How to Calculate Moment of Inertia for Different Door Rotations?

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Homework Help Overview

The discussion revolves around calculating the moment of inertia for a solid door with specific dimensions and mass, focusing on two different axes of rotation: one at the hinges and another 17 cm from one edge. Participants are exploring the application of relevant equations and the implications of the parallel axis theorem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the moment of inertia for two scenarios using established formulas but encounters difficulties with the second part. Some participants question the application of the parallel axis theorem and the coefficients used in the calculations.

Discussion Status

Participants are actively discussing the calculations and the reasoning behind the coefficients for different shapes. There is a recognition of confusion regarding the application of the formulas, and some guidance is offered about the center of mass and the coefficients involved.

Contextual Notes

There is an indication that the original poster may have misapplied the distance from the center of mass in their calculations. The discussion includes clarifications about the definitions and derivations of the coefficients used in the moment of inertia formulas.

tangibleLime
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Moment of Inertia - Unsolved

Homework Statement


A 22 kg solid door is 220 cm tall, 92 cm wide.
a) What is the door's moment of inertia for rotation on its hinges?
b) What is the door's moment of inertia for rotation about a vertical axis inside the door, 17 cm from one edge?


Homework Equations


[tex]I = \Sigma m_i*r^{2}_i[/tex]
[tex]I_{parallel axis} = I_{cm} + md^{2}[/tex]

The Attempt at a Solution


I was able to get part a by using a given formula for a plane with an axis of rotation around it's edge: [tex]\frac{1}{3}Ma^2 = \frac{1}{3}(22)(.93)^2 = 6.2 kg * m^2[/tex].

For part b, I cannot seem to get the correct answer. I applied the parallel axis theorem, which I understand to be the second equation I listed above. From a given list, I used the equation for a plane with the axis of rotation around it's center: [tex]\frac{1}{12}Ma^2[/tex]. My resulting formula ended up as this: [tex]I_{parallel axis} = (\frac{1}{12}(22)(.92)^2) + ((22)(.17^2))[/tex], which results in [tex]2.2 kg * m^2[/tex], but it was determined to be incorrect.

What am I doing wrong here? Any help would be greatly appreciated.
 
Last edited:
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Hi tangibleLime! :smile:
tangibleLime said:
b) What is the door's moment of inertia for rotation about a vertical axis inside the door, 17 cm from one edge?

You've used 17 cm from the centre of mass. :wink:
 
Hm, I guess I'm a little bit confused as to how these coefficients are calculated (the 1/3, 1/12, etc..).

I tried to use [tex]I_{parallel axis} = (\frac{1}{3}(22)(.92)^2) + ((22)(.17^2))[/tex], which I think would be taking the moment of inertia from the edge, as in part a, and moving it 17cm to the right. This results in [tex]6.8[/tex], which is apparently incorrect.

I assume that my final answer is going to be less than 6.2 (the answer from part a) since the axis of rotation is being moved towards the center. That being said, the answer of 6.8 doesn't make sense anyways.
 
Hi tangibleLime! :smile:

(just got up :zzz: …)
tangibleLime said:
Hm, I guess I'm a little bit confused as to how these coefficients are calculated (the 1/3, 1/12, etc..).

The "d" in Icm + md2 must be from the centre of mass.

That's where the 1/3 for the end of a rod or plate comes from …

Icm for a rod or plate has 1/12, you add d2 = 1/4,

and 1/12 + 1/4 = 4/12 = 1/3. :wink:
 

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