Came up with a problem that I can't solve

[golfrmyx]

The angle is equivalent to that position, see the picture in post 3.

Post #3 does not have a picture. Thus, angle to what? Horizontal radian at point of release? Or vertical radian? An angle of 2 what then? 2 degrees above or below horizontal or what?

 

[JorisL]

Alright I made a mistake, post 4.
But have you read the entire thread?

As to finding a solution, perhaps going purely mathematical could work (understanding it).
First try to find the EOM for $\phi$ and $\psi$ (at first sight they look equivalent) using the Lagrangian formalism.
Next suppose you have an initial condition with angle $\phi=\delta \phi$ where $\delta\phi$ is infinitesimal.
Then one could check using an iterative method what the solution could be (and if it's feasible).

 

[golfrmyx]

Alright I made a mistake, post 4.
But have you read the entire thread?

As to finding a solution, perhaps going purely mathematical could work (understanding it).
First try to find the EOM for $\phi$ and $\psi$ (at first sight they look equivalent) using the Lagrangian formalism.
Next suppose you have an initial condition with angle $\phi=\delta \phi$ where $\delta\phi$ is infinitesimal.
Then one could check using an iterative method what the solution could be (and if it's feasible).

A little confused because you are now communicating under a different name. Does not matter, Assume associate. My education is old and not a complete college degree. But, I do know and used extensive math on several different professions. Know nothing of Lagrangian and never learned the complete greek alphabet nor used it much. I use what I know from experience combined with education and experimentation.

Regardless of the comparative density and mass weights and comparative sizes of m and M, without "push off" force, I cannot believe the speed will increase enough for centrifugal force to enter the equation in a calculable amount. Thus my answer in an earlier post of point of leave being at gravitational level radiant. If m is too heavy, M will slip. If M is too heavy or large in diameter it will not continue from even a push off.

Any way you look at the problem with only presented stats, it is not solvable mathematically nor with just establishing an equation.

 

[JorisL]

I wasn't arguing against your explanation, it's how I would explain it as well.

The proposition was how I would approach the problem if I was trying to find an explicit solution :-)
I might follow through on the approach if I find the time (not before the end of June)

Odd that you see my name changing, I've had it since way back in 2011.

 

[golfrmyx]

I wasn't arguing against your explanation, it's how I would explain it as well.

The proposition was how I would approach the problem if I was trying to find an explicit solution :-)
I might follow through on the approach if I find the time (not before the end of June)

Odd that you see my name changing, I've had it since way back in 2011.

I guess you replied to a reply of mine to another poster and I did not recognize that since the system does not show who the replies are to.

 

[mfb]

The friction would not be the same as the weight of gold increases friction in contact with sloping surface.

Friction is also proportional to mass. If we neglect air resistance, all forces and energies are proportional to mass, and acceleration is independent of it. A simple dimensional analysis shows the same.

 

[golfrmyx]

mfb, I see my comment was not clear for some reason. The gold friction on the sloping surface would be more than that of glass that is more than plastic. The increased friction I was referring to would be the air resistance that is too large to ignore. I do not mean increased. Not sure why I used that word. The marbles I was mentioning must be the same diameter so mass would be the same and thus air friction the same. Gravitational pull is the major force.