Determine period for off center disk

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SUMMARY

The discussion focuses on determining the period of a disk suspended horizontally by a rod bored through it at a distance d from the center. The relevant moment of inertia is calculated using the formula I = 1/2 ma^2 + md^2. The user proposes using the relationship alpha = T/I and relates it to the period, ultimately arriving at the expression for the period as π((a^2) + d^2)/g*d. This formula provides an approximate period for the oscillation of the disk, ignoring dissipative effects.

PREREQUISITES
  • Understanding of rotational dynamics and Newtonian mechanics
  • Familiarity with moment of inertia calculations
  • Knowledge of angular acceleration and its relation to torque
  • Basic grasp of harmonic motion principles
NEXT STEPS
  • Study the derivation of the moment of inertia for various shapes, focusing on disks
  • Learn about the relationship between torque, angular acceleration, and period in oscillatory systems
  • Explore the effects of damping on the period of oscillation in physical systems
  • Investigate the application of the small-angle approximation in pendulum motion
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Physics students, mechanical engineers, and anyone interested in rotational dynamics and harmonic motion analysis will benefit from this discussion.

boardaddict
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Homework Statement



A disk of mass m and radius a is suspended horizontally by a rod bored through the disk at a location d distance from the center. When in equilibrium the center of the disk is directly below the rod. Obtain an approximate expression for the period, ignore disapative effects

Homework Equations



per Newtonian

I = 1/2 mr^2 + md^2
I*theta: = -mg(rsin(theta))


The Attempt at a Solution


1/2 ma^2 +md^2

Im missing something simple here...If someone could point me in the right direction it would be appreciated
 
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boardaddict said:

Homework Statement



A disk of mass m and radius a is suspended horizontally by a rod bored through the disk at a location d distance from the center. When in equilibrium the center of the disk is directly below the rod. Obtain an approximate expression for the period, ignore disapative effects

Homework Equations



per Newtonian

I = 1/2 mr^2 + md^2
I*theta: = -mg(rsin(theta))


The Attempt at a Solution


1/2 ma^2 +md^2

Im missing something simple here...If someone could point me in the right direction it would be appreciated

Wait, can I use alpha= to T/I , then relate that to 2pi to find period?
 
boardaddict said:
Wait, can I use alpha= to T/I , then relate that to 2pi to find period?

If I did this right I get ;

pi((a^2)+d^2)/g*d

If someone would confirm that would be great
 

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