How Do You Derive the Torsional Pendulum Period Formula?

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The discussion focuses on deriving the formula for the period of a torsional pendulum, T=2*pi*sqrt{2IL/(pi*r^4*G)}. The initial approach involved relating torque and angular acceleration, but confusion arose between angular velocity and angular frequency. Participants clarified that the differential equation for the system resembles that of simple harmonic motion, leading to the conclusion that the solution can be expressed as theta(t) = A*cos(Ωt + φ). The correct identification of angular frequency, Ω, allows for the determination of the period, T = 2π/Ω. Understanding the distinction between angular velocity and frequency is crucial for accurate derivation.
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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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