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.
theta (double dot) = g/l sin theta
T = 2 pi (l/g) ^ (1/2)
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.