Moment of Inertia/Torque about point O of the rod

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SUMMARY

The moment of inertia of a thin rod of length L and mass M, rotating about point O located at a distance L/3 from one end, is calculated using the formula Irod = 1/3ML². The torque due to the rod's weight when displaced by an angle θ is determined by analyzing the gravitational force acting at the center of mass. For small angular displacements, the period of oscillation can be derived using the principles of rotational dynamics and simple harmonic motion.

PREREQUISITES
  • Understanding of moment of inertia and its calculation
  • Familiarity with torque and its relation to angular displacement
  • Knowledge of the Parallel Axis Theorem
  • Basic principles of oscillatory motion and period calculation
NEXT STEPS
  • Study the Parallel Axis Theorem in detail
  • Learn about torque calculations for rigid bodies
  • Explore the derivation of the period of oscillation for simple harmonic motion
  • Investigate the effects of angular displacement on torque and oscillation
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Physics students, mechanical engineers, and anyone studying rotational dynamics and oscillatory motion will benefit from this discussion.

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Homework Statement


A thin rod of length L and mass M is free to rotate at a point O at a distance L/3 from one end.

a.) What is the moment of inertia of the rod about O?
b.) What is the magnitude of the torque due to the rod's own weight about O when it is displaced from the vertical by and angle θ?
c.) For small angular displacement, find the period of oscillation of this rod.

Homework Equations


Irod= 1/3ML2

The Attempt at a Solution


Not really sure what to do here
 
Physics news on Phys.org
First, what is the moment of inertia of a rod about its center? Then look up "Parallel Axis Theorem".
 

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