Rotational Inertia Problem-PLEASE HELP

[dari09]

Rotational Inertia Problem--PLEASE HELP!

Homework Statement

Figure 12-39 shows a rigid structure consisting of a circular hoop, of radius R and mass m, and a square made of four thin bars, each of length R and mass m. The rigid structure rotates at a constant speed about a vertical axis with a period of rotation of 1.0 s. Assuming R = 0.50 m and m = 3.0 kg, calculate the structure's rotational inertia about the axis of rotation

The figure is attached

Homework Equations

I of point mass = mr^2
I of hoop = mr^2 or .5mr^2
I of rod = 1/3 mr^2

The Attempt at a Solution

I tried to add the I of the hoop and two of the rods together (because the other two rods are parallel to the rotating axis) but nothing that I've tried really worked...

 

[keltix]

http://panda.unm.edu/Courses/Price/Phys160/p27-2.pdf"

 

[Moetasim]

The rotational inertia of a rigid body is a measure of its resistance to changes in rotational motion. In this problem, we are given a circular hoop and a square made of four thin bars, rotating at a constant speed about a vertical axis. The rotational inertia of the entire structure can be calculated by adding the individual inertias of each component.

Using the given equations, we can calculate the rotational inertia of the hoop as I = 0.5mr^2 = 0.5(3.0 kg)(0.50 m)^2 = 0.375 kgm^2. However, the rotational inertia of the square cannot be calculated using the same equation, as it is not a point mass. Instead, we need to use the parallel axis theorem, which states that the rotational inertia of an object about an axis is equal to the sum of its rotational inertia about a parallel axis passing through its center of mass and the product of its mass and the square of the distance between the two axes.

Applying this theorem, we can calculate the rotational inertia of each rod about the axis of rotation, which is also the center of the hoop. The distance between the center of the hoop and the center of each rod is R = 0.50 m. Therefore, the rotational inertia of each rod is I = (1/3)mR^2 + mR^2 = (1/3)(3.0 kg)(0.50 m)^2 + (3.0 kg)(0.50 m)^2 = 0.583 kgm^2.

Since there are four rods in total, the total rotational inertia of the square is 4(0.583 kgm^2) = 2.332 kgm^2. Adding this to the rotational inertia of the hoop, we get the total rotational inertia of the structure as I = 0.375 kgm^2 + 2.332 kgm^2 = 2.707 kgm^2.

Therefore, the rotational inertia of the rigid structure about the axis of rotation is 2.707 kgm^2. I hope this helps you solve the problem. Remember to always consider the parallel axis theorem when calculating the rotational inertia of non-point masses.