A solid steel sphere of density 7.86 g/cm3 and mass 1 kg spin on an axis through its center with a period of 2.3 s. given vsphere = 4 3 π r3, what is its angular momentum? answer in units of kg m2 / s.

The solution for this problem is:

density = mass / volume

7860 = 1 / ((4/3) pi r^3)

r^3 = 1 / (7860 * 4/3*pi)

r = (1 / (7860 * 4/3*pi)) ^ (1/3)

= 0.067 m

Inertia = (2/5) mr^2

= (2/5) x 1 x 0.067^2

= 0.0017956 kg-m^2

1/2.3 = 0.4348 rev/s

0.4348 x 2pi = 2.732 rad/s

Angular momentum = Inertia x rad/s

0.0017956 x 2.732 = 0.00490557 kg m^2/s