How to Calculate Moment of Inertia for Different Door Rotations?

[tangibleLime]

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

$$I = \Sigma m_i*r^{2}_i$$
$$I_{parallel axis} = I_{cm} + md^{2}$$

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: $$\frac{1}{3}Ma^2 = \frac{1}{3}(22)(.93)^2 = 6.2 kg * m^2$$.

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: $$\frac{1}{12}Ma^2$$. My resulting formula ended up as this: $$I_{parallel axis} = (\frac{1}{12}(22)(.92)^2) + ((22)(.17^2))$$, which results in $$2.2 kg * m^2$$, but it was determined to be incorrect.

What am I doing wrong here? Any help would be greatly appreciated.

 

[tiny-tim]

Hi tangibleLime!

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.

 

[tangibleLime]

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 $$I_{parallel axis} = (\frac{1}{3}(22)(.92)^2) + ((22)(.17^2))$$, 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 $$6.8$$, 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.

 

[tiny-tim]

Hi tangibleLime!

(just got up :zzz: …)

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 I_cm + md² must be from the centre of mass.

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

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

and 1/12 + 1/4 = 4/12 = 1/3.

 

[rdl]

I would first like to commend you on your attempt at solving this problem. Moment of inertia is an important concept in rotational dynamics and it is essential to understand it in order to solve problems involving rotation.

For part b, you are correct in using the parallel axis theorem. However, the equation you used for a plane with the axis of rotation around its center, \frac{1}{12}Ma^2, is not applicable here as the axis of rotation is not at the center of the door. Instead, you should use the equation for a thin rod with the axis of rotation at one end, \frac{1}{3}ML^2.

Using this equation, the moment of inertia for rotation about a vertical axis inside the door, 17 cm from one edge, would be:

I_{parallel axis} = (\frac{1}{3}(22)(.92)^2) + (22)(.17)^2) = 7.26 kg * m^2.

I hope this helps clarify your understanding of the parallel axis theorem and its application in solving rotational problems. Keep up the good work!