Moment of inertia about center of mass with four holes?

[tshah321]

Homework Statement

What is the moment of inertia of the 4 kg disk below when it rotates about the center of mass? The mass of the disk without the holes is 6 kg

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Homework Equations

Center of mass equation: (1/M)(∫xdm)
Moment of inertia of a disk: I = 0.5mr²
Parallel axis theorem: I = Io + md²

The Attempt at a Solution

I tried to find the moment of inertia of the disk without the holes using m = 6 kg and r = 0.12 m.

I = 0.5(6 kg)(0.12 m)² = 0.0432 kgm²

I think I'm supposed to subtract each hole's moment of inertia from 0.0432, but am unsure how to find each circle's radius. Is it simply r = 0.05m? Or is it 0.06m? Or is my line of thinking totally off?

 

[SteamKing]

Homework Statement

What is the moment of inertia of the 4 kg disk below when it rotates about the center of mass? The mass of the disk without the holes is 6 kg

.View attachment 92585

Homework Equations

Center of mass equation: (1/M)(∫xdm)
Moment of inertia of a disk: I = 0.5mr²
Parallel axis theorem: I = Io + md²

The Attempt at a Solution

I tried to find the moment of inertia of the disk without the holes using m = 6 kg and r = 0.12 m.

I = 0.5(6 kg)(0.12 m)² = 0.0432 kgm²

I think I'm supposed to subtract each hole's moment of inertia from 0.0432, but am unsure how to find each circle's radius. Is it simply r = 0.05m? Or is it 0.06m? Or is my line of thinking totally off?

The figure shows that the diameter of one of the circular cutouts is 6 cm. What's the radius of the circle then?

BTW, the 5 cm radius represents an imaginary circle, presumably on which the centers of the cutouts are located.

 

[tshah321]

Definitely missed that. So r = 0.03m, and then the mass of each is 0.5kg.

The moment of inertia of each is I = mr², so I = (0.5)(0.03)² = 0.00045 kgm²

So I would just multiply that by 4, then subtract it from 0.0432? @SteamKing

 

[SteamKing]

Definitely missed that. So r = 0.03m, and then the mass of each is 0.5kg.

You might want to check that mass for each cutout.

The moment of inertia of each is I = mr², so I = (0.5)(0.03)² = 0.00045 kgm²

The MOI of each cutout is I = mr² about its own centroid.

So I would just multiply that by 4, then subtract it from 0.0432? @SteamKing

How do you calculate the MOI when it is not taken about the centroid of the body?

 

[tshah321]

You might want to check that mass for each cutout.

The mass without the cutouts of the entire disk is 6kg, and with the cutouts it is 4kg. So the mass of all 4 cutouts is 2kg, so each cutout is 2kg/4 = 0.5kg. Is there something wrong there?

The MOI of each cutout is I = mr² about its own centroid.

Yes, so I believe I did that right, unless I missed something when finding the mass of each cutout.

How do you calculate the MOI when it is not taken about the centroid of the body?

I was under the impression that the center of mass of the disk was here: http://s7.postimg.org/hbvdnoygb/screenshot_54.png
So I would use the moment of inertia of the whole disk without the holes, 0.0432 kgm² and then take the moment of inertia of each cutout and subtract it from the total moment of inertia.

 

[SteamKing]

The mass without the cutouts of the entire disk is 6kg, and with the cutouts it is 4kg. So the mass of all 4 cutouts is 2kg, so each cutout is 2kg/4 = 0.5kg. Is there something wrong there?

Yes, so I believe I did that right, unless I missed something when finding the mass of each cutout.

I missed the part of the OP which said that the mass w/cutouts was 4 kg.

I was under the impression that the center of mass of the disk was here: http://s7.postimg.org/hbvdnoygb/screenshot_54.png
So I would use the moment of inertia of the whole disk without the holes, 0.0432 kgm² and then take the moment of inertia of each cutout and subtract it from the total moment of inertia.

The center of mass is indeed where you have marked it in the image. But this location is not the same as the centroid for each individual cutout.

The formula I = mr² is applicable only when calculating the MOI of a circular disk about its own center. When the center of the cutout and the center of the disk do not coincide, what else must you do to calculate the correct MOI? (Hint: Parallel Axis Theorem).

 

[tshah321]

The center of mass is indeed where you have marked it in the image. But this location is not the same as the centroid for each individual cutout.

The formula I = mr² is applicable only when calculating the MOI of a circular disk about its own center. When the center of the cutout and the center of the disk do not coincide, what else must you do to calculate the correct MOI? (Hint: Parallel Axis Theorem).

I don't understand, though. Doesn't each cutout have its own center of mass, therefore making it okay to use I = mr²?

 

[SteamKing]

I don't understand, though. Doesn't each cutout have its own center of mass, therefore making it okay to use I = mr²?

Yes, but the problem is asking you to calculate the MOI of the large disk which contains the 4 cutouts. The center of that large disk and the center of each of the four cutouts is not in the same location. Simply subtracting the MOI of each cutout from the MOI of the large disk does not lead to the correct result.

BTW, did you examine what the parallel axis theorem is used for?

 

[SammyS]

Homework Statement

What is the moment of inertia of the 4 kg disk below when it rotates about the center of mass? The mass of the disk without the holes is 6 kg

.View attachment 92585

Apparent INCONSISTENCY in the problem as stated.

If it's assumed that the original disk has uniform density and thickness, then there appears to be an inconsistency in the parameters given for the disk with the cutouts.

The total area of the disk before cutouts being removed: 144π cm² .

The total area of the cutouts: 4×9π = 36π cm².

Therefore, removing the cutout areas reduces the overall area by 1/4 .

According to the masses given, 1/3 of the material is removed.

This won't be important for solving the problem, but it would be important if you compare the moment of inertia of this disk with cutouts to the moment of inertia of a solid disk having the same mass and radius as the cutout disk.

 

[SteamKing]

Apparent INCONSISTENCY in the problem as stated.

If it's assumed that the original disk has uniform density and thickness, then there appears to be an inconsistency in the parameters given for the disk with the cutouts.

The total area of the disk before cutouts being removed: 144π cm² .

The total area of the cutouts: 4×9π = 36π cm².

Therefore, removing the cutout areas reduces the overall area by 1/4 .

According to the masses given, 1/3 of the material is removed.

This won't be important for solving the problem, but it would be important if you compare the moment of inertia of this disk with cutouts to the moment of inertia of a solid disk having the same mass and radius as the cutout disk.

You are certainly right about this point, SammyS.