# Applying Torque

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Hi, guys. I'm having trouble understanding torque in 3D. In 2D, T = r(perp)*F. Easy enough. When I googled the equation for 3D, I got T = rxF, which makes sense at first glance. However, I don't understand it now. Here is my problem: Say you are hitting a pool ball that looks like this:
   N
_
W / \ E
\_/
S

This is the top view of the ball. Now let's say you hit the ball horizontally on the very top towards N. Clearly, the resultant torque should be pointing North and have the magnitude r*F. However, if you do T = rxF, with r pointing straight up, the cross product gives you a vector that's orthogonal to both r and F. In other words, it's pointing W/E, not N. I know I'm probably misunderstanding something, but I'd really appreciate it if someone cleared this up for me.

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Quote:
 Original post by yaroslavdClearly, the resultant torque should be pointing North and have the magnitude r*F.

Why? It's not clear to me.

Note in rotations vectors are, well, not quite as simple as that. For example, let us assume that the appropriate torque T is North with magnitude rF. Another vector South with magnitude rF would describe the same exact rotation (like this, side view, with E/W going in and out of the screen):

 _ --->north vector/ \_/<--- south vector

This is a contradiction. Ideally each unique rotation will have exactly one unique vector.

How do we do this? The vector is not the "direction" of the rotation (direction is meaningless in rotations anyway; (simplistically speaking) the ball only moves in a "direction" as a result of regular frictional force against the table)

The direction of the rotational vector, therefore, represents the axis of rotation, and the magnitude the amount of rotation.

This is easier to see if you align the axis of rotation with the Z-axis. Then, a rotation with the radius and force vectors in the X/Y plane will produce a rotational vector with only a Z component(which is what we expect)

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The torque generates a rotation around the direction of the torque vector, and with magnitude equal to the magnitude of the torque vector.

In your case, the torque will induce a rotation around the axis (X), or (Est-West). The torque vector will indead be directed toward the axis X.

That seems all right to me :)

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Thanks, guys. That makes sense. I just thought that the torque was represented by a vector that pointed in the direction of rotation. Now I understand that it actually represents the rotation axis.

Now there's another problem. T = Ia. a = T/I. What are the units for T? Radians?

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Newton.meters

it's a force times (when doing the cross product) a distance (-distance to centre of mass).

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