I'm not understanding the very basics of Moment of Inertia

Click For Summary

Homework Help Overview

The discussion revolves around the concept of moment of inertia, specifically calculating it for a CD with a given mass and diameter. The original poster expresses confusion regarding the basic principles and calculations involved in determining the moment of inertia for rotation about a perpendicular axis through the center of the CD.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the moment of inertia formula but struggles with the integration concept and basic calculations. They express a need for foundational understanding and guidance. Other participants provide insights into the nature of moment of inertia and suggest using tables for common shapes, while also addressing the importance of the axis of rotation.

Discussion Status

Participants are exploring the topic with some offering clarifications and additional context. The original poster has made progress in understanding the concept but still seeks further clarification on variations in the axis of rotation. There is an acknowledgment of a calculation error, but the discussion remains open to further questions.

Contextual Notes

The original poster notes a lack of foundational knowledge in calculus, which may be impacting their understanding of the problem. They also mention that their calculus class is not aligning well with their physics coursework.

TA1068
Messages
13
Reaction score
0

Homework Statement


A 13cm diameter CD has a mass of 25 g.
What is the CD's moment of inertia for rotation about a perpendicular axis through its center?

Homework Equations


[tex]I = \int r^2 dm[/tex]

and possibly...

Moment of inertia of cylinder or disk, about center = [tex]\frac{1}{2}MR^2[/tex]


The Attempt at a Solution


I'm given certain values. The mass of the CD is .025 kg and it's radius is .065 m. Other than this, I am very confused. I tried just plugging in (.025)(.065)^2, but that didn't work.

Now I've checked through other threads regarding moment of inertia, but it's not exactly helpful unless I know the very basics. In all honesty, I've never even done integrals before; my calculus class is supposed to parallel what we are doing in physics, but they are doing a very poor job at it. I think I might have the hang of them from looking it up quickly online, though (I'm just a pain, huh?).

I know with such a very limited amount to work with here it can be difficult, but I catch on really quickly (I swear!), but for some reason this is a little tricky. Any pointers in the right direction would be greatly appreciated!
 
Physics news on Phys.org
Wikipedia is your friend.

The moment of inertia represents an object's resistance to angular acceleration about an axis. If you have two wheels of the same dimensions, but one is made out of lead and the other out of styrofoam, and you apply the same moment to each (by, for example, pushing on both with the same force at the same point), the lead wheel will spin much slower. This is because it has a much larger moment of inertia.

It is very important to keep track of what axis you are rotating the object around, as the same object will have a different moment of inertia depending on which axis it is rotating about. If you think about it, this makes sense--it's much easier to spin a dumbbell along the bar's axis than it is to spin it end-over-end.edit: Also, since doing that integral you posted is usually a big pain, there are tables of moments of inertia for commonly-shaped objects. Using these tables along with the Parallel Axis Theorem usually let's you calculate moments without having to resort to ugly integrals.

Also, for the problem you posted, you forgot to multiply by 1/2
 
Last edited:
Also, for the problem you posted, you forgot to multiply by 1/2

That I did. So I have the correct answer now, and I'm pretty sold on the concept of Moment of Inertia. I had already read up a bit on it, but it was a simple calculation error that had me thrown off the whole time.

What if the axis is at the edge of the disk rather than the center of the disk?
 
TA1068 said:
What if the axis is at the edge of the disk rather than the center of the disk?

That's where the Parallel Axis Theorem comes into play.
 
Got it. Thanks.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 13 ·
Replies
13
Views
3K
Replies
11
Views
4K
Replies
25
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
8
Views
2K
Replies
5
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
24
Views
4K