: To start off, note that the direction of torque obtained by the cross product is NOT the same as the direction in which the torque rotates the body(the former case only gives the sense of  axis-whether into the plane or out of it). The angular acceleration must be perpendicular to the force, for otherwise there would would be tangential components of the acceleration, causing the body to 'tear apart'; inconsistent with the definition of a rigid body.
Now, as to the direction of rotation a torque causes- a simple working rule may be identified-
(1) Identify the point/axis about which the rotation occurs.
(2) DRAW (this is essential) the direction of the force on the body. Make an arrow, sort of.
(3) Now, extend this arrow into an arc such that the arc ENCLOSES the point of rotation.( There will be two directions to extend the arc-leftward or rightward. Extend it along the direction in which the axis of rotation lies).
(4) The direction of this arc is the direction your torque rotates the body.
Thus, in the figure (pardon the lousy diagram), the torque rotates the body CLOCKWISE. Note that you MUST keep track of anticlockwise and clockwise torques; and add them algebraically.
Hope this helps.
