# Torque and reference frames

1. Mar 7, 2015

### tomizzo

Hello,

I have a question regarding the concept of torque and reference frames.

Say for example I have a rod of length L and that the rod has it's right side anchored to act as a pivot point. Now let's say that I input a force on the left side of the bar in an upward direction. Assuming that positive torque corresponds to clockwise rotation, the torque is the force x L (and is positive) with respect to the right pivot point..

Now lets say I do this same experiment, but this time, I want to use a frame of reference such that the left side of the rod appears to be anchored. From what I can deduct, the torque as seen by this virtual pivot point will have the same torque magnitude as the last experiment, except it will be negative now...

Is my reasoning correct? I'm kind of just curious and don't know if phenomenon has a specific name.

Thanks for any help!

2. Mar 7, 2015

### Filip Larsen

You are correct in the sense that as long as you consistently designate one direction as positive and the other as negative the results will be consistent up to that change in sign. Usually one designates counter-clockwise rotation as positive (that is, opposite of your choice above) since this corresponds to the right-hand rule [1], but that is just a practical convention that gives a consistent relation between linear direction of the axis with direction of rotation around those axis.

[1] http://en.wikipedia.org/wiki/Right-hand_rule

3. Mar 7, 2015

### PeroK

The motion of the bar from the point of view of the unanchored end will also be clockwise rotation. So, the fictitious torque in this frame will have the magnitude and direction as the true torque in the original frame.

4. Mar 7, 2015

### tomizzo

Hi Perok,

I can't visualize how the torque would be in the same direction... I keep picturing the bar would have to be rotated counter clockwise to keep the left end in place for our reference frame. Could you explain further?

5. Mar 8, 2015

### PeroK

If the bar starts horizontal and you rotate the left end clockwise by 90°, then the left end is above the right end.

If you imagine the left end is fixed, and rotate the right end clockwise by 90°, then again the left end is above the right. Same as before.

I'm not sure what you're picturing.