Rotational Inertia of a hollow sphere(having trouble)

[Vasco]

Hello,mates.i`,ve been struggling to demonstrate the procedure to
calculate the rotational inertia of a hollow/empty sphere around an axe that passes through its diameter.Please,could
you shed me some light or show me the procedure(Even show me any online papers on it)I`d be Glad.Please help me!

The result is I=2/3(M*R^2).

THANK YOU!
Jason,physicist21@yahoo.com

 

[dextercioby]

$$I=\int r^{2}dm$$ In the case of a sphere of radius R,the moment of inertia becomes

$$I=\int R^{2}\sin^{2}\vartheta \left( \frac{M}{4\pi R^{2}}dS\right) =\frac{1}{4\pi}MR^{2}\int_{0}^{2\pi}d\varphi\int_{0}^{\pi}d\vartheta \ \sin^{3}\vartheta$$

Compute the integrals and get the result.

"r" is in the general case the distance between the mass element "dm" & the rotation axis.In this case,i've chosen spherical coordinates & the axis of rotation as Oz...

Daniel.

 

[Vasco]

$$I=\int r^{2}dm$$ In the case of a sphere of radius R,the moment of inertia becomes

$$I=\int R^{2}\sin^{2}\vartheta \left( \frac{M}{4\pi R^{2}}dS\right) =\frac{1}{4\pi}MR^{2}\int_{0}^{2\pi}d\varphi\int_{0}^{\pi}d\vartheta \ \sin^{3}\vartheta$$

Compute the integrals and get the result.

"r" is in the general case the distance between the mass element "dm" & the rotation axis.In this case,i've chosen spherical coordinates & the axis of rotation as Oz...

Daniel.

hey Daniel,thanks!But,i`m still confused about the choice of dm and the use of coordinates to use on the integrals.could you elaborate some more?(my professor wants things with so much detail! ).Once more,i thank you and wait for yer answer. Jason.

 

[dextercioby]

I've chosen a completely arbitrary mass element of the sphere.I assumed that the density of the sphere is constant & equal to the ratio between the sphere's mass & its surface.Then the mass "dm" of the surface element "dS" and the element of the surface "dS" are linked through

$$\mbox{mass}=\mbox{density}\cdot \mbox{surface}$$

$$dm=\frac{M}{4\pi R^{2}} dS$$

That's all there is to it.

The parametrization of the 2 sphere is of course:

$$\left\{ \begin{array}{c}x=R\sin\vartheta\cos\varphi\\y=R\sin\vartheta\sin\varphi\\z=R\cos\vartheta \end{array} \right$$

which induces the surface element

$$dS=R^{2}\sin\vartheta \ d\vartheta \ d\varphi$$

Daniel.

 

[Vasco]

I've chosen a completely arbitrary mass element of the sphere.I assumed that the density of the sphere is constant & equal to the ratio between the sphere's mass & its surface.Then the mass "dm" of the surface element "dS" and the element of the surface "dS" are linked through

$$\mbox{mass}=\mbox{density}\cdot \mbox{surface}$$

$$dm=\frac{M}{4\pi R^{2}} dS$$

That's all there is to it.

The parametrization of the 2 sphere is of course:

$$\left\{ \begin{array}{c}x=R\sin\vartheta\cos\varphi\\y=R\sin\vartheta\sin\varphi\\z=R\cos\vartheta \end{array} \right$$

which induces the surface element

$$dS=R^{2}\sin\vartheta \ d\vartheta \ d\varphi$$

Daniel.

Daniel,i`m almost there.i just have one more doubt.Where the result $$dS=R^{2}\sin\vartheta \ d\vartheta \ d\varphi$$ comes from?I won't bother you anymore;) thanks!

 

[dextercioby]

It's the surface element in spherical coordinates...It's famous and it is deduced in a course on calculus/analysis.It can be even justified by geometrical arguments.

Daniel.

 

[Vasco]

It's the surface element in spherical coordinates...It's famous and it is deduced in a course on calculus/analysis.It can be even justified by geometrical arguments.

Daniel.

Sorry daniel:) i had forgoten for a moment this deduction) which is so easy.Now,i`ve made it.I`m just tired of demonstrating every inertia possible (big homework!).Ya helped a lot! thank ya!