Moment of inertia of a sphere

[Za Kh]

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

 

[Replusz]

You could find out with Steiner's law (parallel axis theorem):
https://en.wikipedia.org/wiki/Parallel_axis_theorem
Although i would rather think there was a mistake.
Someone seemed to have a similar problem: https://www.physicsforums.com/threads/moment-of-inertia-for-a-rectangular-plate.12903/

 

[nrqed]

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?

 

[Replusz]

Was she doing this?
http://imgur.com/dc0ZTvG

 

[nrqed]

Was she doing this?
http://imgur.com/dc0ZTvG

Oh! It is clearly wrong. I hope a teacher did not do that in a class!!

 

[Za Kh]

I want to show you the paper on my notebook but I dont know how to send a pic here

 

[Za Kh]

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 :)

 

[Za Kh]

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

 

[kdv]

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.

 

[cnh1995]

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?

 

[jbriggs444]

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.

 

[Za Kh]

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"

 

[Za Kh]

Do you mean a hemisphere? Solid or hollow?

A solid sphere

 

[Za Kh]

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

 

[vanhees71]

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 $u=\cos \vartheta$, $\mathrm{d} u =\mathrm{d} \vartheta \sin \vartheta$, $\sin^2 \vartheta=1-u^2$.