Vtyjkyuk

I have been given a sphere of radius a, from this sphere a cap of hight h is cut off.

1) What is the centre of mass of the rest of the sphere?

2) What is the moment of inertia regarding the axis of symmetry?

1 - no real Idea how to do it.

2 - I got $$ I = frac12rho pi(frac1615R^5-frac43R^2h^3+h^4R)$$
could this be right?

Thank you very much for your help!

For part (1), assuming the axis of the cap is the $z$ axis,
consider a slice of the remaining part of the sphere perpendicular to the $z$ axis,
such as the part of the sphere between the planes at $z = z_0$ and $z = z_0 + Delta z$.
For very small $Delta z$, the mass of this slice is approximately
$rho pi (R^2 - z^2) ,Delta z$ and its center of mass is at a distance
of approximately $z$ from the origin,
so its moment about the origin is approximately $rho pi (R^2 - z^2) z ,Delta z$.
The range of possible $z$ values within the remaining part of the
sphere is from $-R$ to $R - h$.
The moment of the entire remaining part of the sphere is just the sum of moments
of all its parts, so this suggests that its total moment is:

$$
int_-R^R-h rho pi (R^2 - z^2) z ,dz.
$$

The center of mass is found by dividing the total moment of the remaining part of the sphere by its mass (which can also be computed by an integral, this time just adding up the masses of the disks without considering their distance from the origin).

To check your answer for part (2), try the following substitutions:

1. $h = 0$.




2. $h = R$.




3. $h = 2R$.

These should result in values of $I$ for an entire spherical ball
($I = frac815 pi rho R^5$),
for half a spherical ball around its axis of symmetry ($I = frac415 pi rho R^5$),
and for what remains when you remove the entire spherical ball (that is, nothing,
$I = 0$).

Your formula checks out for $h = 0$ but not for either of the other two cases.
The problem seems to be that you neglected a term in $h^5$ somewhere in your calculations.

This is pretty straightforward. First set up the mass integral:
$$beginalignm&=int_0^2piint_-a^a-hint_0^sqrta^2-z^2rho r,dr,dz,dtheta\
&=pirhoint_-a^a-h(a^2-z^2)dz=pirholeft(frac43a^3-ah^2+frac13h^3right)\
&=fracpirho3(a+h)(2a-h)^2endalign$$
This checks as it gets $frac43pi a^3$ when $h=0$ and $0$ when $h=2a$. Then we have the integral for the first moment
$$beginalignmbar z&=int_0^2piint_-a^a-hint_0^sqrta^2-z^2rho rz,dr,dz,dtheta\
&=pirhoint_-a^a-hz(a^2-z^2)dz\
&=fracpirho h^24(2a-h)^2endalign$$
So
$$bar z=fracmbar zm=frac34frach^2a+h$$
And this checks because it's $0$ when $h=0$ and approaches $-a$ when $hrightarrow2a$. Then there is the moment of inertia about the $z$-axis.
$$beginalignI_z&=int_0^2piint_-a^a-hint_0^sqrta^2-z^2rho r^3,dr,dz,dtheta\
&=frac12pirhoint_-a^a-h(a^2-z^2)^2dz\
&=frac12pirholeft(frac1615a^5-frac43a^2h^3+ah^4-frac15h^5right)\
&=fracpirho30(2a-h)^2left(4a^3+4a^3h+3ah^2-3h^3right)\
&=frac110mfrac4a^3+4a^3h+3ah^2-3h^3a+hendalign$$
And this checks because it yields $frac25ma^2$ both when $h=a$ and when $h=0$.