Mercury-compensated pendulum

  • Thread starter SimonasV
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  • #1
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Homework Statement


Mercury-compensated pendulum

A small part of Nickel tube is fulfilled with mercury.

(Alpha) Linear coefficient of expansion (Nickel)= 1x10^-5
(Beta) Volumetric coefficient of expansion (Mercury)= 18X10^-5

FIND: What part of tube should be fullfilled, that the period of the pendulum would not depend upon temperature. And there is two different cases:
a) When centre of mass coincide with centre of Mercury in that tube.
b) Include misalignment of centres...


Homework Equations



T=2pi(l/g)^1/2

deltaV/V=Beta*deltaT

deltaL/L=Alpha*deltaT

L=L0(1+alpha*deltaT)
V=V0(1+Beta*deltaT)

Should be more but don't know. Maybe someone could help :-)
 
Last edited:

Answers and Replies

  • #2
djeitnstine
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Hello simon. Would you happen to have any form of diagrams? I am having a difficult time imagining this. What is the tube connected to? Am I to assume the tube is the bob? some more details on the setup would be nice
 
  • #3
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attachment.php?attachmentid=18596&stc=1&d=1240575450.jpg


cases: a) the upper level of mercury should coincide with the mass centre of whole pendulum and b) it should not coincide.
 

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  • #4
Redbelly98
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Hello SimonasV,

Think about: what is "l" in the pendulum equation?
 
  • #5
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l=L - length of pendulum
 
  • #6
Redbelly98
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That's true if the pendulum mass is a point-mass.

However, in this example the mass occupies a region of space. So how would we define l in this case? (Hint: it's the distance from the pivot point to ______?)
 

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