A ball of mass m is rolling along an elastic track (2 elastic strings) which are separated by a track width w, and the track is at an angle of elevation θ (making it a ramp). The elastic strings are already stretched by l to form the ramp. What mass is required to separate the elastics and allow the ball to fall through them?

This is for an engineering project where balls of the same size but of different masses must be sorted simultaneously. The idea is that balls will be allowed one by one to roll down this elastic ramp so that the elastics will separate for the heavier ball, but stay together for the lighter ball. Therefore, the heavy ball will fall through the ramp and into a bin, and the light ball will continue along the ramp and end up in a separate bin.

I thought Hooke's Law (F=-kx) and the normal force of a ramp (N=mgcosθ) could be used to solve this problem, but I am unsure.

EDIT: Here are the calculations I came up with, but I am a first year student and do not know much concerning elasticity physics. My impression was that the normal force of the elastics acting on the ball would be equal to the overall change in "spring force" for the elastics combined. Therefore, half of the normal force would equal the change of spring force for a single elastic, and I know that the elastic must stretch horizontally by 1.775mm in order to allow a marble to fall through it (because the width is 12mm and it must become 15.5mm). My calculations:

Included is the mass of the lighter (wooden) ball and the heavier (glass) marble. Both balls have a diameter of 15.5 +/- 1mm. I have not found out the K value for the elastic bands I am using, though I'd like to find it by measuring the change in the stretch of the elastic band when a known mass is attached.

The actual width of the gap is 12mm and the length of the ramp made by the 2 elastic bands is 8cm. The angle has not been determined but it seems that any angle between 10 degrees and 30 degrees gives high success rates.

Endnote: I am not looking for a number value for the optimal angle, and that would not even be possible due to the lack of a k value. Rather, I am trying to understand how I can rationalize my method of sorting the balls with the elastic track using physics.

  • $\begingroup$ I think this cannot be easily calculated. It will depend on the tension in the strings and the friction between the string and the balls. It may also depend on any vibrations which the rolling balls set up in the string. This is a situation which is very difficult to model mathematically. Far easier to build a model and test it, varying the parameters. $\endgroup$ – sammy gerbil Nov 14 '16 at 0:54
  • $\begingroup$ I created a prototype and did testing. It's a design class and we need to back up our claims with physics. I could say the friction is negligible because it wouldn't much affect how the elastics separate, but I might even be able to find the tension in the elastics, can't I? I'm looking for a kind of general explanation (with equations) for why a higher mass would separate the elastics while a lighter mass would not. Maybe like an equation for how much the elastics stretch in the x-direction which would be dependent on the tension of the elastics, mass of the ball, angle of the elastics, etc. $\endgroup$ – Aidan Bruneel Nov 14 '16 at 2:02
  • $\begingroup$ Fair enough, Aidan. I will take a look at this problem and post a response. Can you also show your attempt to solve this? As an engineering student you ought to be able to make a start on the calculation. ... It would also be very useful if you could post some details of your prototype (values of $m, l, \theta$ and if possible tension in strings) and your results (minimum mass to slip between strings?). These questions work best when we can match up theory with your results. $\endgroup$ – sammy gerbil Nov 14 '16 at 2:19
  • $\begingroup$ I edited my original question with more information and my initial calculations (which are undoubtedly incorrect due to my lack of knowledge on the subject). $\endgroup$ – Aidan Bruneel Nov 14 '16 at 4:04

It's an interesting problem.

The elastic band only produces a vertical force on the ball when it is bent at the point of contact. If the angle of deflection between the two parts of the band is $\theta$, and the tension in the band is $T$, then the force on the ball is $F=2T\sin\frac{\theta}{2}$;

enter image description here

this force will be acting on the ball at an angle $\alpha$ (which is a function of the separation of the bands, and the size of the ball). Again, the vertical force will be a function of both $F$ and $\alpha$; if $\alpha$ is the angle between the vertical and the line from the center of the ball to the point of contact, the vertical force due to each elastic band will be $F\cos\alpha$, and since there are two bands, the total force is $2F\cos\alpha$:

enter image description here

Now $\alpha$ relates to the spacing between the bands; as the bands go further apart, $\alpha$ increases and the vertical force decreases. At the same time, the angle of the band increases, so the component of tension increases.

I hope you can do it from here - if you are still stuck, I may give you more details tomorrow.

