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

so easy. just google the electromagnetic waves and it’ll pop up.

Answer from: Quest

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j

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