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March 31, 2015

March 31, 2015

Posted by **bec** on Sunday, November 6, 2011 at 9:54pm.

- physics -
**drwls**, Sunday, November 6, 2011 at 11:32pmThe bob rises to the same height on the opposite sde.

MgL(1 - costheta) + (1/2) M V^2

= M g L[1 -cos(thetamax)]

M cancels out.

At the bottom of the swing,

theta = 0 and

V^2 = 2 g L [1-cos(thetamax)] = Vmax^2

For half the velocity at bottom,

1 - cos(theta)max must be reduced by a factor of 4. Theta becomes 28 degrees.

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