Torsion Pendulum Oscillation (SHM)

  • Thread starter SuperCass
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  • #1
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Homework Statement



A uniform meter stick is hung at its center from a thin wire. It is twisted and oscillates with a period of 5 s. The meter stick is then sawed off to a length of 0.76 m, rebalanced at its center, and set into oscillation.

With what period does it now oscillate?

Homework Equations



T=2pi(sqrt(I/K))
I=(1/12)ML^2

The Attempt at a Solution



I tried solving for m in terms of k and then plugging that into a new equation solving for T with the length being .76, but that didn't work. Any suggestions?

Thanks!
 

Answers and Replies

  • #2
tiny-tim
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Homework Helper
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Hi SuperCass! :smile:

(have a pi: π and a square-root: √ and try using the X2 tag just above the Reply box :wink:)

Hint: by how much is I reduced? :wink:
 
  • #3
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(1-.76)² ?
So since the formula is T=2π√(I/k), would I multiply the original T by (1-.76)?


(Thanks for your signature!)
 
  • #4
tiny-tim
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No, I is proportional to ML2, and both M and L are reduced. :wink:
 
  • #5
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Remember that you've been told that the rod is uniform, that means that its mass is proportional to its length! If we define a mass per unit length, [tex]\mu\equiv \frac{M_0}{L_0}[/tex] then the mass of a rod of length [tex]\ell[/tex] made of the same material would be..?
 
  • #6
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.76 of the original mass?
 
  • #7
671
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.76 of the original mass?

Exactly! That makes for [tex]m \propto \ell[/tex] which implies [tex]I \propto \ell^3[/tex]
 
  • #8
60
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Oh!
I got it now!
So I just do the original period times the square root of .76^3!

Thank you all very very much!!
 

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