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Falling pencil (inverted pendulum)
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[QUOTE="rogeralms, post: 4683424, member: 504728"] [h2]Homework Statement [/h2] 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. [h2]Homework Equations[/h2] theta (double dot) = g/l sin theta T = 2 pi (l/g) ^ (1/2) [h2]The Attempt at a Solution[/h2] 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 d[SUP]2[/SUP]theta = g/l dt[SUP]2[/SUP] 1/sin theta = sin theta/ sin[SUP]2[/SUP]theta= sin theta/(1 - cos[SUP]2[/SUP]theta) let u = cos theta so du=- sin theta d theta Integral( 1/ u[SUP]2[/SUP] - 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. [/QUOTE]
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Falling pencil (inverted pendulum)
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