Posted by me on Friday, December 9, 2011 at 5:13pm.
A pendulum consists of a massless rigid rod with a mass at one end. The other end is pivoted on a frictionless pivot so that it can turn through a complete circle. The pendulum is inverted, so the mass is directly above the pivot point, and then released. The speed of the mass as it passes through the lowest point is 6.0 m/s. If the pendulum undergoes small amplitude oscillations at the bottom of the arc, what will be the frequency of the oscillations?

physics  bobpursley, Friday, December 9, 2011 at 5:18pm
Not 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 5:30pm
Is the formula r = (1/4)(v^2/g)? or (1/4 v^2)/g ?

physics  bobpursley, Friday, December 9, 2011 at 5:58pm
period= 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 6:09pm
I 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 6:10pm
period= 2PI * sqrt(v/g)

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

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

physics  me, Friday, December 9, 2011 at 6:37pm
So 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 6:42pm
There 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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