Track Movement of Spherical Mass with 3 Contact Points

[Kekrozz]

Hello! I'm having a little problem to figure out the rotation of a spherical mass, maybe you have more clear ideas!
Imagine to have a sphere laying (just it weight) on a support with three contact point:
1) one wheel, that has an external torque apply.
2) helping support that touch the sphere in two point with two conical wheels (one shaft mounting two conical wheels).

I would like to rotate the sphere, just with the rotation of the driver wheel.
Also I would like to know the rotation of the test mass, driver wheel and helping wheels.
I know, of course that if the applied torque is less than a certain value there is the pure rotation without sliding, so I am able to relate the rotation of the driver to the rotation of the others.

My question regards the start on of the system.
Is it possible to find equation that describe the dynamics of the start on of the system?
If I have a slide in some contact I'm not able anymore to know the exact rotation of the sphere because I can't not relate it to the driver rotation.
Sorry if it difficult to understand what I wrote, if you have question just tell me!
Thanks!

 

[jack action]

I'm not sure if this is what you are asking, but these are the equations of motion for the system:

For the driver:
T_{d\ in} - T_{d\ out} = I_d\alpha_d
For the sphere:
T_{s\ in} - T_{s\ out} = I_s\alpha_s
For the conical supports (assuming both are identical):
T_{c\ in} = 2I_c\alpha_c
Furthermore:
\alpha_s r_s = \alpha_d r_d
\alpha_c r_c = \alpha_s r_s
\frac{T_{d\ out}} {r_d} = \frac{T_{s\ in}} {r_s}
\frac{T_{s\ out}} {r_s} = \frac{T_{c\ in}} {r_c}
$T$ is torque, $I$ is moment of inertia, $\alpha$ is angular acceleration, $r$ is radius.

That is 7 equations with 7 unknowns ($T_{d\ out}, \alpha_d, T_{s\ in}, T_{s\ out}, \alpha_s, T_{c\ in}, \alpha_c$), so this system of equations can be solved.

Of course, in addition, the angular velocities ($\omega$) relations are:
\omega_s r_s = \omega_d r_d
\omega_c r_c = \omega_s r_s

 

[Kekrozz]

why the αsrs=αdrd should be equal? you are already supposing the no skid with this equation?

 

[jack action]

Because at the contact point, assuming no slipping, the acceleration is the same for both the sphere and the driver ($a_s = a_d$).

 

[jack action]

I realize now that you want to know what happens if there is slipping. Then, you have to consider the friction force between the driver and the sphere ($F_{ds}$):
T_{d\ in} = F_{ds}r_d
Or:
T_{d\ in} = \mu N r_d
Where $N$ is the normal force between the sphere and the driver.

 

[Kekrozz]

Yes of course, but at time zero when the system, is starting to rotate, in the contact point there is the frictional force (the system is initially at rest), so first of all the transmitted force has be bigger than the frictional force to activate the rotation, in that particular instant, for the mathematical model, there is skid, am I right?
If this is right you are not able to track the rotation of the sphere whenever you want.
Is it correct? I'm getting confused about the role of the friction force when a system at rest start to rotate.

 

[jack action]

If $T_{d\ in} \gt (\mu N)_{ds} r_d$, then there is slip between the driver and the sphere:

The driver (1 equation, 1 unknown):
T_{d\ in} - (\mu N)_{ds} r_d = I_d\alpha_d
The sphere + supports, no slip (4 equations, 4 unknowns):
(\mu N)_{ds} r_s - T_{s\ out} = I_s\alpha_s
T_{c\ in} = 2I_c\alpha_c
\alpha_c r_c = \alpha_s r_s
\frac{T_{s\ out}} {r_s} = \frac{T_{c\ in}} {r_c}
If $T_{s\ out} \gt (\mu N)_{ds} r_s$, then there is slip between the sphere and the supports:

The sphere (1 equation, 1 unknown):
(\mu N)_{ds} r_s - (\mu N)_{sc} r_s = I_s\alpha_s
The supports (1 equation, 1 unknown):
(\mu N)_{sc} r_c = 2I_c\alpha_c

 

[Kekrozz]

I think this is not feasible, in with way you are saying that the Conical support receive
(μN)scrc moment, but actually it has to receive the tangential force due to the rotation of the test mass, also I think that this model is able to explain what happens in steady rotation, but is not describing the "shut on" of the system, so the little transitory time in which the system is settling down and came to an dynamic equilibrium.