Question

The moment of inertia of a thin ring of mass ????M and radius ????R about its symmetry axis is ????????????=????????2ICM=MR2. Kira is working the ring‑toss booth at a local carnival. While waiting for customers, Kira occupies her time by twirling one of the plastic rings of mass ????M and radius ????R about her finger. Model the motion of the plastic ring as a thin ring rotating about a point on its circumference. What is the moment of inertia of the plastic ring?

Answer 1

The moment of inertia of a thin ring about an axis through its center of mass (center of the ring) is given by:

$I_{CM}$ = MR²

$I_{CM}$ = moment of inertia about center of mass, M = total mass, R = radius of ring

We want to find the moment of inertia of the ring about an axis (parallel to the axis running through the center of mass) running through any point on its circumference. We'll use the parallel axis theorem to find this quantity:

I = $I_{CM}$ + Md²

d = distance between center of mass axis and the parallel axis

Known values:

$I_{CM}$ = MR²

d = R

Plug in the known values and solve for I:

I = MR² + MR²

I = 2MR²

Answer 2

The moment of inertia of the plastic ring will be after calculation I = 2MR²

What is moment of inertia?

The moment of inertia of a thin ring about an axis through its center of mass (center of the ring) is given by:

$I_{cm}= MR^2$

$I_{cm}$= moment of inertia about center of mass,

M = total mass,

R = radius of ring

We want to find the moment of inertia of the ring about an axis (parallel to the axis running through the center of mass) running through any point on its circumference. We'll use the parallel axis theorem to find this quantity:

$I = I_{cm}+ Md^2$

d = distance between center of mass axis and the parallel axis

Known values:

$I_{cm}= MR^2$

d = R

Plug in the known values and solve for I:

I = MR² + MR²

I = 2MR²

Thus the moment of inertia of the plastic ring will be after calculation I = 2MR²

To know more about moment of inertia follow

https://brainly.com/question/14015345