Physical pendulum and meter stick

  • Thread starter bearhug
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  • #1
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A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance x from the 50. cm mark. The period of oscillation is observed to be 1.656 s. Find the distance x.

I know I have to find the moment of Inertia and I believe the equation for this is I=1/2mL^2 + mx^2 is this the right equation? Then use the equation T= 2pie(I/mgx)^1/2.

This is what I did but it's not right so if any one can point out any errors will be greatly appreciated:
I=1/2mL^2 + mx^2
I=1/2m(0.5)^2 + mx^2
I=0.125m + mx^2

T=2pie(I/mgx)^1/2
1.656=2pie{(0.125m + mx^2)/(9.8mx)}^1/2
0.263= {(0.125+x)/9.8}^1/2
0.0692(9.8)=0.125+x
0.553=x

Is this right?
 

Answers and Replies

  • #2
I believe the factor for your moment of inertia is off. I=(1/2)mL^2 is the moment of inertia of a disc. I think you should be using the moment of inertia for a rod, which turns out is just a different factor... it may be 1/3 instead of 1/2. You should look this up though, I'm not entirely positive.
 

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