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

[writelu]

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

 

[writelu]

find I and plug it into T=2pi(sqrtof I/(mgx)?

 

[kerimek]

I would like to first point out that the equation for the moment of inertia of a hoop is actually I=MR^2, without the additional offset term. This error may have led to the incorrect calculation of the moment of inertia in the attempted solution.

To find the period of oscillation, we can use the equation T=2pi*sqrt(I/mgh), where I is the moment of inertia, m is the mass, g is the acceleration due to gravity, and h is the distance from the point of suspension to the center of mass. In this case, h=R/2.

Substituting the given values, we get T=2pi*sqrt(2(.44)(.18^2)/(.44)(9.8)(.18/2))=0.596s.

Therefore, the period of oscillation for the given physical pendulum hoop is approximately 0.596 seconds.