# When the gyro moves down, does it spin faster?

First, I have a gyro, $$\omega$$ is its angular velocity

Then, I simple it to

When it moves down, there is

Gravity does work differently for A and B. Therefore, the angular velocity should be changed. But I am not sure it is right since I haven't noticed anything similar in reality.

I'm just a hobbyist in physics, so I might be loose in terms of description. Besides, there is not a tablet to draw on, so I can only draw on paper and then take pictures. So it looks a little ugly. Sorry about that.

## 3 Answers

You are correct that there is more gravitational work done on A than on B in the setup that you made. The other answer is correct that the gyro is not going to spin faster just as a skateboard wheel will not either when you drop the skateboard.

The problem in your analysis is that what causes a gyro (or any wheel as mmesser314 says) to spin faster (or slower) is the torque about the axis of rotation. In this case, there is no torque about the axis because the two masses are the same and the moment arm is the same (and obviously the gravitational field is assumed uniform everywhere). The rate of rotation is not a function of the difference in work on each mass.

If you feel unsatisfied because this seems like a bit of a puzzle, that is because it is. The difference in work on each mass leads to a feeling (an incorrect one) that one mass should move faster than the other. This does not happen because when you consider the masses individually, you must consider the force of the rod connecting the two. When the entire gyro is considered, the tension in the rod is an internal force and so can be ignored, as it cancels out acting oppositely on each mass. However, the motion of each mass individually requires this internal tensile force, which is equal and opposite between the two.

This is especially tricky, because people are used to the idea that centripetal force does not to any work, but it does do work when viewed from a reference frame where the center of rotation is moving, such as this one.

I promise that if you consider the work of the tension provided by that rod on each mass in addition to the work done by gravity on each mass, you will find that the system can "keep its shape" (meaning the center is colinear with the two masses) and continue rotating at the same rotational velocity. The extra energy due to the whole system moving down through the gravitational field will be accounted for with acceleration of the center of rotation, just like when you drop the skateboard in the example from mmesser314.

If you don't believe me, I can work this out for you, but if that seems a bit too much and you are happy with this explanation, I will leave it there.

EDIT: Since you asked, here is an energy conservation check. We will assume, as stated, that the rotation is independent of the free fall in gravity. The gyro will start at the origin as shown in your first picture, with X right and Y up. Thus, the center of the gyro moves as a point particle with:

$$\vec{r}_{G} = -\frac{1}{2}gt^{2}\hat{e}_{y}$$

and the individual masses are given by:

$$\vec{r}_{A} = \vec{r}_{G}-L\hat{e}_{r}$$ $$\vec{r}_{A} = -L\cos(\theta)\hat{e}_{x} -\left(L\sin(\theta) + \frac{1}{2}gt^{2}\right)\hat{e}_{y}$$

and we note that since the rotation rate is assumed constant, we can substitute $$\theta = \omega t$$ at any time. I will do the calculation for mass $$A$$, but you can repeat it for $$B$$ if you want using:

$$\vec{r}_{B} = \vec{r}_{G}+L\vec{e}_{r}$$ $$\vec{r}_{B} = +L\cos(\theta)\hat{e}_{x} +\left(L\sin(\theta) - \frac{1}{2}gt^{2}\right)\hat{e}_{y}$$

The total force on $$A$$ (both gravity and the rod, where the rod must be standard centripetal), is

$$\vec{F}_{A} = mL\omega^{2}\hat{e}_{x} + m\left(L\omega^{2}-g\right)\hat{e}_{y}$$

Differentiating the position of $$A$$ from above, we find its velocity:

$$\vec{v}_{A} = L\omega\sin(\theta)\hat{e}_{x}-\left(L\omega\cos(\theta)+gt\right)\hat{e}_{y}$$

With this expression for $$\vec{v}$$, we find

$$v^{2} = L^{2}\omega^{2}+g^{2}t^{2}+2gL\omega t\cos(\theta)$$

Remembering that $$\theta = \omega t$$ we can get the kinetic energy as a function of time:

$$KE = \frac{1}{2}mL^{2}\omega^{2}+\frac{1}{2}ml^{2}g^{2}t^{2}+mgL\omega t\cos(\omega t)$$

You should note that the first term is just the gravitational free fall of the whole gyro as a particle, the second is the rotation, and the third term is purely oscillatory. So the kinetic energy oscillates around the sum of the energies of the two motions.

