Answer:
[tex]14.8\; {\rm kg \cdot m^{2} \cdot s^{-1}}[/tex].
Explanation:
The angular momentum [tex]L[/tex] of a rotating object is the product of angular velocity [tex]\omega[/tex] and moment of inertia [tex]I[/tex]:
[tex]L = I\, \omega[/tex].
For a sphere of uniform density rotating about its diameter, the moment of inertia is proportional to the square of its radius:
[tex]\displaystyle I = \frac{2}{5}\, m\, r^{2}[/tex],
Where [tex]m[/tex] is the mass of the sphere and [tex]r[/tex] is the radius of the sphere.
Substitute the expression for angular momentum [tex]I[/tex] into the equation [tex]L = I\, \omega[/tex] to express angular momentum [tex]L[/tex] in terms of the radius of the sphere:
[tex]\displaystyle L = \frac{2}{5}\, m\, r^{2} \, \omega[/tex].
In other words, if the mass [tex]m[/tex] and angular velocity [tex]\omega[/tex] of this sphere stays unchanged, the angular momentum [tex]L[/tex] of the sphere would be proportional to the square of the radius [tex]r[/tex] of the sphere. Reducing [tex]r[/tex] to [tex](1/2)[/tex] of the initial value would reduce angular momentum [tex]L[/tex] to [tex](1/2)^{2}[/tex] of the initial value:
[tex]\displaystyle \left(\frac{1}{2}\right)^{2} \left(59.2\; {\rm kg\cdot m^{2} \cdot s^{-1}\right) = 14.8\; {\rm kg\cdot m^{2}\cdot s^{-1}}[/tex].