How does the Cavendish Experiment relate moment of inertia to elasticity?

[Cyrus]

Hi, I have been reading about the cavendish experiment and in one equation they say that

T = 2 \pi \sqrt{\frac I k}

where I is the moment of inertia and k is the elastic constant.

in my physics books i only find T = 2 \pi \sqrt{\frac m k}

where m is the mass, not the moment of inertia, anyone know how they got to this equation?

 

[kurious]

Since moment of inertia is usually something like mass x radius squared,
I think they have kept the radius constant and included it in the constant k.

 

[jamesrc]

The k in your first equation is a torsional stiffness with units (Nm)/rad

The k in your second equation is a linear stiffness with units N/m

If you work out the units, you will see that they both work out properly. Both equations come from the equations of motion for each system (meaning the differential equation that describes the motion of the system). We can get into that if you like; just let me know.

 

[Cyrus]

Hey James, Id sure like it if you could explain that in more detail please. I will look forward to your post. I will try to find some more info in this via the net in the meantime.

Look forward to reading your post.


Cyrus Abdollahi

 

[jamesrc]

Consider a simple physical system composed of a mass attached to a (grounded) spring. If you were to pull the mass a distance x, the restoring force from the spring with a magnitude equal to |kx|, where k is the stiffness of the spring and has units of N/m. This comes from Hooke's law.

Say you want to write the equation of motion for this simple system (this should be in your textbook). Draw a free body diagram and remember Newton's second law so that you can write:

m\ddot x = -kx
\ddot x + \frac{k}{m}x = 0

The solution to this differential equation is:

x = A\cos\left(\omega_nt + \phi \right)

(I hope this looks familiar.) In this equation,
ω_n = √(k/m) and the period of the motion is given by:
T = \frac{2\pi}{\omega_n} = 2\pi\sqrt{\frac m k}

Now all of this is analagous to the torsional situation in Cavendish's experiment. Now, Hooke's law looks like this:

\tau = K_t \theta

where K_t is the torsional stiffness (of the wire) that we're talking about, with units [Nm/rad].

The analagous equation of motion becomes:
\ddot \theta + \frac{K_t}{I}\theta = 0
\theta = A\cos\left(\omega_nt + \phi \right)

and in this case, ω_n = √(K_t/I)

so

T = \frac{2\pi}{\omega_n} = 2\pi\sqrt{\frac{I}{K_t}}

So I hope that made things clearer rather than muddier. Let me know if I can help any more.

 

[Cyrus]

I am taking physics 2 right now and we did this as the first chapter! WOO HOO! Thanks it all makes sense to me now. WOW what a clever way of making linear oscialtion into rotational through the use of that little picture. What a clever guy that hooke was.

 

[jfrusciante]

James or anyone else, is there an equation to derive the torsional stiffness (k), involving say young's modulus? Thanks.

 

[jamesrc]

Yes. It is a function the shear modulus, G (which is related to Young's modulus by Poisson's ratio for isotropic materials) and the geometry of the cross section. K_T = GJ(x)
(J(x) is the polar moment of inertia of the cross section). For example, for a cylinder, the cross section is a circle, for which J = πd⁴/32.

 

[jfrusciante]

Thanks James...that was very helpful.

The investigation i am doing now is to do with the factors that affect the time period and damping effects of a torsional pendulum. Recommend any good places for research?

 

[joep]

How does length affect the time period of a torsional pendulum?

 

[jfrusciante]

Joep...come on Joe I've told you before! k=GPi d^4/32L

T=2Pi(SQRT(I/K))

Substitute all that together and you'll have it.

 

[joep]

yeah by doing that i got

Gd^4
T= -------
64 l sqrt I

I was looking for something like T = k * l^1/2
or something, as I wasn't banking on working
out moment of inertia etc. ah well... cheers JF.