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Hi from one of my lab classes I had to calculate the moment of inertia of an object by turning it into a torsional pendulum, and measuring its period with a known constant A, using the following equation:

$$I=\frac{A}{4\pi^{2}}T^{2}$$

However, this expression comes from a small angle approximation, and in the experiment I am suposed to rotate the whole system (resembling a torsional pendulum) at 90º, which clearly isn't a small angle. Can someone help me to get an expression of the motion of an undamped torsional pendulum without approximation, so I can run some simulations to prove the validity of the small angle approximation? That is, discover at what angles I get an error below 1%.

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    $\begingroup$ Why does a regular pendulum need the small angle approximation? Does it apply to the geometry of a torsion pendulum? $\endgroup$ – JEB May 2 '18 at 17:33
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The small angle refers to the distortion angle of a small part of the wire. Refer to the figure below for an illustration of the distortion angle $\theta$ as a function of the twist angle $\psi$

pic

By equating the tangential movement at one end due to twisting to the movement due to distortion, one arrives as

$$ r \psi = \ell \theta \; \\ \boxed{\theta = \frac{r}{\ell} \psi} $$

So even if $\psi = 90^\circ$, the distrotion angle $\theta \ll \psi$ since $r \ll \ell$ for any wire.

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  • $\begingroup$ I think I got it, though, just to make sure, the apparatus I used was similar to the one in thsi image:thumb7.shutterstock.com/display_pic_with_logo/1033249/600358694/…, so I'm assuming the angle $\theta$ is relative to the spring and not to the object, thus even if I rotate the body in $\psi$=90º, the formula I presented for the moment of inertia stays valid. Is that it? $\endgroup$ – Bidon May 2 '18 at 18:27
  • $\begingroup$ Yes. The distortion in the spring is due to bending (curvature change) and the corresponding angle is small because the thickness is $\ll$ length. $\endgroup$ – ja72 May 2 '18 at 19:21
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For a torsional pendulum the wire from which the object is suspended exerts a torque $\tau =- \kappa \, \theta$ where $\kappa$ is the torsional constant and $\theta$ is the angle of twist.
This expression holds for large angles of twist and so in the derivation of the period of the torsional pendulum no small angle approximation has to be applied.


In derivations where the approximation $\sin \theta \approx \theta$ is made where $\theta$ is the angle in radians if the value of $\theta$ is less than about $0.39$ radian or $22^\circ$ the the approximation is good to about $1\%$.

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  • $\begingroup$ Isn't $\sin(1/11)-1/11\sim10^{-4}$, which is much better than 1%, right? $\endgroup$ – Kyle Kanos May 2 '18 at 21:16
  • $\begingroup$ @KyleKanos Thank you for pointing out my error which I have now corrected. $\endgroup$ – Farcher May 2 '18 at 21:46

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