# Gyroscope experiment

yuiop
The axis of a spinning gyroscope is observed to be parallel to the x,y plane and at 45 degrees to both the x and y-axis by an inertial observer at rest with the gyroscope axis. (frame S)

To an inertial observer in frame S' moving parallel to the x-axis of frame S, the axis of the gyroscope appears to be at an angle of less than 45 degrees to the y axis. Now if he considers the change in position of the gyroscope axis in x,y plane to be due to a torque applied to the gyroscope then the gyroscope should have reacted by precessing out of the x,y plane. Obviously, that is not the case because the gyroscope always remains in the x,y plane according to observer S and both observers should agree on that. How is this resolved?

Mentz114
Hi Kev:

I've only done a quick sketch, but is seems that the the angle isn't the only thing the moving oberver sees differently.

If S marks out a triangle with sides 1 (x),1(y),sqrt(2), then S' sees the triangle with sides gamma, 1, sqrt(1+gamma^2).

The distance between the gyroscope and any origin we choose, must have decreased also. Could this apparent movement cancel the ghost torque ?

yuiop
Hi Kev:

I've only done a quick sketch, but is seems that the the angle isn't the only thing the moving oberver sees differently.

If S marks out a triangle with sides 1 (x),1(y),sqrt(2), then S' sees the triangle with sides gamma, 1, sqrt(1+gamma^2).

The distance between the gyroscope and any origin we choose, must have decreased also. Could this apparent movement cancel the ghost torque ?

For simplicity can we assume one end of the gyroscope axis is at the origin and is a pivot that can move in any direction.

I probably should have also stated that the experiment is carried out in flat space far away from any massive bodies to keep things simple. (I'm not getting into a Mach's principle argument here)

Mentz114
Kev:
For simplicity can we assume one end of the gyroscope axis is at the origin and is a pivot that can move in any direction.
Don't understand that. The axis is any line through the center of rotation and perpendicular to it. It has no end unless we assign one.

Can you draw a diagram, please ?

Mentz114
Please have a look at the pic. Is this what you mean ?

Am I wrong in thinking that $$L=\gamma ^{-1}$$ and $$H=\sqrt{ 1 + \gamma ^{-2}}$$ for S'

My suggestion is that the apparent shortening of H may also be relevant.

Is it ?

#### Attachments

• Gyros.jpg
6.7 KB · Views: 352
yuiop
Kev:

Don't understand that. The axis is any line through the center of rotation and perpendicular to it. It has no end unless we assign one.

I was thinking of a physical axle like on a toy gyroscope that you can stand on a table more than a geometrical line.

Please have a look at the pic. Is this what you mean ?

Am I wrong in thinking that $$L=\gamma ^{-1}$$ and $$H=\sqrt{ 1 + \gamma ^{-2}}$$ for S'

My suggestion is that the apparent shortening of H may also be relevant.

Is it ?

I was thinking more like the diagram I have attached to this post with $$L_x' = L_x \gamma ^{-1}$$ and $$L_y' = L_y$$ with the x-axis of both reference frames parallel. (i.e. relative linear motion along the x axis). I am not sure how you got apparent shortening of H.

Note in the diagram it can be seen that the mass of the spinning gyroscope is longer symmetrically orthogonal to its own axis in the S' frame.

#### Attachments

• Gyros2.JPG
11.1 KB · Views: 383
Mentz114
Thanks, Kev. I'll think some more about it.

yuiop
Please have a look at the pic. Is this what you mean ?

Am I wrong in thinking that $$L=\gamma ^{-1}$$ and $$H=\sqrt{ 1 + \gamma ^{-2}}$$ for S'

My suggestion is that the apparent shortening of H may also be relevant.

Is it ?

Ah! OK, I get your observation about H. For some reaon I was think H for height and not hypotenuse. It may be relevant, but it can also be elliminated by placing the centre of mass of the gyrosope at the origin. There will still be a ghost torque in the x.y plane and no corresponding precession of the gyro spin axis out of that plane, which is normally how you would expect a gyroscope to respond when its orientation is altered.

Mentz114
Yes, I wasn't too assiduous in my nomeclature, sorry. The thing that bothers me now, is that I used Pythagoras to get H, but wouldn't H be seen to have contraction

$$\left( 1 - v^2cos^2(\theta)\right)^{-\frac{1}{2}}$$

which would mean the triangle is no longer right angled ?

Anyhow, it's an interesting case and I'll give it more thought later.

Mentz114
Kev, I don't think that there is a contradiction here. There are no balanced forces in the rest frame, so seeing the axis at a different angle is a normal length contraction effect.

This situation is not like the right-angle lever, or compressed gas where forces are present and balanced in the rest frame.

yuiop
Kev, I don't think that there is a contradiction here. There are no balanced forces in the rest frame, so seeing the axis at a different angle is a normal length contraction effect.

This situation is not like the right-angle lever, or compressed gas where forces are present and balanced in the rest frame.

I wasn't looking for a contradiction. I just thought something interesting might be going on and I think I have discovered what it is. In the tranformed pic i uploaded if you look at the shape of the main rotation mass of the gyroscope and imagine a virtual axis orthogonal to a line joining two masses on the rim of the gyro you will see the virtual axis has rotated in the opposite direction (and to a greater extent) than the visible spindle of the gyro. Now when you consider that part of the mass is moving in the same direction as the linear motion of the frame and the part on the opposite side is moving in the opposite direction the effective centre of mass of the gyro HAS moved out of the x,y plane due to differential mass transformation. This shift in the centre of mass appears to result in a precession reaction that causes the rotation of the virtual spin axis in the x,y plane. Its a bit difficult to visualise and I will try and explain better if that does not make any sense.

Mentz114
I understand. Thanks.