Rolling Ball Friction Calculationby denkmall Tags: force, friction, gravity acceleration, rolling ball 

#1
Mar3011, 06:19 AM

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Hi All,
I have a question concerning a rolling ball on a metal track. Basically I am designing a rolling ball sculpture and want to lay a bunch of elevated track (2 wires) up and down and around turns. I know that as the ball is rolling it is losing speed from friction, so as the track drops the ball picks up speed ,and when the track inclines again it needs to level out at a lower elevation. Is there just a straight formula to calculate the drop in height needed based on the length of track? I have a statistics backround so am good with math but have never attempted to engineer something on paper. Thanks for the help! 



#2
Mar3011, 06:43 AM

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There's not enough to go on here. But the best thing would be to experiment yourself with a length of straight track and find what slope you need in order to 'just' get the ball rolling.
You could make sure that your sculpture never had less slope than that. But there's more . . . . This is not the same simple problem as having a block sliding down an inclined plane. The friction in this case will be due to the rotation of the finite area of the ball in contact with the wire as it rolls. There will always be more friction than for a ball rolling down a flat track. There is a potential problem with the actual spacing of your two wires. The friction will very much depend on the spacing. The further apart they are, the greater force there is to push them apart. As the friction force is proportional to the 'normal' force between the two surfaces (i.e. at right angles), there will be less loss with the wires close together. BUT close together will compromise the cornering  and the stability of the ball in its track. If you want your balls to have the longest possible time of travel, you will need to treat almost every section of the track separately. You ask for a formula. A simple one that would make sense would be based on a frictionless surface. For a drop of s metres, the final velocity v (either falling directly or taking a long winding path) would would be v=√(2gs) v = √(2 X 9.81 X s) or √(19.6s) You can substitute (horrid) Imperial units if you want  your distance will be in feet and g would be 32. I have seen a demo in which a ball rolls along this kind of track and appears to go uphill as the rails diverge a bit. That's another potential (pun ha ha) for your sculpture. 



#3
Apr311, 06:33 AM

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Thanks for the reply! I didn't have the turns in mind when asking the question but there is a lot more to consider and especially to play around with.
Cheers! 



#4
Apr311, 04:11 PM

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Rolling Ball Friction Calculation
Ignoring any losses (rolling resistance) or sliding, you'd need to know how much of the gravitational potential energy is converted into angular energy of the ball. I think there was a previous thread about a ball rolling on a pair of rails, but couldn't find it. Ignoring losses, the formula for acceleration of a rolling (and not sliding) object is:
a = g sin(θ) / (1 + I/(m R^{2})) Where I is the effective angular inertia of the ball. For a ball rolling down a plane, it's angular inertia is I = 2/5 m R^{2} where m is mass of the ball, and R is radius of ball. It's acceleration due to gravity would be: a = (5/7) g sin(θ) For a ball rolling on rails, call the distance from the horizontal axis perpendicular to the rails to where the ball touches the rails r. Then (if I did my math corrrectly, see my next post) the acceleration down hill or deceleration up hill should be: a = g sin(θ) / ( 1 + (2/5) (R/r)^{2} ) The other factor mentioned above is if the rails get narrower or wider. During transitions, the rails apply a force in the direction of travel if they're getting wider, and an opposing force if they're getting narrower (related to gravity and/or centripetal acceleration), work is done, so the total energy of the ball is changed, but I haven't figured out the math for this. Assuming no losses, total energy (kinetic and potential) and momentums are conserved only if you consider what the track is attached to (usually the earth). Take the case where the rails are horizontal but either get wider or narrower, and the ball has some initial velocity. If the rails get wider, the ball's altitude (with respect to gravity) decreases and vice versa. Assuming total energy is constant, then a change in gravitational potential energy will be offset by the change in total kinetic (angular and linear) energy of the ball. As an example of this, here's a toy where the rails are initially widened to cause a ball to start rolling forwards due to gravity and the fact that the rails are getting wider (being angled outwards, there's a net forwards force), then narrowed in a smooth manner so that the ball rolls upwards towards the player: 



#5
Apr311, 05:04 PM

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@rcgldr
I'm not too sure about your analysis. Your formula for the acceleration of the ball should surely include the geometry of the rails and the ball because the torque will depend on the radius of the rolling contact with the rail. The limits would be rolling down a flat plane, at one extreme and, at the other extreme, with the ball very low between the rails (i.e. separation of the rails being only just greater than the ball diameter. The latter would accelerate very slowly. Analysis of the toy in your picture would need this to be factored in. But it is not necessary to consider the rotation of the ball unless you want to know the time taken for the journey. A simple consideration of the Energy transfer, using some experimental measurement of friction / loss will tell you whether the ball will make it to the bottom. All the Kinetic Energy (translational plus rotational) must come from the lost Gravitational Potential on the way down and this also applies to any uphill bits of track. I reckon the suck it and see method would yield quickest results. 



#6
Apr311, 08:05 PM

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Let: a = linear acceleration, T = torque, α = angular accelertion, F = friction force related to angular acceleration, R = radius of ball, r = effective radius from axis of rotation to contact point with rails: α r = a α = a / r T = I α = I (a / r) = (2/5) m R^{2} (a / r) F r = T = (2/5) m R^{2} (a / r) F = (2/5) m R^{2} (a / r^{2}) F = (2/5) (R/r)^{2} m a m a = m g sin(θ)  F = m g sin(θ)  (2/5) (R/r)^{2} m a a = g sin(θ)  (2/5) (R/r)^{2} a a + (2/5) (R/r)^{2} a = g sin(θ) (1 + (2/5) (R/r)^{2}) a = g sin(θ) a = g sin(θ) / (1 + (2/5) (R/r)^{2}) 



#7
Apr411, 05:14 AM

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Nice. I get it now.
Do you get my point about the Energy considerations in assessing whether the ball will make it over a hump in the track? I remember seeing a similar exhibit to your Shoot the Moon game which consisted of a roller, tapered to a point at each end. This rolled along two diverging rails with a slope. The motion was very slow and ponderous but impressive. It was counterintuitive (you tend not to spot how the centre of mass is actually falling as the rails rise) but calculations like yours would suitably describe the motion. 



#8
Apr411, 10:33 AM

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I mentioned the case where horizontal rails get wider generating a forwards force due to gravity, and that the total energy of the ball (kinetic and potential) remains a constan during the transition. For the toy / puzzle you mentioned, the spinning wheel with the tapered axis is actually descending, even though the rails are ascending as they get wider. The shoot the moon toy is allowing the player to increase the energy by moving the rails together, which increases the forwards force and also probably generates an upwards force since the rails are "squeezing" the ball, which in turn increases the total amount of energy of the ball, and in that case the ball's center of gravity is actually rising (if the player is skilled with the toy). 



#9
Apr411, 10:39 AM

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