Buoyancy correction in a Kater's pendulum

[JulienB]

Homework Statement

Hi everybody! While preparing my next experiment (Kater's pendulum), I was given for homework to derive an equation to correct the buoyancy when calculating $g$. I am given the result:

$g_c = (\frac{2 \pi}{T(\varphi_0)})^2 l_r (1 + \frac{\varphi_0^2}{8} + \frac{\rho_L}{\rho})$

where $\varphi_0$ is the amplitude, $l_r$ is the length of the equivalent simple pendulum, $\rho_L$ is the density of the air, $\rho$ is the density of the pendulum.

2. The attempt at a solution

I manage to get the $1 + \frac{\varphi_0^2}{8})$ from the Taylor development of $T(\varphi_0)$ and taking the approximation $\sin \varphi \approx \varphi$.

Then I thought I was going to derive the buoyancy effect pretty easily from the torque, but that's what I get:

$\tau = m g l_r \sin \varphi - l_r \rho_L V g \sin \varphi = I \ddot{\varphi}$
$\implies \ddot{\varphi} - \frac{l_r g (m - \rho_L V)}{I} \varphi = 0$
$\implies T^2 = (2 \pi)^2 \frac{I}{l_r g (m - \rho_L V)}$
$\implies g = (\frac{2 \pi}{T})^2 \frac{I}{l_r (m - \rho_L V)}$
$= (\frac{2 \pi}{T})^2 \frac{m l_r^2}{l_r (m - \rho_L V)}$
$= (\frac{2 \pi}{T})^2 l_r \frac{\rho}{\rho - \rho_L}$

Mmm... That's not too far but not quite it! I've been thinking a lot about it now, and I don't get why it's not working. Does anyone have an idea? (Note that the Taylor development is added by putting $T(\varphi_0)$ inside the equation)

Thanks a lot in advance for your answers!Julien.

 

[TSny]

$= (\frac{2 \pi}{T})^2 l_r \frac{\rho}{\rho - \rho_L}$

Mmm... That's not too far but not quite it! I've been thinking a lot about it now, and I don't get why it's not working. Does anyone have an idea?

Can you expand $\frac{\rho}{\rho - \rho_L}$ to first order in a small quantity?

 

[JulienB]

@TSny Hi and thanks for your answer. There is no other indication than what I have written. The problem with expansion is also that I get an extra $+1$...

Julien.

 

[TSny]

You should have something proportional to $(1+\frac{ \varphi_0^2}{ 8})(\frac{\rho}{\rho - \rho_L})$. As you say, if you expand the second quantity to first order, the second quantity will be of the form 1 + ε, where ε is a small quantity. But, you can then multiply the whole thing out and keep only terms up to first order in small quantities.

 

[JulienB]

@TSny I see. Thanks for your answer, our teacher explained something similar today.Julien.