Moment of Inertia for a door

In summary, the moment of inertia for a 22 kg solid door of dimensions 220 cm x 92 cm is 6.2 kg*m^2 for rotation on its hinges. For rotation about a vertical axis inside the door, 17 cm from one edge, the correct calculation using the parallel axis theorem is I = (1/12)(22)(.92)^2 + (22)(.17)^2 = 2.2 kg*m^2. The mistake in the incorrect answer of 6.8 kg*m^2 was using the distance from the center of mass instead of the axis of rotation. The coefficients 1/3 and 1/12 come from the parallel axis theorem and the distance from the center
  • #1
tangibleLime
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0
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.
 
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  • #2
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:
 
  • #3
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.
 
  • #4
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:
 
  • #5


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!
 

What is Moment of Inertia for a door?

Moment of Inertia for a door is a physical property that describes the distribution of mass and how it is located in relation to the axis of rotation of the door. It is a measure of the door's resistance to changes in its rotation.

How is Moment of Inertia for a door calculated?

Moment of Inertia for a door can be calculated by multiplying the mass of the door by the square of the distance from the axis of rotation to the center of mass of the door.

Why is Moment of Inertia important for a door?

Moment of Inertia is important for a door because it determines the amount of torque needed to rotate the door and the door's angular acceleration. It also affects the door's stability and how easily it can be opened or closed.

How does the shape of a door affect its Moment of Inertia?

The shape of a door can greatly affect its Moment of Inertia. For example, a door with a wider width or longer height will have a larger Moment of Inertia compared to a door with a smaller width or height, assuming they have the same mass. This is because the distance from the axis of rotation to the center of mass is larger for the wider or taller door.

Can Moment of Inertia be changed for a door?

Yes, Moment of Inertia for a door can be changed by altering the distribution of mass or the door's shape. For example, adding weight to one side of the door will increase its Moment of Inertia, while making the door thinner will decrease it. This can have an impact on the door's overall performance and stability.

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