Moment of inertia of a sphere about an axis

1. Sep 20, 2007

azila

1. The problem statement, all variables and given/known data
A sphere consists of a solid wooden ball of uniform density 800 kg/m^3 and radius .20 m and is covered with a thin coating of lead full with area density 20 kg/m^2.

A. calculate the moment of inertia of this sphere about an axis through the center.

2. Relevant equations
For a sphere: I = (2/5)MR^2
Volume of a sphere: (4/3)pir^3
Area of a sphere: 4pir^2
D = m/v

3. The attempt at a solution

Ok, So this is what I did. I don't know the mass, so I have to find the mass through the density. So, for the uniform sphere itself, I did 800 kg/m^3 * (the volume of a sphere) and got the mass of the sphere without the lead covering to be 26.8083. Then I did the lead covering, I did 20 kg/m^2 * (volume of the sphere) and got 10.0531. So then, I added 26.8083 + 10.0531 and got 36.86. I then plugged this mass into the equation of inertia: I= (2/5)(36.86) (.20 (radius))^2 and got .590. However, that is not the answer. So if someone can tell me where I am going wrong, I would appreciate it. Thanks.

2. Sep 20, 2007

mjsd

the lead covering is a spherical shell, not a sphere! so you need the volume of the spherical shell... so how thin is that coating?
EDIT: oh hang on, they give you the area density so... you just assume it is infinitesimally thin and just need surface area.

3. Sep 20, 2007

azila

but how would I calculate the volume of the spherical shell??

4. Sep 20, 2007

azila

i am only given a area density for the shell...so how would i calculate the volume??

5. Sep 20, 2007

mjsd

6. Sep 20, 2007

azila

i did calculate the surface area of the lead covering to be 10.0531. But when I combined the masses, and used the moment of inertia equation it didn't work. What am i doing wrong?

7. Sep 20, 2007

mjsd

there is a separate formula for moment of inertial for a spherical shell
the problem here is that you no longer have uniform density as soon as you add in the lead layer....

8. Sep 20, 2007

azila

so...........like would i do the I= MR^2 for the shell and the I=(2/5)MR^2 for the sphere and add them together... I am so stressed because I just can't seem to solve the problem..Please help. Thanks .

9. Sep 20, 2007

mjsd

since you seem to have the correct answer given to compare, just try it and see. by the way the moment of inertial of a spherical shell is NOT I=MR^2
look it up or dervie it!