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Find g (gravity) on a compound pendulum

  • Thread starter GTdan
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1. A compound pendulum is arranged to swing about either of two parallel axes through two points O, O' located on a line through the center of mass. The distances h, h' from O, O' to the center of mass, and the periods [tex]\tau[/tex], [tex]\tau'[/tex]of small amplitude vibrations about the axes through O and O' are measured. O and O' are arranged so that each is approximately he center of oscillation relative to the other. GIven [tex]\tau=\tau'[/tex], find a formula for g in terms of measured quantities. Given that [tex]\tau'=\tau(1+\delta)[/tex], where [tex]\delta\ll1[/tex], find a correction to be added to your previous formula so that it will be correct to terms of order [tex]\delta[/tex]



2. [tex]\omega=\frac{2\pi}{\tau}=\sqrt{\frac{g}{l}}[/tex]

[tex]l=\frac{k_{o}^2}{h}[/tex]

[tex]l=h+h'[/tex]

[tex]hh'=k_{o}^2-h^2=k_{G}^2 [/tex]

[tex]k_{o}^2=k_{G}^2+h^2[/tex]

Where k is the center of oscillation at O, O', or the center of mass G (depending on the subscript).




3.

I am pasting and attaching a picture of my work because it's going to take a long time to type all that work down in Latex. I find g just fine when the period of oscillation is the same for when it's suspended at either point but when they are not equal (part b). The book's answer in the back is this:

[tex]g=[\frac{4\pi^2(h+h')}{\tau^2}][\frac{1+2h'\delta}{h-h'}[/tex]

As you can see, the last parts are guesses. I'm off by a strange factor somewhere and I don't know how. Help?


http://home.comcast.net/~dannyrod/1.13.jpg [Broken]
http://home.comcast.net/~dannyrod/1.13b.jpg [Broken]
 
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Answers and Replies

Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
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You make reference to equation numbers in your solution. Could you please tell us what book you're using?
 

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