Moment of Inertia Questions

[jjiimmyy101]

**The problem and solution is attached as moment of inertia.**
**The equations for moment of inertia of known homogeneous solids is also attached**

I'm having some trouble grasping this concept. I hope someone can help clarify some things for me.

Looking at the problem I realized I would have to use the parallel axis theorem, but after that the trouble begins. (I = Ig + m*d^2)

"Deterimine the wheel's M of I about an axis perpendicular to the page and passing thru pt A."

What does this mean? What is it saying?

And a related question to the above: Why when calculating M of I for the thin ring (Homogeneous solids jpeg) do you use Izz = m*r^2 instead of Ixx or Iyy? Same question for the slender rod. How do you know which one to use?

I realize I've said a lot, but I hope someone can give me some pointers.
Thanks.

 

[vsage]

Have you identified what the axis passing through A looks like?

The axis A appears parallel to the z axis judging from the orientation of the second picture so yes you are correct. If you look at the orienation of the rods they are perpendicular to the A axis so judging from the second picture again you should use the y M of I.

My personal opinion is that for simple objects like that it is easier to derive the M of I so perhaps just remembering that the M of I is in integral form (r^2dm)

 

[wais]

Hi there,

I can definitely understand your confusion with the concept of moment of inertia. It can be a bit tricky to grasp at first, but with some practice and understanding of the equations, it will become much clearer.

First, let's address the problem at hand. The problem is asking you to find the moment of inertia of a wheel about an axis that is perpendicular to the page and passing through point A. Moment of inertia is a measure of an object's resistance to changes in its rotational motion. In simple terms, it is a measure of how difficult it is to make an object rotate. In this case, the wheel is rotating about an axis that is perpendicular to the page, which means it is rotating in a plane that is parallel to the page. Point A is simply a point on that axis. So, the problem is asking you to find the moment of inertia of the wheel about this specific axis.

Now, to answer your second question about the thin ring and slender rod, the equations for moment of inertia of known homogeneous solids (such as a thin ring or slender rod) are specific to the shape of the object. For example, the equation for a thin ring is Izz = m*r^2, where m is the mass of the ring and r is the radius of the ring. This equation only works for a thin ring, and not for other shapes. Similarly, the equation for a slender rod is Izz = m*l^2/12, where m is the mass of the rod and l is the length of the rod. This equation only works for a slender rod.

To know which moment of inertia equation to use, you have to understand the shape of the object and which equation applies to it. For example, if you have a thin ring, you would use the equation Izz = m*r^2, but if you have a solid cylinder, you would use the equation Izz = m*r^2/2. It all depends on the shape of the object.

I hope this helps clarify some of your confusion. Keep practicing and don't hesitate to ask for help if you need it. Best of luck!