Another moment of inertia problem. help

[elsternj]

Homework Statement

Find the moment of inertia I_x of particle a with respect to the x-axis (that is, if the x-axis is the axis of rotation), the moment of inertia I_y of particle a with respect to the y axis, and the moment of inertia I_z of particle a with respect to the z axis (the axis that passes through the origin perpendicular to both the x and y axes).
Express your answers in terms of and separated by commas.

Homework Equations

I = mr²

The Attempt at a Solution

Well the x-axis as the axis of rotation was the only one i got correctly in which the answer was mr²

The answer to the y-axis is 9mr²
The answer to the z axis is 10mr² ... can someone explain to me why?

To me, it looks like the distance of particle a from the y-axis is 3r... where is the 9 coming from? As for the z axis, can someone explain that to me also?

 

[Doc Al]

To me, it looks like the distance of particle a from the y-axis is 3r... where is the 9 coming from?

In computing the moment of inertia, the distance is squared. The general formula for point masses is I = mD², where D is the distance to the axis.

The same principle applies to each case.

 

[elsternj]

so it gets squared twice? because i thought that by having 3mr² that took care of the squaring. I must have been wrong somewhere here. I have gotten answers to other moment of inertia problems without squaring my distance. When do I know to do this and when not to?

For an example in this problem:
Small blocks, each with mass m , are clamped at the ends and at the center of a rod of length L and negligible mass.

Compute the moment of inertia of the system about an axis perpendicular to the rod and passing through the center of the rod.
Express your answer in terms of the given quantities.

my answer was I = m(.5L)²+m(.5L)² and this answer was RIGHT. I did not square those distances.

 

[Doc Al]

so it gets squared twice?

The distance gets squared once.

because i thought that by having 3mr² that took care of the squaring. I must have been wrong somewhere here. I have gotten answers to other moment of inertia problems without squaring my distance. When do I know to do this and when not to?

You always square the distance.

For an example in this problem:
Small blocks, each with mass m , are clamped at the ends and at the center of a rod of length L and negligible mass.

Compute the moment of inertia of the system about an axis perpendicular to the rod and passing through the center of the rod.
Express your answer in terms of the given quantities.

my answer was I = m(.5L)²+m(.5L)² and this answer was RIGHT. I did not square those distances.

What do you mean you didn't square the distances? Sure you did! Look at your equation. The distance is (.5L), which you squared to get (.5L)².

 

[elsternj]

okay let me try and explain my confusion the best I can. The distance in that example problem was half the length, (.5L) so yes, I did square it and made it (.5L)². Now the distance in the problem from the original post is 3r. so i made it 3mr². I also squared the distance here. If the right answer is 9mr² then the 3 is squared and it STILL is squared in the final answer even after the 3 gets squared. Do you see what I mean?

 

[cepheid]

okay let me try and explain my confusion the best I can. The distance in that example problem was half the length, (.5L) so yes, I did square it and made it (.5L)². Now the distance in the problem from the original post is 3r. so i made it 3mr². I also squared the distance here. If the right answer is 9mr² then the 3 is squared and it STILL is squared in the final answer even after the 3 gets squared. Do you see what I mean?

Not really. You're not making any sense. The general formula for the moment of inertia of a point mass is md² where d is the perpendicular distance to the axis of rotation. In the case of rotation about the y-axis, the distance is d = 3r. Hence

I = md² = m(3r)² = 9mr²

 

[cepheid]

As for the moment of inertia for rotation around the z-axis, consider this. If the point rotates around the z-axis, then it basically traces a circle around the origin in the x-y plane. What is the distance of point a from the origin?

 

[Doc Al]

okay let me try and explain my confusion the best I can. The distance in that example problem was half the length, (.5L) so yes, I did square it and made it (.5L)². Now the distance in the problem from the original post is 3r. so i made it 3mr². I also squared the distance here.

The distance is 3r, not just r. So to square the distance means (3r)² = 9r².

If the right answer is 9mr² then the 3 is squared and it STILL is squared in the final answer even after the 3 gets squared. Do you see what I mean?

No.