Question

The moment of inertial of the hoop yo-yo when it is: (a) rotating about its center of mass, and (b) rotating about the point where the tension force is applied
mass=332g
diamater=35.9cm
thickness.95cm

Answer 1

Answer:

a) The moment of inertia of the hoop yo-yo rotating about its center of mass is $I_{g} = 0.0108\,kg\cdot m^{2}$.

b) The moment of inertial of the hoop yo-yo rotating about its center of mass is $I_{O} = 0.0216\,kg\cdot m^{2}$.

Explanation:

a) The hoop yo-yo can be modelled as a tours with a minor radius $a$, related with the thickness, and with a major radius $b$, related with the diameter, and with an uniform mass. The momentum of inertia about its center of mass ($I_{g}$), measured in kilogram-square meters, which is located at the geometrical center of the element, is determined by the following formula:

$I_{g} = \frac{1}{4}\cdot m \cdot (4\cdot b^{2}+3\cdot a^{2})$ (1)

$b = 0.5\cdot D$ (2)

$a = 0.5\cdot t$ (3)

Where:

$D$ - Diameter, measured in meters.

$t$ - Thickness, measured in meters.

$m$ - Mass, measured in kilograms.

If we know that $m = 0.332\,kg$, $D = 0.359\,m$ and $t = 9.5\times 10^{-3}\,m$, then the moment of inertia of the hoop yo-yo is:

$a = 0.5\cdot (9.5\times 10^{-3}\,m)$

$a = 4.75\times 10^{-3}\,m$

$b = 0.5\cdot (0.359\,m)$

$b = 0.180\,m$

$I_{g} = \frac{1}{4}\cdot (0.332\,kg)\cdot [4\cdot (0.180\,m)^{2}+3\cdot (4.75\times 10^{-3}\,m)^{2}]$

$I_{g} = 0.0108\,kg\cdot m^{2}$

The moment of inertia of the hoop yo-yo rotating about its center of mass is $I_{g} = 0.0108\,kg\cdot m^{2}$.

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:

$I_{O} = I_{g} +m\cdot r^{2}$ (4)

Where:

$r$ - Distance between parallel axes, measured in meters.

If we know that $I_{g} = 0.0108\,kg\cdot m^{2}$, $m = 0.332\,kg$ and $r = 0.180\,m$, then the moment of inertial of the hoop yo-yo rotating about the point where the tension force is applied is:

$I_{O} = 0.0108\,kg\cdot m^{2}+(0.332\,kg)\cdot (0.180\,m)^{2}$

$I_{O} = 0.0216\,kg\cdot m^{2}$

The moment of inertial of the hoop yo-yo rotating about its center of mass is $I_{O} = 0.0216\,kg\cdot m^{2}$.