Ripping Toilet Paper Rolls

1. May 26, 2007

mindboggling

I'm extremely excited to embark on my physics project and it has all to do with ripping toilet paper rolls.

As it is commonly known, when a full toilet paper roll is given a fast jerk, a sheet breaks off. However, as the roll is gradually used up, a faster jerk has to be applied or the toilet paper will unravel like crazy haha.

Anyway, I'm interested in finding a relationship between the velocity at which the toilet paper roll has to be tugged and the radius of the roll ( as in how much it is used up).

Now, i have to come up with the physics behind this:

sum torque = (radius of roll) x ( Force of pull) - (distance from center to the core of the toilet roll) x ( Friction)

and this is also equals to

sum torque = (moment of inertia of roll) x ( angular acceleration)

This is where i'm stuck at, i don't know about the physics that explains whether the roll will break or continue unraveling.

Here are some thoughts, but they are most probably faulty:

To simplify the problem, i've decided to consider an inertial ball ( ball hangs from a string, and another string hangs from the ball)

at the lower string
F = (change in momentum) / time

thus if given a fast jerk

F > (F required to break string) and therefore lower string breaks

but why doesn't the upper string bring? is it because according to F = ma, the mass is so much greater than the string that it doesn't accelerate much?

Please show me the physics behind why when a slow pull is applied, the upper string breaks while the lower string breaks if given a fast jerk. I think once i've understand this, i can apply my knowledge to ripping toilet paper.

Thanks, i'll add more of my thoughts on this later when i come up with more ideas.

2. May 26, 2007

Danger

You lost me on most of that post; math ain't my thing. To the best of my knowledge, a full roll will rip because of the inertial mass. It takes a lot more force to get it moving. If that force is greater than the structural strength of the perforated lines, then it'll give way.

edit: I'm pretty sure that I used some wrong terminology here, but I'll leave it up to someone else to sort it out.

Last edited: May 26, 2007
3. May 26, 2007

FredGarvin

You might want to look into the change in polar moment of inertia of the roll as the paper runs out. A full roll will have a larger inertia that will resist angular acceleration. That inertia provides enough of a resistance to allow for the breakage of the sheets. As the roll runs out, that inertia is lower and the roll accelerates easier and thus not providing the resistance for the sheet tearing.

4. May 26, 2007

mindboggling

now i understand what's happening conceptually? but how do i represent this using math?

net torque = (angular momentum) / (time) = (moment of inertia) x (angular acceleration)

using math, how can i show that, if the roll is pulled at a certain velocity, the sheet will break, and if otherwise, the roll will keep unraveling?

5. May 26, 2007

FredGarvin

That's tough. You'll need to correlate friction between the roll and the rod that holds it. In this case, you may need to do some actual data collection to create/validate any model you come up with.

6. May 26, 2007

Danger

Remember, too, that this is entirely dependent upon the brand (and in some cases the batch of that brand) of paper involved. There is no consistency.

7. May 27, 2007

KingNothing

Does friction really matter that much? Isn't the mass of the roll much more important? If you had a holder that allowed it to spin freely, could one not still rip a sheet off this way?

8. May 27, 2007

FredGarvin

I would think that friction between sheets and the friction between the tube and the holder would have an effect. In any case, I would include them in my model until testing proved otherwise. I can't say I have spent much time testing this theory.

9. May 27, 2007

AlephZero

Your "ball and two strings" problem can be modeled fairly simply. You need to consider the two strings are flexible, like springs.

To see why the different strings break: if the force appled to the lower string is F, the force in the upper string is F + Mg - Ma where M is the mass of the ball, g the accel due to gravity, and a the acceleration of the mass (positive downwards).

If you increase the force slowly, the acceleration is small, the force in the upper string is bigger than F, and the upper string will tend to break first. If you increase the force quickly, the accleration is large, the force in the upper string is less than F, and the lower string will break first.

To convert that logic into numbers, you need to know the flexibility of the string, and how to set up and solve the equations of motion. The important parameter is the vibration frequency of the system (similar to a mass on a spring). The period of one cycle of vibration defines a natural timescale for the system to respond. (call it Tr sec). Suppose the time it takes for the force to get large enough to break the strings is Tf. If Tf << Tr the bottom string will break. If Tf >> Tr the top string will break. If Tf is fairly close to Tr you have to work through the maths to find out.

The toilet roll problem is similar in principle, but much more complicated because the "mass" can do a lot of dynamics on its own. It can rotate, swing on its support like a pendulum, slide on its support, deform from circular to elliptical, etc, etc. I've worked as an engineer on "hard" dynamics problems and my instinct is there isn't a "simple" model that will predict anything very interesting about all that.

10. May 27, 2007

wiredGuy

I;m sure the angle at which you pull is also a factor. If the component of the force tangent to the surface of the roll is greater, then more force is delivered to the ripping of the sheet than to the rotation of the roll.

Also consider the speed it takes to pull a sheet out from underneath dishes on a table. the dishes don't move because they have inertia. The fast pull on the smaller roll tears the sheet, just as it would on the larger roll. The larger roll has more inertia so less of a tug is necessary.

11. May 27, 2007

Danger

The angle does indeed make a difference. Whether or not the math supports it, experience does. If you pull straight down, it's far more likely to tear than if you pull it horizontally.

12. May 27, 2007

Loren Booda

By its loose end, hold a paper roll over a ledge and let it the roll fall. How long will it take for the roll to completely unravel, assuming it does not break and the paper is spirally packed to the center? (This is a tough problem, involving nonlinear mathematics.)

13. May 28, 2007

Danger

It can't be that tough; my cat does it on a regular basis. :grumpy:

14. May 28, 2007

wiredGuy

Sounds somewhat simple. The mass changes over the drop, but thats the only complexity. The mass must change with respect to how much comes off the roll; thus, the change in mass has nothing to do with the radius and everything to do with the speed...

d=v1t + 1/2gt^2 -- the mas changing does not affect the acceleration. And I don't see anything resisting the drop of the roll except perhaps friction on the roll; however, if you consider that the mathematical model becomes overly complex for the scenario - making it a bad model. So the answer would seem to be: t=(2d/g)^(1/2)

15. May 28, 2007

birulami

Just forget the roll and all. Instead consider a thin rope attached to the moon and you want to move the moon be tugging at the rope. The rope is constructed to hold a force no greater than F. Below that force, you can accelerate the moon by a=F/m where m is the mass of the moon. If you apply a greater force, the rope rips in two. No friction involved.

Cheers,
Harald.

16. May 28, 2007

pixel01

Lets do some maths before arguing.
Call v the velocity needed your hand tugs the paper (or acceleration a ?)
F : the tearing force of the perforated paper band
R: radius of the remaining paper roll
M: mass of the remaining paper roll, which can be calculated via R and density d (kg/m3) and the width H of the roll.
Then we can calculate v as a function of R, F, d and H.
I do not think friction (between the roll and the axis) should be considered.
There's one thing i am still not so sure which is when we tug the paper, how can we identify the action : acceleration or velocity?
If v is used, there should be some thing like stretch factor of the paper. Am i right?

Last edited: May 28, 2007
17. May 28, 2007

Loren Booda

wiredGuy,

Remember the moment of inertia (a function of mass and radius)

18. May 28, 2007

KingNothing

It doesn't seem like the moment of inertia would play a very large role at all compared to gravity. What I mean is, gravity will impart a certain acceleration that is resisted by the moment of inertia...but this resisting force could not be any greater than that required to rip a single sheet of toilet paper.

In other words, if the roll had some steel core that made the moment of inertia much greater, the paper would probably just rip.