What Is the Period of Oscillation for a Hoop Suspended by Its Perimeter?

AI Thread Summary
The discussion focuses on calculating the period of oscillation for a hoop suspended by its perimeter, with given parameters of radius 0.18 m and mass 0.44 kg. The moment of inertia is calculated as I = 2(MR^2), resulting in I = 0.028512 kg·m². The relationship between potential energy and rotational kinetic energy is established, leading to the equation mgh = 1/2Iw². By substituting values and solving for angular frequency, the period of oscillation is determined to be approximately T = 0.60213 seconds. The calculations demonstrate the application of physics principles to find the period of oscillation for the given system.
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Homework Statement



a hoop of radius R=.18m and mass=.44kg is suspended by a point on its perimeter. If the hoop is allowed to oscillate side to side, what is the period of oscillation?

Homework Equations



I=MR^2 plus an offset? MR^2
so I=2(MR^2)?
PE=1/2kx^2
w=(2pi)/T
KErot=1/2Iw^2

The Attempt at a Solution


I=2(MR^2)
I=2(.44)(.18^2)
I=.028512
PEi=KErotfinal
mgh=1/2Iw^2
(.44)(9.8)(.36)=1/29.o28512)w^2
10.43498=w
10.43498=(2pi)/T
T=.60213s
 
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find I and plug it into T=2pi(sqrtof I/(mgx)?
 

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