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Buoyancy correction in a Kater's pendulum

  • Thread starter JulienB
  • Start date
  • #1
408
12

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 gonna 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.
 

Answers and Replies

  • #2
TSny
Homework Helper
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##= (\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?
 
  • #3
408
12
@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.
 
  • #4
TSny
Homework Helper
Gold Member
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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.
 
  • #5
408
12
@TSny I see. Thanks for your answer, our teacher explained something similar today.


Julien.
 

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