Tilting Pencils 

If a pencil is carefully balanced on its point, how long will it take to fall over? Neglecting all external disturbances and nonideal effects, this can be modeled as an inverted pendulum of length L and mass m, released from an initial position making an angle of q_{0} relative to vertical, as illustrated below. 



The simple equation of motion for this tilting pencil can be found by either equating the torque with the product of the moment of inertia and the angular acceleration, or by an energy balance. If the mass per unit length is r = m/L, then the kinetic energy of an incremental segment dr of the rod is dK = (1/2) r dr v^{2} where v = wr. Hence the total kinetic energy of the rod is 

_{} 

The potential energy P can be expressed as the product of the acceleration of gravity times the total mass of the pencil times the height of the midpoint of the pencil, so we have P = rLg(L/2)cos(q) where g is the acceleration of gravity. Setting the sum of the kinetic and potential energies to a constant C gives the equation of motion 

_{} 

Dividing through by rL^{3}/6, we arrive at 

_{} 

This implies 

_{} 

so we have 

_{} 

Consequently the time for the pencil to fall to the horizontal position from an initial angle of q_{0} beginning with an initial angular speed of _{} is given by the integral 

_{} 

There is no simple closedform expression for this integral, although it can be converted to the standard form of an elliptic integral. However, for extremely small initial angles q_{0} (e.g., 10^{10} degrees) the integral is very illconditioned and not easily evaluated. 

To arrive at a more easily evaluated expression, we can first consider only the initial interval of time, during which the pencil just begins to tip from nearly a vertical position. Differentiating (1) and dividing through by 2(dq/dt) gives 

_{} 

This is the same equation of motion we would derive directly from equating the torque to the product of angular acceleration and moment of inertia mL^{2}/3 of a slender rod about its end point. Note that as long as the angle q is extremely small, the value of sin(q) is essentially equal to q itself. (This is the same "small angle approximation" that we commonly use for an ordinary noninverted pendulum to predict simple harmonic motion.) Therefore, as long as the pencil is very near vertical its motion essentially satisfies the linear equation 

_{} 

This simple secondorder linear equation has the solution 

_{} 

where k is the square root of 3g/(2L), and the constants A,B are determined by the initial conditions. If the pencil is initially at rest at an angle q_{0} from vertical, then A = B = q_{0}/2, and the solution is 

_{} 

Solving for the time t_{1} when the pencil has reached an angle of q_{1}, we find that 

_{} 

The angular velocity of the pencil at this instant is given by differentiating the expression for q(t) and substituting this value for t. The result is 

_{} 

Now we can use equation (2) to give the time required to tip from q_{1} to q_{2} = p/2, and we can add this to the time t_{1} required to tip from q_{0} to q_{1} to give the total time. By equation (2) we have 

_{} 

For very small values of q_{0} we can neglect the q_{0}^{2} in this expression, and we can expand the expression under the square root as 

_{} 

Since we are taking a small value of q_{1}, the higher order terms can be neglected, and the argument of the square root is simply 1  cos(q) = 2sin(q/2)^{2}. Making this substitution, the integral can be evaluated in closed form as 

_{} 

Returning to the expression for t_{1}, notice that for arguments much greater than 1 the inverse of the hyperbolic cosine is essentially equal to the natural log of twice the argument, so we have 

_{} 

Adding this to the previous expression gives a very close approximation to the total time for the pencil to tip from q_{0} to p/2 

_{} 

Recall that q_{1} is really just an intermediate small angle, which can be chosen small enough so that sin(q_{1}/2) is essentially equal to q_{1}/2, and so that 1 + cos(q_{1}/2) is essentially equal to 2. Therefore the above expression can be simplified to 

_{} 

As an example, if the pencil is L = 0.2 m long, and if we take g = 9.8 m/sec^{2}, then the total time for the pencil to tip over from an initial stationary position of q_{0} = 10^{10} degrees (which equals (p/180)10^{10} radians) is 3.29773... seconds. On the other hand, if we imagine that the initial pencil position is just q_{0} = 10^{100} degrees, then the time to tip over is 27.469838205... seconds. In general, the time to tip over from an initial stationary position of 10^{n} degrees is 

_{} 

This idealized pencil would take 1 year = 31557600 seconds to tip over (theoretically) if its initial stationary position is q_{0} = 10^{117498396} degrees from vertical. 
