Ball rolling down an incline between 2 rails

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SUMMARY

The discussion focuses on modeling a solid ball of radius R rolling down an incline at an angle Theta, specifically between two rails. The derived equations include F_x = mg sin(Theta) and a transfer function of (5/7g)/s^2 for small Theta. Key questions raised involve determining the effective radius of the ball when constrained by the rails and the discrepancies between experimental data and the theoretical model, attributed to un-modeled friction and the need for precise track width. The analysis suggests using total energy methods for a more accurate model of the ball's motion.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with rotational dynamics and moment of inertia (I = 2/5mR^2)
  • Knowledge of energy conservation principles in mechanics
  • Basic proficiency in calculus for deriving equations of motion
NEXT STEPS
  • Explore the concept of effective radius in constrained motion scenarios
  • Research the effects of friction on rolling motion and how to model it
  • Study the total energy approach to analyze rolling objects
  • Investigate the impact of aerodynamic drag and rolling resistance on motion
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Mechanical engineers, physics students, and researchers in dynamics who are interested in the behavior of rolling objects and the effects of constraints on motion.

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The problem is modeling a solid ball or radius R rolling down an incline of angle Theta. Note that I am an Electrical Engineer who is somewhat flaky with rotational concepts (but not completely useless :p). If the ball was rolling down a road (or any solid, flat surface), I get the following equations (which I believe neglect friction EXCEPT for causing the ball to roll):

F_x = mg sin (Theta)

T = I * (d^2x/dt^2)/R

Which, after subbing in for I (2/5*mR^2) and converting to linear displacement by multiplying by R, I get a rotational displacement force of F_rx = 2/5m (d^2x/dt^2).

Combining these two equations, we get:

m(d^2x/dt^2) = mg sin (Theta) - 2/5m(d^2x/dt^2). which can be rearranged as

(d^2x/dt^2) = 5/7gsin(Theta).

We then linearize about small Theta, and get a transfer function between position and Theta of (5/7g)/s^2.

Now, I have two questions:

#1 If instead of a solid surface, we instead (and this is the actual experimental setup) we have a ball placed between 2 rails a distance d apart. Do I need to find the "effective radius of the ball", i.e., h from the following diagram and use that to calculate both the displacement of the ball (so converting the Torque equation to position), and through use of the // axis Theorem for finding I?
attachment.php?attachmentid=16512&stc=1&d=1227308894.jpg


#2 When comparing experimental to simulated data with the above model, no "gain value" for the transfer function fits the data (i.e, the curve has a fundamentally different shape than what our model predicts). I thought that this was due to un-modeled friction effects. Is that correct or is this the result of something else?
 

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Both the torque and the rolling speed would be relative to the effective radius. The track width would have to be very precise, not vary with position, and not expand due to the balls weight in order for the effective radius to not have a dynamic effect. Fiction could be an issue with a softer ball, since the rails "squeeze" the ball. Steel on steel might work if the ball doesn't slide. Aerodynamic drag and normal rolling resistance is another issue.
 
You're missing a term for the translation inertia of the sphere.Also T is wrong, it has to be angular acceleration. There's an easier way to do this analysis using total energy.Say initial position is at a height z0=0 so that deltaz is z. (z is vertical direction, deltaz=z-z0). At any z value, the energy balance is mgz=1/2mv^2+1/2Iw^2 (v is the velocity in z direction, w is the angular speed). Now there's a geometric relationship between v and w if we assume there's no sliding. v=w*R*sin(teta). (you can calculate it like this, if the ball rolls down some angle in radians teta2, it goes (teta2/2pi)*2*pi*R*sin(teta) distance in the z direction.) I is 2/5mR^2. So you put everything together and solve v in terms of z.
I think it becomes v=sqrt(gz/(0.5+0.2(sinteta)^2)). if you're trying to find velocity in the xprime direction-on the wedge, z=xprime*sinteta. you can recollect terms to get this
v=sqrt(10*g*xprime*(sinteta)^3/(5*(sinteta)^2+2)) .Use xprime/t=v.Then you can find xprime and/or v as a function of t and teta.
As long as you don't have sliding on the rails, it doesn't matter if you have a solid block or rails.
 

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