Period of Small Oscillations for Rigid Body Pendulum

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SUMMARY

The discussion focuses on calculating the period of small oscillations for a rigid body pendulum consisting of a 250mm light rod with two identical solid spheres of 50mm radius attached. The formula used is Period = 2*pi*sqrt(I/MgR), where I is the moment of inertia. The parallel axis theorem is applied incorrectly, leading to confusion in obtaining different results for oscillations perpendicular and along the line of centers. The correct application of the parallel axis theorem is crucial for accurate calculations.

PREREQUISITES
  • Understanding of rigid body dynamics
  • Familiarity with the parallel axis theorem
  • Knowledge of moment of inertia calculations
  • Basic principles of pendulum motion
NEXT STEPS
  • Review the application of the parallel axis theorem in rigid body dynamics
  • Study the derivation of the moment of inertia for composite bodies
  • Explore the effects of pivot points on pendulum motion
  • Learn about small angle approximations in oscillatory motion
USEFUL FOR

Physics students, mechanical engineers, and anyone studying oscillatory motion in rigid body systems will benefit from this discussion.

willisverynic
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Homework Statement


A pendulum consists of a light rigid rod of length 250mm, with two identical uniform solid spheres of radius 50mm attached one on either side of its lower end. Find the period of small oscillations (a) perpendicular to the line of centers, and (b) along it.

Homework Equations


Period = 2*pi*sqrt(I/MgR)
Parallel axis theorem: I = Icm + Md^2

The Attempt at a Solution


I tried simply applying the parallel axis theorem to a ball using Icm = 2/5mr^2 and d = sqrt(.25^2+.05+2) but this yields and incorrect answer. I believe this method would be wrong because it would give the same answer for both parts, which is not the case

Thanks

ps not sure if this belongs in advanced or not
 
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hi willisverynic! welcome to pf! :smile:

(try using the X2 icon just above the Reply box :wink:)
willisverynic said:
I tried simply applying the parallel axis theorem to a ball using Icm = 2/5mr^2 and d = sqrt(.25^2+.05+2) but this yields and incorrect answer. I believe this method would be wrong because it would give the same answer for both parts, which is not the case

i suspect you're measuring from the pivot point rather from than the axis

the axis of rotation is through the pivot point, either North or East …

the parallel axis is through the centre of one sphere, and the distance between the two axes does depend on which way round they are :wink:
 

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