Answer:
Part a)
[tex]I = 17.4 \times 10^{-3} kg m^2[/tex]
Part b)
[tex]I = 14.9 \times 10^{-3} kg m^2[/tex]
Explanation:
Part a)
Moment of inertia of the core of the ball
[tex]I_1 = \frac{2}{5}m_1r_1^2[/tex]
[tex]I_1 = \frac{2}{5}(1.6)((\frac{0.196}{2})^2)[/tex]
[tex]I_1 = 6.14 \times 10^{-3} kg m^2[/tex]
now the moment of inertia for thin shell
[tex]I_2 = \frac{2}{3} m_2r_2^2[/tex]
[tex]I_2 = \frac{2}{3}(1.6)((\frac{0.206}{2})^2)[/tex]
[tex]I_2 = 11.3 \times 10^{-3} kg m^2[/tex]
now total inertia of the ball is given as
[tex]I = I_1 + I_2[/tex]
[tex]I = 17.4 \times 10^{-3} kg m^2[/tex]
Part b)
Moment of inertia of uniform ball of mass 3.2 kg
[tex]I = \frac{2}{5} mr^2[/tex]
[tex]I = \frac{2}{5}(3.2)((\frac{0.216}{2})^2)[/tex]
[tex]I = 14.9 \times 10^{-3} kg m^2[/tex]