How Do You Derive the Torsional Pendulum Period Formula?

In summary: My apologies.In summary, the formula derivation for T=2*pi*sqrt{(2IL)/(pi*r^4*G)} can be shown using the equation tau=I*alpha=pi*G*r^4/(2L)*theta and the fact that T=2pi/omega. By setting theta(t) as f(t) in the differential equation of SHM, \ddot \theta=-Omega^2 \theta, the solution can be found to be \theta (t) = A*cos{(\Omega t + \phi)}, with a period of T=2pi/Omega.
  • #1
Melawrghk
145
0

Homework Statement


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

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


Homework Equations


See above


The Attempt at a Solution



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

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

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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  • #2
You mixed up two VERY different [tex]\omega[/tex]'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 [tex]\ddot x=-kx[/tex] (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 [tex]\Omega[/tex] for angular velocity instead, to prevent further mixups.
 
  • #3
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?
 
  • #4
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:

[tex]\ddot \theta=-\Omega^2 \theta[/tex]

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

[tex]\theta (t) = A\cos{(\Omega t +\phi)}[/tex]

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

[tex]T=\frac{2\pi}{\Omega}[/tex]

On a side note, when you tried to integrate:

[tex]d\omega = -k\theta \cdot dt[/tex]

You overlooked the fact that [tex]\theta[/tex] 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.
 

1. What is a torsional pendulum?

A torsional pendulum is a type of pendulum that consists of a mass suspended from a wire or string that can twist or rotate. It is used to study the effects of torsional force and the period of oscillation of the pendulum.

2. How is the period of a torsional pendulum calculated?

The period of a torsional pendulum is calculated using the equation T = 2π√(I/k), where T is the period of oscillation, I is the moment of inertia of the pendulum, and k is the torsional constant of the wire or string.

3. What factors affect the period of a torsional pendulum?

The period of a torsional pendulum is affected by the length and thickness of the wire or string, the mass of the pendulum, and the force of gravity.

4. How is a torsional pendulum used in scientific research?

A torsional pendulum is used in scientific research to study the physical properties of materials, such as their elasticity and shear modulus. It is also used to measure small changes in gravitational force and to calibrate other instruments.

5. How can the period of a torsional pendulum be increased?

The period of a torsional pendulum can be increased by increasing the length and thickness of the wire or string, increasing the mass of the pendulum, or decreasing the force of gravity. Additionally, using a material with a higher shear modulus can also increase the period of the pendulum.

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