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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.

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  • $\begingroup$ The direction of angular acceleration (and angular velocity) is indeed perpendicular to the surface. If you know the direction of the torque, you know the direction of the angular acceleration. From the direction of the angular acceleration, you can obtain the direction of motion. $\endgroup$ – Yashas Mar 17 '17 at 13:52
  • $\begingroup$ @YashasSamaga How do I obtain the direction of motion? $\endgroup$ – Tommi Mar 17 '17 at 13:52
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    $\begingroup$ You have to clearly distinguish between the angular acceleration of the bar and the linear acceleration of a point at the end of the bar. You are confusing the two. The angular acceleration points "out of the page", but the linear acceleration of a point at the end of the bar accelerates down. They are two different but related quantities. $\endgroup$ – garyp Mar 17 '17 at 14:02
  • $\begingroup$ @garyp Yes, I was confusing the two and noticing this confusion is what prompted this question. $\endgroup$ – Tommi Mar 17 '17 at 14:06
  • $\begingroup$ So ... are you ok now, or are you still confused? $\endgroup$ – garyp Mar 17 '17 at 14:07
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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.

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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 $$\dot {\vec L}=\vec\tau=\mathbf I \dot{ \vec \omega}$$

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Suppose the force $\vec F$ and displacement $\vec r$ are in the xy plane.

enter image description here

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 $\vec \alpha = \alpha \hat z$.

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  • $\begingroup$ The only thing missing is to point out that the linear acceleration of a point at the end of the bar is in the direction of $\vec{F}$. So $\vec{a}$ and $\vec{\alpha}$ are perpendicular to each other. $\endgroup$ – garyp Mar 17 '17 at 14:09
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    $\begingroup$ @garyp Thank you for the advice. It is much appreciated. $\endgroup$ – Farcher Mar 17 '17 at 14:32
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enter image description here

enter image description here

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.

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  • $\begingroup$ True, but this does not answer the question. Take another close look at the question. $\endgroup$ – garyp Mar 17 '17 at 14:03
  • $\begingroup$ The OP wants to find out what the direction of the tangential acceleration is (OP did not ask for tangential acceleration rather, they asked for the direction of motion; this answer is for a general case where the torque can cause a tangential deceleration instead of starting a body from rest). This answer tells him to use the right-hand thumb rule backward to obtain the direction of the tangential acceleration from the direction of the torque. What did I exactly miss? $\endgroup$ – Yashas Mar 17 '17 at 14:06
  • $\begingroup$ I guess I'm concerned that the connection between tangential acceleration and the acceleration of a point at the end of the bar might not be obvious. But that could just be me. $\endgroup$ – garyp Mar 17 '17 at 14:12
  • $\begingroup$ @garyp I'm fine with that. $\endgroup$ – Tommi Mar 17 '17 at 14:13
  • $\begingroup$ @TommiBrander ok, good. I'm being too cautious. $\endgroup$ – garyp Mar 17 '17 at 14:15
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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.enter image description here

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.

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  • $\begingroup$ This gives a way of determining the linear acceleration from the force; I'd like to determine it when given torque, without necessarily knowing the precise forces involved. $\endgroup$ – Tommi Mar 17 '17 at 14:13
  • $\begingroup$ @TommiBrander when given torque, you must obviously know which force causes it, don't you? It is a kind of circular argument here. Or am I missing something? $\endgroup$ – GRrocks Mar 17 '17 at 15:34

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