How Do You Derive the Torsional Pendulum Period Formula?

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SUMMARY

The derivation of the torsional pendulum period formula is established as T=2*pi*sqrt{(2IL)/(pi*r^4*G)}. The discussion highlights the importance of distinguishing between angular frequency (Ω) and angular velocity (ω) to avoid confusion during calculations. The differential equation governing the motion is identified as d''θ/dt'' = -Ω²θ, leading to the solution θ(t) = A*cos(Ωt + φ). The period of oscillation is derived as T = 2π/Ω, confirming the relationship between angular frequency and the period.

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  • Understanding of differential equations, specifically second-order linear equations.
  • Familiarity with angular frequency and angular velocity concepts.
  • Knowledge of simple harmonic motion (SHM) principles.
  • Basic grasp of torque and moment of inertia in rotational dynamics.
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  • Study the derivation of the simple harmonic motion equations.
  • Learn about the applications of differential equations in physics.
  • Explore the relationship between torque, angular momentum, and rotational motion.
  • Investigate the effects of varying moment of inertia on the period of oscillation.
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Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators seeking to clarify concepts related to torsional pendulums and 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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