Determine period for off center disk

AI Thread Summary
The discussion focuses on determining the period of a horizontally suspended disk with a rod bored through it at a distance from the center. The moment of inertia is calculated using the formula I = 1/2 ma^2 + md^2. A participant suggests using the relationship between angular acceleration (alpha), torque (T), and moment of inertia (I) to derive the period. They propose that the period can be approximated as π√((a^2 + d^2)/g*d). Confirmation of this expression is requested from others in the discussion.
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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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