Moment of Inertia for three spherical masses

AI Thread Summary
To calculate the moment of inertia for three spherical masses located in a plane, the correct approach involves treating them as point masses and using the formula I = Σmr². The initial calculations mistakenly set all distances to 0.5, which led to incorrect results. The accurate distances from the z-axis for each mass must be determined using the distance formula, resulting in values such as 3.202 for mass A, 1.118 for mass B, and 4.924 for mass C. After correctly applying the moment of inertia formula with these distances, the final result should reflect the sum of the individual contributions. The discussion confirms that using the point mass approximation is appropriate, and the calculations can yield a reasonable moment of inertia.
MissPenguins
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Homework Statement



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.
a5.jpg

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.

Homework Equations


I = sum of mr2
I = (2/5)mr2

The Attempt at a Solution


Homework Statement


(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. ;(
PLEASE HELP!
 
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MissPenguins said:
(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
What's the distance of each mass to the axis? You have them all set to 0.5. Why?


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
You were told to "neglect the contribution due to moments of inertia about their center
of mass". So forget about this and just treat them as point masses.
 
Doc Al said:
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
of mass". So forget about this and just treat them as point masses.

What do you mean treat them as point masses? How do I do that? Thanks.
 
Use this:
MissPenguins said:
I = sum of mr2

Not this:
I = (2/5)mr2
 
Doc Al said:
Use this:

Not this:

so my answer would just be 23.2? What's wrong with my radius? Isn't it 1/2 = 0.5? Thanks.
 
MissPenguins said:
so my answer would just be 23.2?
No.
What's wrong with my radius? Isn't it 1/2 = 0.5?
r is the distance from the axis for each mass. Where do you get 0.5 from?
 
Would you use the Distance formula to find the radius between the Z Axis and the Point?
 
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+...
 
nazerofsun said:
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+...

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
Mantello said:
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.

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
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
MissPenguins said:
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
Good!
then I used
I = sum of mv2 = m1r1+ m2r2 =m3r3
Too many typos here. I assume you did I = Σmr².
so I added them all up and got 874.50
Is this reasonable? Am I doing it right?
Looks like you did the right thing.
 
  • #13


Doc Al said:
Good!

Too many typos here. I assume you did I = Σmr².

Looks like you did the right thing.



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

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