Falling pencil (inverted pendulum)

1. Mar 8, 2014

rogeralms

1. The problem statement, all variables and given/known data
A pencil of length l = 0.2 m is balanced on its point. How much time does it take to fall? Assume that the pencil is a massless rod and all of its mass is at the tip. To make the math easier, assume the small angle approximation.

2. Relevant equations

theta (double dot) = g/l sin theta

T = 2 pi (l/g) ^ (1/2)

3. The attempt at a solution

Since the pencil seems to be an inverted pendulum and the fall would be only 1/4 of the period, I first tried T/4 = pi/2 (.2/9.8) ^ (1/2). But that is not a correct solution.

Next I tried to rearrange and integrate

1/sin theta d2theta = g/l dt2

1/sin theta = sin theta/ sin2theta= sin theta/(1 - cos2theta)

let u = cos theta so du=- sin theta d theta

Integral( 1/ u2 - 1) du = 1/2 ln(u - 1) - 1/2 ln (u + 1)

[ 1/2 ln (cos theta -1) - 1/2 ln (cos theta + 1)] d theta = gt/l dt

I get lost here because the cos pi/2 is zero which makes the above meaningless.

The way I interpret the problem, all the mass is at the balancing point. So how can you use energy to calculate potential difference or KE?

If someone could just give a hint as to the approach, I would be greatful.

Thank you.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 11, 2014

Arkavo

if the whole pencil can fall then θ will be π/2 how can the 'small angle approximation' hold?

3. Mar 11, 2014

rogeralms

Sorry, I agree, but that was the problem that I was given.

I think as in a lot of physics problems the assumption is made to make a good approximation.

As in how to design an automatic chicken plucker: First you assume a perfectly symmetrical, spherical chicken...

4. Mar 11, 2014

BvU

The small angle approximation helps you to get started. $\theta(0)=0$ doesn't.

5. Mar 14, 2014

rogeralms

Attached is the answer for your falling pencil pleasure. Hope this helps someone.

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