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April 20, 2014

Posted by **me** on Friday, December 9, 2011 at 5:13pm.

- physics -
**bobpursley**, Friday, December 9, 2011 at 5:18pmNot knowing the center of gravity, lets look at energy.

Assume the distance from the piviot point to the center of mass is r.

Then the distance the mass fell is 2r, and the PE lost is 2rmg.the KE at the bottom is 1/2 m v^2, and they are equal

1/2 m v^2=2rmg

r= 1/4 v^2/g

so, period of pendulum= 2PI sqrt(r/g)

and frequncey is 1/Period

- physics -
**me**, Friday, December 9, 2011 at 5:30pmIs the formula r = (1/4)(v^2/g)? or (1/4 v^2)/g ?

- physics -
**bobpursley**, Friday, December 9, 2011 at 5:58pmperiod= 2PI sqrt (1/4 v^2/g^2)=2PI v/2g=PI v/g

frequency= 1/period= g/(PIv)

check that.

- physics -
**me**, Friday, December 9, 2011 at 6:09pmI am just confused on the way some of these are written out like in the period equation you just gave, is pi multiplied by v and then divided by g or is the whole quantity given by v/g multiplied by pi?

- physics -
**bobpursley**, Friday, December 9, 2011 at 6:10pmperiod= 2PI * sqrt(v/g)

- physics -
**me**, Friday, December 9, 2011 at 6:21pmSo the first formula period=2pi*sqrt(r/g) is incorrect? I am just confused I am sorry

- physics -
**bobpursley**, Friday, December 9, 2011 at 6:28pmsorry about that, it is r/g, I am trying to clear this board too quickly.

- physics -
**me**, Friday, December 9, 2011 at 6:37pmSo is the formula for r: r = (1/4)(v^2/g)? or (1/4 v^2)/g ? sorry I am asking so much

- physics -
**bobpursley**, Friday, December 9, 2011 at 6:42pmThere is no difference.

a/b * c/d= ac/bd= ac/b * 1/d

1/4 v^2/g= 1/4 v^2/g= 1/(4g) * v^2

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