Moment of inertia of a sphere

  • #1
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I know the moment of inertia for both a solid sphere and a hollow sphere is , but my teacher has derived a moment of inertia of the sphere but am not sure about what axis she was deriving it , and she got this answer 3/5 MR^2
 

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  • #3
nrqed
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I know the moment of inertia for both a solid sphere and a hollow sphere is , but my teacher has derived a moment of inertia of the sphere but am not sure about what axis she was deriving it , and she got this answer 3/5 MR^2
I also think that there must be a mistake in her derivation. Do you have it? Do you agree with all the steps?
 
  • #6
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I want to show you the paper on my notebook but I dont know how to send a pic here
 
  • #7
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Now I understand , it has turned out that she meant the moment of inertia of the center of sphere and I thought it were the moment of an axis passing through the center , just because she hasn't specified clearly what she meant , thank you everyone :)
 
  • #8
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How could we calculate the moment of inertia of a sphere , cut into half by the xoy plane ,, its supposed that moment of ineria about the xy and zy axes is zero , i want to know why
 
  • #9
kdv
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Now I understand , it has turned out that she meant the moment of inertia of the center of sphere and I thought it were the moment of an axis passing through the center , just because she hasn't specified clearly what she meant , thank you everyone :)
But what does that mean exactly? There is no way to make a sphere rotate in such a way that it will have that moment of inertia, so it is actually a completely unphysical result. That's why it is never quoted as a moment of inertia of a sphere.
 
  • #10
cnh1995
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How could we calculate the moment of inertia of a sphere , cut into half by the xoy plane
Do you mean a hemisphere? Solid or hollow?
 
  • #11
jbriggs444
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How could we calculate the moment of inertia of a sphere , cut into half by the xoy plane ,, its supposed that moment of ineria about the xy and zy axes is zero , i want to know why
The "xy" axis is a synonym for an axis in the z direction? So if the moment of inertia about this axis is zero, every point within the sphere must be somewhere on the z axis.

The "zy" axis is a synonym for an axis in the x direction? So if the moment of inertia about this axis is zero, every point within the sphere must be somewhere on the x axis.

It follows that every point within the sphere must be at the intersection of the x and z axes. i.e. at the origin. Well, yeah, that moment of inertia is fairly easy to calculate.
 
  • #12
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I haven't sensed a physical meaning to it yet in my head , but there's a formula that states tge summation of moments of inertia wrt x-axis ,y axis and z axis , respectively is equal to two times the moment of inertia at the origin . I think so .. Unfortunately i dont have enough time to understand this lesson we've taken lately , my exam is after tomorrow , but am sure that she wrote this title "moment of inertia w.r.t a point"
 
  • #13
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Do you mean a hemisphere? Solid or hollow?
A solid sphere
 
  • #14
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The "xy" axis is a synonym for an axis in the z direction? So if the moment of inertia about this axis is zero, every point within the sphere must be somewhere on the z axis.

The "zy" axis is a synonym for an axis in the x direction? So if the moment of inertia about this axis is zero, every point within the sphere must be somewhere on the x axis.

It follows that every point within the sphere must be at the intersection of the x and z axes. i.e. at the origin. Well, yeah, that moment of inertia is fairly easy to calculate.
They are not given zero to us , they should be proved by calculation , but am not knowing how
 
  • #15
vanhees71
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Why so complicated? I guess we assume a homogeneous sphere of density ##\rho=3m/(4 \pi R^3)##. The rotation axis is through the center (all other cases can be evaluated with Steiner's law). Take spherical coordinates and the rotation axis around the polar axis. Then we have
$$\Theta=\rho \int_0^{2 \pi} \mathrm{d} \varphi \int_0^{\pi} \mathrm{d} \vartheta \int_0^{R} \mathrm{d} r r^2 \sin \vartheta r^2 \sin^2 \vartheta = 2 \pi \rho \frac{R^5}{5} \int_{-1}^1 \mathrm{d} u (1-u^2) = \frac{8\pi }{15} \rho R^5=\frac{2}{5} m R^2.$$
In the last step, I've substituted ##u=\cos \vartheta##, ##\mathrm{d} u =\mathrm{d} \vartheta \sin \vartheta##, ##\sin^2 \vartheta=1-u^2##.
 

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