The change in KE between any two times is then:

$$\Delta KE = \frac{1}{2}mg^{2}(t_{2}^{2}-t_{1}^{2}) + mgL\omega\left(t_{2}\cos(\omega t_{2})-t_{1}\cos(\omega t_{1})\right)$$

Now, the trick is calculating the work, which is what you were asking. This is by definition:

$$dW = \int_{\vec{r}_{1}}^{\vec{r}_{2}}\vec{F}\cdot d\vec{r}_{A}$$ $$dW = \int_{t_{1}}^{t_{2}}\vec{F}\cdot\vec{v}_{A}dt$$ $$dW = \int_{t_{1}}^{t_{2}}\left(mL\omega^{2}\hat{e}_{x} + m\left(L\omega^{2}-g\right)\hat{e}_{y}\right)\cdot\left(L\omega\sin(\theta)\hat{e}_{x}-\left(L\omega\cos(\theta)+g\right)\hat{e}_{y}\right)dt$$ which will clean up to $$dW = mg\int_{t_{1}}^{t_{2}}\left(gt+L\omega\cos(\theta)-L\omega^{2}t\sin(\theta)\right)dt$$

Upon evaluating the integral we get

$$W_{12} = \left.\left(\frac{1}{2}mg^{2}t^{2}+Lg\sin(\omega t)-Lg\omega^{2}\left(-\frac{t}{\omega}\cos(\omega t)+\frac{1}{\omega^{2}}\sin(\omega t)\right)\right)\right|_{t_{1}}^{t_{2}}$$

which cleans up to

$$W_{12} = \frac{1}{2}mg^{2}\left(t_{2}^{2}-t_{1}^{2}\right)+mgL\omega\left(t_{2}\cos(\omega t_{2})-t_{1}\cos(\omega t_{1})\right)$$

It is evident that work done on mass $$A$$ between any arbitrary times $$t_{1}$$ and $$t_{2}$$ is equal to the change in kinetic energy between those two times. This means that the assumed motion is consistent with the work energy theorem. The assumed motion was, of course, simple constant rotational motion and simple point particle free fall in gravity, decoupled from one another.

• Thanks your detail answer. Do you mean that the work of the tension provided by that rod on each mass is equal to the work done by gravity on each mass ? I also feel that centripetal forces do work, but I didn't know how to calculate it. If possible, could you tell me how to calculate it. Just write on paper, and give me a picture is enough. Thanks again. Commented Jan 13 at 12:22
• The work done by the tension will not in general equal the work of gravity, but it will be the right amount to make the equations all work out. I am not going to write it on paper and photograph it, I can write up the equations here. Also, regarding your picture due to lack of a tablet, I may be dating myself here, but in my day there was a program called Paint. Commented Jan 13 at 16:24

I have in the past handled a bicycle wheel to try and get a feel for gyro effects. I recommend it. I believe that in this case you can use it to come closer to feeling an answer to your question.

A bicycle front wheel is just the right weight and size to allow you to hold in your hands and feel what is going on. Equally important: a bicycle wheel has for its mass a very large moment of inertia. That means you don't have to spin it very fast to obtain a gyroscopic effect that you can clearly feel in your hands; a rate of spin that is still safe is already sufficient.

The standard demonstration is of course that you have an assistent spin up the wheel while you are supporting both ends of the axle. When the wheel is up to speed: gingerly release support on one side: the wheel then proceeds to gyroscopic precession.

There is a specific reason to release gingerly: if you remove support all of a sudden then the wheel goes in a combination of two motion patterns: nutation and gyroscopic precession. Releasing gingerly suppresses onset of nutation, while still allowing gyroscopic precession.

For observation of the phenomenon of nutation the bicycle wheel is the best. That's because the rate of nutation is proportional to the rate of spin. A small gyroscope is spun very rapidly, and at such a high rate of spin the nutation is extremely fast, it has a very small amplitude, and it dampens out very rapidly. Those factors tend to hide the nutation. In the case of a bicycle wheel demonstration the slow rate of spin allows you to experience the phenomenon of nutation.

Try the following:
As the wheel is spun up you are supporting both ends of the axle. Then, exerting a mild force, transition to a motion pattern where you make the axle sweep out a cone. That motion pattern is called nutation.

Now: a necessary condition for nutation to occur is that the spinning object is very rigid. The reason for that: during nutation momentum is transferred internally. That is: during nutation there are significant internal stresses, transfering momentum.

Decades ago when I tried that axle-sweeping-out-a-cone I had the impression that during large nutation the instantaneous rate of spin of the wheel was slower. Of course, when going back to nutation-free spin the wheel would go back to its original rate of spin.

(It just occurred to me: I can now put a marker on a bicycle wheel, and record a video with my smartphone. That should settle it.)

Further reading:

Svilen Kostov and Daniel Hammer, 2010, It has to go down a little, in order to go around Table top experiment conducted by Svilen Kostov and Daniel Hammer to confirm: for any gyroscope, in order to transition to precessing motion the center of mass must drop down.

2012 answer by me, to a question titled What determines the direction of precession of a gyroscope In that answer I discuss how nutation is essential in the process of onset of gyroscopic precession.

• Thanks. I'll try it when I get a bike wheel. The paper you mentioned seems interesting. It just so happens that I came up with this question while reading Feynman's book. Commented Jan 13 at 12:39

There is nothing special about a gyroscope. It is just a thing that can spin. Any wheel can do that.

If you drop a skateboard, the wheels don't change how they spin.

You are right, the gyroscope does gain energy. If falls faster and faster while continuing to spin at the same rate.