Physics Help Forum Compound pendulum

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 May 28th 2019, 02:03 AM #1 Junior Member   Join Date: May 2019 Posts: 2 Compound pendulum a compound pendulum is formed by suspending a heavy ring from a point on its circumference . determine the time period of oscillation of radius of the ring is 1m. Time period T =2π √I/mgl I took I= mR^2/2 and l=r which gave me an answer of 1.41s . But the actual answer is 1.28s what am I doing wrong?
May 28th 2019, 11:03 AM   #2
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 Originally Posted by McKenziemc799 a compound pendulum is formed by suspending a heavy ring from a point on its circumference . determine the time period of oscillation of radius of the ring is 1m. Time period T =2π √I/mgl I took I= mR^2/2 and l=r which gave me an answer of 1.41s . But the actual answer is 1.28s what am I doing wrong?
From your verbal description, I gather the compound pendulum looks something like the attached figure. Would that be correct, or no?
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 May 28th 2019, 11:16 AM #3 Junior Member   Join Date: May 2019 Posts: 2 Yes.
May 28th 2019, 02:08 PM   #4
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 Originally Posted by McKenziemc799 Yes.
Unless I'm missing some other information, I disagree with both your calculation for period and the "actual answer" value.

Using the parallel axis theorem,

$I = I_{com} + md^2$

for a uniform ring of mass $m$, $I_{com} = mr^2$ and $d = r$

so, $I = mr^2 + mr^2 = 2mr^2$

$T = 2\pi \sqrt{\dfrac{I}{mgd}} = 2\pi \sqrt{\dfrac{2mr^2}{mgr}} = 2\pi \sqrt{\dfrac{2r}{g}}$

For $r= 1 \text{ m } \implies T = 2\pi \sqrt{\dfrac{2}{g}} \approx 2.84 \text{ sec}$

 Tags compound, pendulum, time period

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