Homework Help: Moment of Inertia for three spherical masses

1. Apr 9, 2010

MissPenguins

1. The problem statement, all variables and given/known data

2) Three spherical masses are located in a plane at the positions shown in the ﬁgure below. A
has mass 39.3 kg, B has mass 35.9 kg, and C has mass 17.6 kg.

Calculate the moment of inertia (of the three masses) with respect to the z-axis perpendicular to the xy plane and passing through the origin. Assume the masses are point particles; e.g., neglect the contribution due to moments of inertia about their center
of mass. Answer in units of kg X m.

2. Relevant equations
I = sum of mr2
I = (2/5)mr2

3. The attempt at a solution
1. The problem statement, all variables and given/known data
(39.3)(0.52) = 9.825 kg$$\dot{}.$$m2
(35.9)(0.52) = 8.975 kg$$\dot{}.$$m2
(17.6)(0.52) = 4.4 kg$$\dot{}.$$m2
(9.825+8.975+4.4) = 23.2

I = (2/5)mr2
I = (2/5)9.8252 = 3.93
I = (2/5)8.9752=3.59
I = (2/5)4.42=1.76
Add them all up, I got 9.28 kg$$\dot{}.$$m2

AND I got it WRONG. ;(
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 9, 2010

Staff: Mentor

What's the distance of each mass to the axis? You have them all set to 0.5. Why?

You were told to "neglect the contribution due to moments of inertia about their center

3. Apr 9, 2010

MissPenguins

What do you mean treat them as point masses? How do I do that? Thanks.

4. Apr 9, 2010

Use this:
Not this:

5. Apr 9, 2010

MissPenguins

so my answer would just be 23.2? What's wrong with my radius? Isn't it 1/2 = 0.5? Thanks.

6. Apr 9, 2010

Staff: Mentor

No.
r is the distance from the axis for each mass. Where do you get 0.5 from?

7. Apr 9, 2010

Mantello

Would you use the Distance formula to find the radius between the Z Axis and the Point?

8. Apr 9, 2010

nazerofsun

don't know if this is right but have you thought of finding the center of mass then from that (x,y) coordinate use the parallel axis theorem Inew= Icm+h^2*(total mass) where h is the distance between the new axis of rotation and the center of mass.

but then again it did say, "neglect the contribution due to moments of inertia about their center of mass"
and the problem says to treat the particles as point masses so no need for the sphere thing,

so then basically, sumI= mr^2+m2r2^2+....

9. Apr 9, 2010

Mantello

I don't think you need to worry about using the parallel axis theorem, since there wasn't an actual change of axis. I think you second part of your answer is correct, because you are just finding the moment of inertia of the system.

10. Apr 9, 2010

nazerofsun

agreed :) just make sure for the r to use the sqrt(x^2+y^2) for the radius since the question would be to rotate everything around the origin, so they will be following a constant radius.

11. Apr 10, 2010

MissPenguins

I did d or r = sqrt of x2+y2 for A, B, and C
A = sqrt of -22+2.52 = 3.202
B = sqrt of 12+.52 = 1.118
C = sqrt of 4.52+22 = 4.924
then I used
I = sum of mv2 = m1r1+ m2r2 =m3r3
so I added them all up and got 874.50
Is this reasonable? Am I doing it right?
Thanks guys. ;)

12. Apr 10, 2010

Staff: Mentor

Good!
Too many typos here. I assume you did I = Σmr².
Looks like you did the right thing.

13. Apr 10, 2010

MissPenguins

Re: [SOLVED]Moment of Inertia

opps, sorry for the typos. YAY, I got it right.