Answer:
a) The moment of inertia of the hoop yo-yo rotating about its center of mass is [tex]I_{g} = 0.0108\,kg\cdot m^{2}[/tex].
b) The moment of inertial of the hoop yo-yo rotating about its center of mass is [tex]I_{O} = 0.0216\,kg\cdot m^{2}[/tex].
Explanation:
a) The hoop yo-yo can be modelled as a tours with a minor radius [tex]a[/tex], related with the thickness, and with a major radius [tex]b[/tex], related with the diameter, and with an uniform mass. The momentum of inertia about its center of mass ([tex]I_{g}[/tex]), measured in kilogram-square meters, which is located at the geometrical center of the element, is determined by the following formula:
[tex]I_{g} = \frac{1}{4}\cdot m \cdot (4\cdot b^{2}+3\cdot a^{2})[/tex] (1)
[tex]b = 0.5\cdot D[/tex] (2)
[tex]a = 0.5\cdot t[/tex] (3)
Where:
[tex]D[/tex] - Diameter, measured in meters.
[tex]t[/tex] - Thickness, measured in meters.
[tex]m[/tex] - Mass, measured in kilograms.
If we know that [tex]m = 0.332\,kg[/tex], [tex]D = 0.359\,m[/tex] and [tex]t = 9.5\times 10^{-3}\,m[/tex], then the moment of inertia of the hoop yo-yo is:
[tex]a = 0.5\cdot (9.5\times 10^{-3}\,m)[/tex]
[tex]a = 4.75\times 10^{-3}\,m[/tex]
[tex]b = 0.5\cdot (0.359\,m)[/tex]
[tex]b = 0.180\,m[/tex]
[tex]I_{g} = \frac{1}{4}\cdot (0.332\,kg)\cdot [4\cdot (0.180\,m)^{2}+3\cdot (4.75\times 10^{-3}\,m)^{2}][/tex]
[tex]I_{g} = 0.0108\,kg\cdot m^{2}[/tex]
The moment of inertia of the hoop yo-yo rotating about its center of mass is [tex]I_{g} = 0.0108\,kg\cdot m^{2}[/tex].
b) The hoop yo-yo rotate at a point located at a distance of half diameter from the center of mass of the element, whose moment of inertia is determined by the Theorem of Parallel Axes:
[tex]I_{O} = I_{g} +m\cdot r^{2}[/tex] (4)
Where:
[tex]r[/tex] - Distance between parallel axes, measured in meters.
If we know that [tex]I_{g} = 0.0108\,kg\cdot m^{2}[/tex], [tex]m = 0.332\,kg[/tex] and [tex]r = 0.180\,m[/tex], then the moment of inertial of the hoop yo-yo rotating about the point where the tension force is applied is:
[tex]I_{O} = 0.0108\,kg\cdot m^{2}+(0.332\,kg)\cdot (0.180\,m)^{2}[/tex]
[tex]I_{O} = 0.0216\,kg\cdot m^{2}[/tex]
The moment of inertial of the hoop yo-yo rotating about its center of mass is [tex]I_{O} = 0.0216\,kg\cdot m^{2}[/tex].
