Rocking frequency of half-cylinder

[smolloy]

Hi all,
I'm trying to calculate the rocking frequency of a half-cylindrical arch. That is, a half-cylinder, that has had a smaller half-cylinder "bitten" out of it. If placed with the curved surface on the floor, it can be made to rock from side to side (sort of like an inverted pendulum).

In calculating this, I have come up with the following differential equation:
$$\theta'' = -g\frac{3}{\pi}R\theta$$
Where $$\theta$$ is the angle of the arch from its nominal position, g is the gravitational constant, and R is the radius of the arch.

The problem is that I interpret this as meaning that the frequency is proportional to the square root of the radius, when observation tells me that it should be an inverse relationship (i.e. large radius leads to low frequency).

Have I interpreted the differential equation incorrectly, or have I made an error somewhere in my derivation?

 

[tiny-tim]

hi smolloy!

(have a theta: θ and a pi: π )

your equation is wrong …

you can use either conservation of energy or τ = Iα …

show us what you get ​

 

[smolloy]

Hmmm... that's what I thought.

I started by calculating the position of the centre of mass for an arch, and then using the limiting case when the inner radius approaches the outer radius. I calculated the limiting value for this as (3R)/π (i.e. approximately 95% of the full radius).

When you rock the arch to some angle, θ, you can then draw a line connecting the centre of mass to the point of contact with the floor, and you can resolve the gravitational force perpendicular to this line. Multiplying this resolved force by the length of the line (found by a bit of trig with triangles), gives the torque.

Since torque is the rate of change of angular momentum, you can then use this result to write down the previous equation (you also need to use the small angle approximations to sin(x)-->x and cos(x)-->1).

Sigh... I *know* I'm wrong, but I can't figure out why!

 

[tiny-tim]

well, your torque is proportional to R, and your moment of inertia is proportional to … ?

 

[smolloy]

Ah! :)

The equation I wrote down was as follows:
$$m\theta'' = -mg\frac{3}{\pi}R\theta$$

But perhaps that is incorrect? Perhaps the m on the left hand side should be $$mR^2k^2$$ (where R.k is the distance between the centre of mass and the point of contact with the floor)?

This would imply that the mass cancels out (as in the case of a pendulum), and that $$R^{-1}$$ appears on the right hand side (as required)...

Is this what you meant?

 

[tiny-tim]

yes, you were using τ = mα instead of τ = Iα !

happy new year! ​