Physics, 22.12.2020 14:00 smkw04p3ao0n

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

The moment of inertial of the hoop yo-yo when it is: (a) rotating about its center of mass, and (b)

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Answer from: VamPL

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 , related with the thickness, and with a major radius , 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:

- Diameter, measured in meters.

- Thickness, measured in meters.

- 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:

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

Answer from: Quest

1) when you rub your hands together quickly, the kinetic energy is converted to mechanical energy.2) thermal energy from the sun travels to earth as infrared waves.3) energy is transferred through the matter of space by electromagnetic waves. 4) when a match is burned, chemical potential energy is converted to thermal and radiant energy. 5) seismicwaves are caused by back and forth movement , or vibration, of an object. 6) mechanicalenergy in an engine changes thermal energy into kinetic energy to make a car move.

Answer from: Quest

c. newton's third law of motion

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