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MAPgirl23
Jun5-05, 10:29 AM
A sphere consists of a solid wooden ball of uniform density 800 kg/m^3 and radius 0.20 m and is covered with a thin coating of lead foil with area density 20 kg/m^2.

How do I calculate the moment of inertia of this sphere about an axis passing through its center?

** The equation I'm using for the solid sphere is I = 2/5 * M * R^2
Here again I know the mass of the inner sphere is the density times the volume of the sphere. They gave me the area density for the layer of lead, so you can find the mass of the lead using the area density times the surface area of the sphere. By definition of moment of inertia, the total moment of inertia of the sphere is the sum of the moment of inertia of the solid wood part, plus the moment of inertia of the lead shell. Is this method wrong? Do you not get the moment of density this way without being given mass but instead given density?
Please help.

arildno
Jun5-05, 11:06 AM
A sphere consists of a solid wooden ball of uniform density 800 kg/m^3 and radius 0.20 m and is covered with a thin coating of lead foil with area density 20 kg/m^2.

How do I calculate the moment of inertia of this sphere about an axis passing through its center?

** The equation I'm using for the solid sphere is I = 2/5 * M * R^2
Here again I know the mass of the inner sphere is the density times the volume of the sphere. They gave me the area density for the layer of lead, so you can find the mass of the lead using the area density times the surface area of the sphere. By definition of moment of inertia, the total moment of inertia of the sphere is the sum of the moment of inertia of the solid wood part, plus the moment of inertia of the lead shell. Is this method wrong?

No, this is a correct method.
Do you not get the moment of density this way without being given mass but instead given density?
I'm sorry, I don't understand your question.

Dr.Brain
Jun5-05, 11:20 AM
The answer= MI of wooden solid sphere+ MI of lead shell

MAPgirl23
Jun5-05, 11:58 AM
volume of sphere: 4/3 * pi * r^3
surface area sphere: 4 * pi *r^2

mass of lead: 20 * (4 * pi * 0.2^2) = 10 kg
mass of sphere: 800 * (4 * pi * 0.2^2) = 26.8 kg

I solid sphere: 2/5 * 26.8 * 0.2^2 = 0.429
I lead shell: what do I use?

Nylex
Jun5-05, 12:03 PM
I lead shell: what do I use?

I = \frac{2}{3}MR^2

dextercioby
Jun5-05, 01:30 PM
Incidentally,the sphere has zero volume and its moment of inertia wrt an axis passing through its center is given in post #5.

BTW,what do you mean by solid sphere,is it a special sphere,i dunno,similar to a plasma or liquid one...?

Daniel.

Nylex
Jun5-05, 02:37 PM
:rolleyes:.