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January 31, 2015

January 31, 2015

Posted by **Anonymous** on Tuesday, November 22, 2011 at 10:37pm.

- physics -
**Anonymous**, Tuesday, November 22, 2011 at 10:48pmSuppose the simple pendulum shown below were released from an angle of θ = 74°, with L = 0.73 m and m = 0.25 kg..What angle of release would give half the speed of that for the 74° release angle at the bottom of the swing?

- physics -
**Fizzylicous**, Tuesday, November 22, 2011 at 10:50pmYou must have been given a formula to calculate this problem. What formulas do you have? You can't just expect to recieve an answer without you asking what you need help on and jotting down the formulas, that's being pure lazy.

- physics -
**drwls**, Wednesday, November 23, 2011 at 7:25amFor half the speed at the bottom, you would need 1/4 the initial potential energy.

The initial P.E. is proportional to

(1 - cos A)

where A is the initial angle from vertical.

1 - cos74 = 0.7244

For 1/4 of that value,

1 - cosA = 0.1811

cosA = 0.8189

A = 35.0 degrees

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