Help with force analysis in the precession of a bicycle wheel

[Orya]

For a volenteering project im doing with high school student with difficult backround- its been a while since i saw that mateial and i need to explain it to them. Could someone please help me do the forces analysis in order to undestand the phenomenon in the image, where the chair in turning in certain velocity (i want to find it) because the spining wheel is not vetical to the ground-

I understand it happened because the balace of forces in the z axis should remain z, but can someone sho me how to get it please?

 

[kuruman]

This is a demonstration of angular momentum conservation. The system of man + spinning wheel are isolated from the rest of the world in the sense that the rest cannot exert a torque about a vertical axis. That's because the stool, which is the only connection of the system to the rest of the world, can rotate freely.

Now suppose the wheel is initially spinning horizontally in the x-direction and the man rotates it so that it spins vertically in the z-direction. Initially, the angular momentum of man + wheel is in the x-direction and its magnitude, call it $L_0$ is entirely that of the wheel. This means that the z-component of the angular momentum of man + wheel is zero.

When the wheel is turned to the vertical direction, the angular momentum of man + wheel in the z-direction must still be zero because angular momentum is conserved in that direction as already stated. We know that the wheel contributes spin angular momentum $L_0$ in the z-direction. For the total angular momentum to be zero, the stool and the man sitting on it must acquire angular momentum $-L_0$. In other words if the wheel spins clockwise when its axis is vertical, the man and the stool will spin counterclockwise and vice-versa.

 

[Orya]

Thank you so much! And when I calculate the magnitude of the angular momentum of the spining man I use L=wmr^2 ,whem m is the combine mass of the wheel and the man (and the chair?) and r is the radius of the stool, and for the angular momentum of the wheel I use L=Iw ?
But if the wheel is in an angle i should acount for gravity as well, no?

 

[kuruman]

The expression $L=mr^2\omega$ is applicable to point masses at distance $r$ from the axis of rotation. For the angular momentum of man + stool (abbreviated $ms$), I would use $L_{ms}=I_{ms}\omega$ where $I_{ms}$ is the moment of inertia of man + stool about the axis of rotation of the stool. It's not a calculable quantity but can be measured experimentally.

Gravity does not enter the picture. If the angular momentum of the wheel with its axis horizontal is $L_w$, then when it's at angle $\theta$ above the horizontal, it will have horizontal component $L_{w,x}=L_w\cos\theta$ and vertical component $L_{w,z}=L_w\sin\theta.$

Let's do a quick calculation to see how this is put together. I will write the angular momentum in the z-direction before and after the rotation of the wheel to the vertical position and conserve angular momentum to find the final angular velocity of rotation of the man + stool in terms of the angular velocity of the wheel.

In the z-direction the total angular momentum of the two component system is
$L_{\text{before}}=I\omega_{\text{w}}\sin\theta+I_{\text{ms}}*0=I\omega_{\text{w}}\sin\theta.$
$L_{\text{after}}=I_{\text{w}}\omega_{\text{w}}+I_{\text{ms}}\omega_{\text{ms}}.$

Angular momentum conservation in the z-direction says $~L_{\text{after}}=L_{\text{before}}~$ which gives
$I_{\text{w}}\omega_{\text{w}}+I_{\text{ms}}\omega_{\text{ms}}=I_{\text{w}}\omega_{\text{w}}\sin\theta.$
This can be solved for the angular velocity of the man + stool to give

$\omega_{\text{ms}}=\dfrac{I_{\text{w}}(\sin\theta-1)}{I_{\text{ms}}}\omega_{\text{w}}.$

 

[Orya]

I understand ! Thank-you so much it has been so helpfull!!