Period of oscillations of the disk

AI Thread Summary
The discussion focuses on calculating the period of oscillations for a circular disk pivoting about a point on its circumference. The initial approach incorrectly applies uniform circular motion equations, leading to confusion about the moment of inertia. Participants clarify that the parallel axis theorem is necessary to find the correct moment of inertia for the disk in this scenario. It is emphasized that the disk does not undergo uniform circular motion due to varying angular velocity during oscillations. The correct formula for the period of oscillation is T = 2π√(R/g) after accounting for the disk's pivot point.
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Homework Statement



A circular disk of radius R and uniform density is free to pivot about a fixed point P on its circumference. Calculate the period of oscillations of the disk, in the plane of Figure I, when it is displaced by a small angle about its pivot and released.


Homework Equations



F=ma
a=v^2/R
v=2*R*pi/T

The Attempt at a Solution



F=mv^2/R
mg=mv^2/R
g=v^2/R

where v= 2*R*pi*/T

thus, T= 2*pi*root(R/g)

can anyone advise?
 

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I am afraid its not correct. You have to calculate the new moment of inertia. Use parallel axis theorem to help you.
 
And the problem here is, this is not a uniform circular motion.

What is the characteristic of a uniform circular motion? And how does it NOT fit into this situation?
 
why this is not a uniform circular motion?
is it because the disk is not moving in constant speed?
 
You are right. Because it is not moving at constant linear velocity. So angular velocity is always changing too, v=rw.
 

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