Period of Small Amplitude Oscillation for Hoop of Radius 50cm

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SUMMARY

The period of small amplitude oscillation for a circular hoop of radius 50 cm is determined using the formula T = 2π√(I/(MgD)). In this scenario, the mass (M) of the hoop and rod is not provided, complicating the calculation. The distance (D) is defined as the distance from the center of mass to the pivot point. The moment of inertia (I) must account for both the hoop and the rod, necessitating further information about the rod's length for accurate calculations.

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Homework Statement


"A circular hoop of radiusm 50 cm is hung on a narrow horizontal rod and allowed to swing in the plane of the hoop. What is the period of its oscillation, assuming that the amplitude is small?

Homework Equations



T= 2*pi(I/(MgD))^.5

The Attempt at a Solution



Ok... so if M is not given for the hoop and the rod, how do I go about figured it out? What is the value of D? And do I factor the moment of inertia for both the rod and hoop, and if so... how do I figure out the length of the rod, if it is not given.

Thanks.
 
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Find the moment of inertia- it will have a factor of M in it.
I believe D is the distance from the center of mass to the pivot point.
Is the rod rotating?
 
robb_ said:
Find the moment of inertia- it will have a factor of M in it.
I believe D is the distance from the center of mass to the pivot point.
Is the rod rotating?

Yes.. it is swinging from the hoop
 

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