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

a) The moment of inertia of the hoop yo-yo rotating about its center of mass is .

b) The moment of inertial of the hoop yo-yo rotating about its center of mass is .

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 (), measured in kilogram-square meters, which is located at the geometrical center of the element, is determined by the following formula:

(1)

(2)

(3)

Where:

- Diameter, measured in meters.

- Thickness, measured in meters.

- Mass, measured in kilograms.

If we know that , and , then the moment of inertia of the hoop yo-yo is:

The moment of inertia of the hoop yo-yo rotating about its center of mass is .

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:

(4)

Where:

- Distance between parallel axes, measured in meters.

If we know that , and , then the moment of inertial of the hoop yo-yo rotating about the point where the tension force is applied is:

The moment of inertial of the hoop yo-yo rotating about its center of mass is .

in order to follow the instructions and complete the statements, i need to use data from table c of your student guide. can you think of a way to let me see your student guide for a while ? i can't answer the question without the data in table c.

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm