Question

QUESTION 3 A solid ball of mass M and radius R is connected to a rod of length L as shown. What is the moment of inertia of just the ballo rotations about an axis perpendicular to the other end of the rod? Svel

Answer 1

If we divide the solid sphere into infinetesimal thin disks on the z axis and add the moments of inertia of each disk, the moment of inertia of the disk would be given by the following equation:

$dI=\frac{1}{2}y^{2}dm$

where

$dm=\rho dV$,

$dI=\frac{1}{2}y^{2}\rho dV$

where

$dV=y^{2}\pi dz$,

$dI=\frac{1}{2}y^{2}\rho \pi y^{2}dz$

$I=\frac{1}{2}\rho \pi \int_{-R}^{R}y^{4}dz$

determining the value of y in terms of the variable z, we have:

$y^{2}=R^{2}-z^{2}$.

$I=\frac{1}{2}\rho \pi \int_{-R}^{R}(R^{2}-z^{2})^{2}dz$

Writing in polynomial form and solving the integral, we obtain:

$I=\frac{1}{2}\rho \pi \left [ R^{4}z-\frac{2R^{2}z^{3}}{3}+\frac{z^{5}}{5} \right ]_{-R}^{R}$

$I=\frac{8}{15}\rho \pi R^{5}$

where the density is given by:

$\rho =\frac{M}{V}$

and the volume of the sphere is given by:

$V=\frac{4}{3}\pi R^{3}$

substituting, you get

$I=\frac{8}{15}(\frac{M}{\frac{4}{3}\pi R^{3}}) \pi R^{5}$

MR2