Solve Torsional Pendulum Homework Statement

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SUMMARY

The discussion centers on calculating the oscillation period of a torsional pendulum after modifying its length. Initially, a uniform meter stick oscillates with a period of 5 seconds. After being shortened to 0.76 meters, the new period is calculated using the moment of inertia formula and the relationship between the periods and lengths of the pendulum. The final period is determined to be 3.8 seconds, confirming the accuracy of the calculations presented.

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  • Understanding of torsional pendulum mechanics
  • Familiarity with moment of inertia calculations
  • Knowledge of oscillation period formulas
  • Basic algebra for manipulating equations
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Homework Statement



https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/phys2111/fall/homework/Ch-15-SHM/torsion-pendulum/4.gif


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?

L = 1m
L(final) = 0.76m
T = 5s

Homework Equations



Moment of Inertia: I = (1/12)*M*L^2
period: T = 2pi*sqrt(I/K) Where K is the torsional constant

The Attempt at a Solution



I first found the torsional constant:

K = [(1/12)*M*(1^2)]/[(T/(2pi))^2] = 0.13459M

So no I have K = 0.13459M

T(final) = 2pi*[sqrt((1/12)*M*(0.76^2)/0.13159M)] = 3.8s

The answer looks correct but I'm not sure, any ideas? Thanks!
 
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instead of going for so much calculation...
u could have done it like this...

becauz..

T=2*pi*(I/K)^0.5

and u have I=(M*L^2)/12

so jus take the ratio

T1/T2=L1/L2 :smile:

where T2 is the time period when the length is 0.76
directly arrives at your answer ...
 
Oh yeah it does...cool.
 

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