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April 16, 2014

April 16, 2014

Posted by **how to do this?** on Monday, January 30, 2012 at 1:11pm.

- physics -
**Damon**, Monday, January 30, 2012 at 4:24pmpotential energy = m g h = m g L (1-cosA)

if angle A is small, cos A = 1 - A^2/2 +...

so

Pe = m g L A^2/2

kinetic energy = (1/2) m v^2 = (1/2) m L^2 [dA/dt]^2)

if A = sin wt

dA/dt = -w cos wt

max potential energy at top = max Ke at bottom

(1/2) m L^2 w^2 = (1/2) m g L

w^2 = g/L which by the way is in your book

g = L w^2

but w = 2 pi /T since when t = T, the angle w t must be 2 pi, full circle

g = L (2 pi)^2 / T^2

here L = .6 and T = 1.55

so g = 9.85 m/s^2

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