How Do You Derive the Torsional Pendulum Period Formula?

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Homework Help Overview

The discussion revolves around deriving the formula for the period of a torsional pendulum, specifically the expression T=2*pi*sqrt{(2IL)/(pi*r^{4}*G)}. The original poster presents their understanding of the relationship between torque, moment of inertia, and angular acceleration, and attempts to manipulate these equations to find an expression for angular frequency.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive the period by relating torque and angular acceleration, leading to an expression for angular frequency. They express confusion over integrating their equation and the distinction between angular frequency and angular velocity.

Discussion Status

Some participants provide feedback on the original poster's approach, suggesting that they have mixed up angular frequency and angular velocity. Others confirm the correctness of relating the differential equation to simple harmonic motion and provide insights into the general solution for such equations. There is a recognition of the need to clarify terminology and the implications of the equations being discussed.

Contextual Notes

Participants are navigating the complexities of angular motion and the specific definitions of terms like angular frequency and angular velocity. There is an emphasis on ensuring the correct application of mathematical principles related to harmonic motion.

Melawrghk
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Homework Statement


I have to show the formula derivation of this:
T=2*pi*\sqrt{\frac{2IL}{pi*r^{4}*G}}

based on the fact that I know this:
\tau=I\alpha=\frac{pi*G*r^{4}}{2L}\theta


Homework Equations


See above


The Attempt at a Solution



Well, I know T=2pi/\omega and \alpha=\Delta\omega/\Delta(t)

So I decided to just get an equation for omega from the expression for tau.
So I had:
d\omega/dt=\frac{pi*G*r^{4}}{2IL}\theta
Which looked promising until I integrated both sides wrt 't' and got:
\omega=\frac{pi*G*r^{4}}{2IL}\theta*t

And this really gets me nowhere and I don't know what else to do. Thanks in advance for the help!
 
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You mixed up two VERY different \omega's

One is angular frequency and the other is angular velocity. They are completely unrelated.

Look at your net torque equation, it is a differential equation of the form \ddot x=-kx (Remember that it is a restoring torque, so you missed a negative sign)

You should be very familiar with the general solution to that equation.

I suggest that you use \Omega for angular velocity instead, to prevent further mixups.
 
Well, it's familiar and it's from SHM.

There was the d''f(t)/dt''=-omega^2 * f(t) when f(t)=Asin(omega*t+phi)

If I make theta(t)=f(t), then that k would equal omega^2 (the angular frequency omega)

Is that correct though?
 
Melawrghk said:
Well, it's familiar and it's from SHM.

There was the d''f(t)/dt''=-omega^2 * f(t) when f(t)=Asin(omega*t+phi)

If I make theta(t)=f(t), then that k would equal omega^2 (the angular frequency omega)

Is that correct though?

That is 100% correct. :)

Once you have the differential equation:

\ddot \theta=-\Omega^2 \theta

The solution should be something that immediately pops into your head:

\theta (t) = A\cos{(\Omega t +\phi)}

And the period for a harmonic function is something you can easily find,

T=\frac{2\pi}{\Omega}

On a side note, when you tried to integrate:

d\omega = -k\theta \cdot dt

You overlooked the fact that \theta is a function of time. That was the source of your error. I was mistaken in thinking you got angular velocity and frequency mixed up.
 

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