Estimate from the uncertainty principle how long a lead penc

Estimate from the uncertainty principle how long a lead pencil can be balanced upright on its point.

Solution

Modeling the pencil as a uniform rod with mass m, length , moment of inertia I=(1/3)m², the torque when it is at an angle to the vertical is

=(1/2)mgsin

For small deflections, sin= and we will use that assumption below. Then the equation of motion becomes

I¨=(1/2)mg

(1/3)m²¨=(1/2)mg¨=(3g/2)

This looks a lot like the equation for a simple harmonic oscillator, but with the wrong sign. We do indeed get a very similar solution, but with hyperbolic functions.

Putting 3g/2=², we can integrate this twice to get a

=C₁e^t+C₂e^t

If the center of mass is off by x, then the angle is =(2x/).

Using the same equations as before, we find C1+C2=. Let\'s assume the initial velocity is zero - because \"on average\" it will be, given that the direction of initial velocity is equally likely to point back towards equilibrium and away from it - so we get C1=C2, and the solution is a cosh function:

=2C₁cosh(t)

Where C₁=/2=x/.

We now have the time to fall (time to reach a certain ) as

t=(1/)cosh¹(/2x)

The equation we derived earlier for the time taken with a given initial velocity can be rewritten as

t=(1/)sinh¹(/v)

and we know that

xp=

Obviously the longest time to balance will be reached when the two times are the same - otherwise, one will be longer and the other shorter, and it\'s the shorter time that will dominate. To solve, we substitute for x=/mv and get

cosh¹(mv/)=sinh¹(/v)

If the term in the parentheses is sufficiently large, then we can (for the purpose of estimating) set

(mv/)=(/v)

From which it follows that

v= sqrt(/m)     = /m

Substituting the values for the pencil, we find

v=4*10¹⁶m/s

The time to fall is then

t=(1/)sinh¹(/v)3.7s

So a perfectly balanced, \"theoretical\" pencil that stands on its mono-atomic tip, will fall on average in a handful of seconds because of the uncertainty principle.