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Hollow Sphere Moment of Inertia Help

  • Thread starter George3
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  • #1
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Homework Statement


A hollow sphere has a mass of 15 kg, an inner radius of 12 cm and an outer radius of 18 cm. What is the rotational inertia (moment of inertia) of the sphere about an axis passing through its center?


Homework Equations





The Attempt at a Solution


I = 2/3 MR^2 for a hollow sphere so i did this:
2/3 (15) (.18^2) = .32 kg m^2

But this is wrong the answer is .24 kg m^2.
Any thoughts?
 

Answers and Replies

  • #2
AEM
360
0

Homework Statement


A hollow sphere has a mass of 15 kg, an inner radius of 12 cm and an outer radius of 18 cm. What is the rotational inertia (moment of inertia) of the sphere about an axis passing through its center?


Homework Equations





The Attempt at a Solution


I = 2/3 MR^2 for a hollow sphere so i did this:
2/3 (15) (.18^2) = .32 kg m^2

But this is wrong the answer is .24 kg m^2.
Any thoughts?
Have you tried computing the moment of inertia of a solid sphere of radius 18 cm and subtracting from it the moment of inertia of a solid sphere with a radius of 12 cm? In other words "scooping" out the center of the original sphere to create your object?
 
  • #3
31
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BUMP... FINAL TOMORROW NEED HELP ON THIS
I really need help on this. And to the previous poster the moment of inertia of a hollow sphere is bigger than that of a solid so your method would not work...right?
 
  • #4
AEM
360
0
BUMP... FINAL TOMORROW NEED HELP ON THIS
I really need help on this. And to the previous poster the moment of inertia of a hollow sphere is bigger than that of a solid so your method would not work...right?
Well, I just computed your moment of inertia by doing what I told you to do and got the right answer. Your mistake is confusing a sphere with a spherical portion of the interior removed with a spherical shell.

By the way, you will have to compute the density of the material out of which the object is made.
 
  • #5
31
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Well, I just computed your moment of inertia by doing what I told you to do and got the right answer. Your mistake is confusing a sphere with a spherical portion of the interior removed with a spherical shell.

By the way, you will have to compute the density of the material out of which the object is made.
So did you take (2/5)MR^2 for both radii and then subtract the two moments of inertia? Which formula for I did you use??
 
Last edited:
  • #6
AEM
360
0
So did you take (2/5)MR^2 for both radii and then subtract the two moments of inertia? Which formula for I did you use??
What you do is use [itex] I = \frac{2}{5}Mr^2 [/itex] for each sphere. This is why you need the density of the material. You have to know the mass of the smaller sphere and the mass of the larger sphere. You also have to use the appropriate radius for each. And, yes you subtract the moments of inertia that you compute.
 
  • #7
31
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I was able to get it thanks for the insight it helped a lot.
 

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