# The falling pencil

1. Dec 23, 2011

### malicor

hi,

i've got a little complicated question, let's see if i can formulate it in an understandable way.

imagine there's an (ideal) pencil, standing on its (ideal) tip. if it would stand at 90degree to the (ideal) table, it would never fall.

now the question is, if it doesn't stand at 90degrees, and therefor does fall after an amount of time, and that amount is 15 minutes, how big is the difference of the degree to the table to 90degrees.
(it should be pretty small, i said, it's less than 10^-1000000, a friend of me bets it's more.)

can anyone help ? (needn't be an exact calculation, an estimation would do)

regards, malicor

2. Dec 23, 2011

### Simon Bridge

An object balanced on a point is "unstable" - any change at all from exactly balanced will cause it to start falling immediately. The amount of time to hit the ground can be worked out for ideal conditions - gravity provides a torque at the pencil's center of mass which depends on the current angle. The pencil rotates about it's point.

This is Newton's Laws for a rotating system - write it out and solve for initial angular displacement.

IRL: even an ideal pencil, perfectly balanced, will be subject to small random forces - vibrations, air-gusts etc. How long it takes to begin falling is a matter for statistics.

3. Dec 23, 2011

### DaveC426913

You're asking if an object, ideally balanced, takes 15 minutes before it falls over, how off true was it?

There is no way to answer this.

Under ideal conditions, it will either fall immediately or never. Anything less than ideal conditions comes down to the margin of error in the object, the environment or the accuracy of setup.

4. Dec 23, 2011

### Simon Bridge

There are several ways to treat it - it's just not clear which one applies here. It looks like malicor and friend have a misconception to overcome first.

The pencil described would start to fall immediately - because it is off center.
But there would be a hang-time effect (15mins seems a tad long for this).

If it starts perfectly balanced, then it is a stability-analysis problem and you can ask about the mean-time to start falling under a very small perturbation - then ask how big the perturbation has to be for the mean-time to be 15mins.

The exercise normally given to students is for an irregular object spinning about it's middle axis - how long before it tumbles?

But as you say - IRL, for a pencil: all bets are off.

5. Dec 23, 2011

### rcgldr

In the case of a pencil, it would start to slide and fall more quickly about half way over.

If the pencil couldn't slide, and assuming you want the time it takes for the pencil to fall down on its side, then the problem is similar to an inverted pendulum made of a solid uniform rod, which would be the equivalent of a simple inverted pendulum (all of the mass at the end of the pendulum) 2/3rd the length of the rod. The math involves solving an integral, defining θ = 0 when a pendulum is pointed straight down, you'd want to know the time it takes for the pendulum to fall from θ0 to π/2, and adjust or solve for θ0 close enough to π to get the time you're looking for, using the math from this wiki article:

Pendulum_Arbitrary-amplitude_period.htm

Last edited: Dec 23, 2011
6. Dec 23, 2011

### Spinnor

If we took the ideal case of a pencil of length l, = .1 m, and mass m then ?,

m g l sin(θ) = 1/3 m l^2 θ,tt = m g l θ for small θ -->

θ,tt/θ = 3g/l = c^2

This is solved here,

http://www.wolframalpha.com/

with input,

d^2x/dt^2 * 1/x = c^2

and output,

x(t) = d_1*exp[c*t] + d_2*exp[-c*t] or

θ(t) = d_1*exp[c*t] + d_2*exp[-c*t]

I think we can throw out the second term.

For t = 0, θ(0) = d_1 = θ_o

θ(t) = θ_o*exp[c*t] For t = 15 minutes θ becomes of order unity -->

exp[c*t] = about exp[15,400] which implies θ_o is pretty small?

Any mistakes?

7. Dec 23, 2011

### AlephZero

This is moot, even in Newtonian mechanics, for an arbitrary balanced object. Google for "Norton's dome". There was a recent PF thread about it.

Note, this does NOT apply to an idealised pencil toppling - Norton's dome is not spherical.

8. Dec 24, 2011

### malicor

hello

there's some misunderstandings here, which i will try to remove

first: the whole (theoretical) experiment shall be under ideal conditions, no wind, no air gusts, (not even a real pencil, but just a mass).

second: we're looking at the amount of time from the start situation to where the 'pencil' hits the ground. of course it starts falling immediately, but the time until it 'lands' is what we're after, and this time shall be 15 minutes.

third: fifteen minutes is an awefull high amount of time, the difference of the starting degree to 90degree will be like 0,000000000000000000000000000000000000000000something

in fact we suspect it's that small, that it's less than 0,000000[... insert one million more '0']0000001

this is actual the question, is it less than 1/10^1000000 or not. (that it's small we know already)

a formula wouldn't help me much, since i'm in no way a physician (but a computer scientist), and even if i had a formula i couldnt solve it.
we're pretty sure that it'll be hard to solve exactly, because it's a that small number, but quite sure that it'll be possible to solve the question "is it less than 1/10^1000000 due to some approximation strategies)

merry xmas to all of you :)

9. Dec 24, 2011

### Simon Bridge

http://demonstrations.wolfram.com/FallingStick/
... would be one interpretation of what you've said.
At the bottom is the equation of motion ... see how complex it gets?

In a simpler model, where you just call it a point mass on a massless but rigid pole, and the pivot is not allowed to slide, you get something more like:
$$r\frac{d\theta}{dt} = g\sin\theta$$... where theta is measured from upright.

Solutions go something like this.

- indeed - nor a physicist. (A "physician" is a medical doctor.) As a computer scientist you should be able to solve it numerically using a computer program xD

Also look at a mass sliding over the top of a sphere.
... as you look through these you'll notice that time is not often asked about.

One strategy would be to divide the angle of the fall into N small parts size $\Delta \theta = \pi/2N$, and treat the acceleration as constant over each part. This will give you N data points, for which you can get N accelerations - you can find the change in speed from acceleration and Δθ (knowing that the initial speed is zero), and so find the time to traverse each section. Add up the times for the overall time to fall.

Get a computer to do it.

10. Dec 24, 2011

### Spinnor

Again using,

http://www.wolframalpha.com/

input,

exp[15400],

output,

1.36X10^6688 --> θ_o = [1.36X10^6688]^-1 = 7.35X10^-6689 radians give or take θ_o *10^100 .

I think you win the bet (but as others have said you could never do the experiment).