  • $\begingroup$ Seeing how complicated it must be, I do not know if I will be able to rationalize my method with that approach. You seem very knowledgeable about this kind of thing but I almost need a dumbed down version where there are not angles which are functions of displacement and whatnot. Is there any easier way of even just describing why an object's weight would move the elastics apart? I do not need a full-proof equation to optimize for the angle of the ramp or anything like that, I just need to show that my method of sorting the 2 masses of balls by using elastics is viable according to physics. $\endgroup$ – Aidan Bruneel Nov 14 '16 at 4:03


The diagram below left looks at a cross-section along the length of the elastic rails. The ball (blue) rests on the 2 wires (red). Assuming there is no friction between ball and rails, the contact forces (N) which the ball exerts on the rails are normal to the surface of the ball and radially outwards. There are thus horizontal components of N, which push the elastic wires outwards and downwards. The wires exert equal and opposite forces N on the ball, pushing it inwards and upwards.

The condition for balance is that $2N\cos B=W=mg\cos\theta$, where $\theta$ is the local angle of inclination of the rails - either due to local deformation or the global slope between supports. When the net tension $F$ in the rail normal to the surface of the ball then $N$ equals Floris' force $F$. If the ball is only just heavy enough to slip between the rails, this will happen at the midpoint, where $\theta$ is the inclination between supports.

enter image description here

The middle figure is the same view as on the left but focussing on the movement of the rails. The rails and supports are initially at X before the ball reaches the midpoint. As the elastic rail extends and sags the contact point moves outward while the support move vertically upwards relative to the ball.

If the contact point is at Y while the support is at X" then the rail is in the plane YX" with net tension force $F$ directed along YX"; there is a component of this force $F$ which tends to move the rail further outward round the ball' surface. If the support is at X' on radius OY then the force $F$ from the rail is normal to the surface of the ball; $F=N$ and there is no tendency for the rail to move any further around the ball, either inwards or outwards.

The rightmost figure illustrates the elongation of the rails and its connection with z=YX' in the middle figure. Angle $A=\frac12\theta$ in Floris' 1st diagram.

From the geometry in the middle and right figures we have :
$(a-z)\sin B=w$
$L^2-L_0^2=(L-L_0)(L+L_0) \approx z^2$
$x=L-L_0 \approx \frac{z^2}{2L_0}$
because $L \approx L_0$. Here $a$ is the radius of the ball, $2w$ is the lateral separation of the supports and $2L_0$ the longitudinal separation.

When $N=F$ then by substitution, using z as a convenient parameter, this becomes :
$$T=\frac{mg\cos\theta}{4\sin A\cos B} \approx \frac{L_0 mg\cos\theta}{4z\sqrt{1-(\frac{w}{a-z})^2}}$$

With the values for $m, a, w, L_0$ which you supplied, and taking a mean value of $\theta=15^{\circ}$, I have plotted the RHS (red) for the 2 values of $m$ in the graph below.

enter image description here

The remaining task is to find the pre-loaded tension $T_0$ which will give the required value of $T$ for optimum separation of the balls when they reach the midpoint. The difficulty is that Hooke's Law may not apply for the elastic strings. You will need to examine load $T$ vs extension $x$ experimentally to see if it does apply and under what conditions. According to Do Rubber Bands act like Springs?, if the rubber band is not stretched too quickly and it has exceeded an initial minimum extension, then it does obey a modified version of Hooke's Law :
where $k$ and $c$ are constants which you can find from your graph of load vs extension. When $z=0$ then the pre-loaded tension in the rails is $T_0$ so we can write
This can be plotted on the same graph (blue) using an arbitrary values of $T_0$ to shift the blue line vertically until it intersects the red lines with the largest separation, so that a small change in tension gives a large change in the mass which falls through. (I have assumed an arbitrary value of $k=500$).

The best value for this is $T_0 \approx 2$. From this and your equation $T=kx+c$ you can work out what length of unstretched elastic your need.

I assume that $a$ and $m$ are constants, things you cannot vary. Values you can vary are $w, L_0, \theta, T_0$. One question to ask is : What combination of these variables allows you the maximum discrimination between the two masses given? You could explore the predictions for various combinations of parameters, particularly those which have most influence.

Some limitations of the above analysis :

  • It does not take account of friction. While this may be small for the marble, probably it is significant for wood.

  • The rolling balls may cause vibrations in the strings which help them slip around the balls more easily. This may compensate somewhat for friction, or perhaps over-compensate and cause balls to fall with more tension that this theory predicts they will need.


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