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a 121 kg horizontal platform is a uniform disk of radius 1.71 m and can rotate about the vertical axis through its center. a 66.9 kg person stands on the platform at a distance of 1.19 m from the center, and a 25.5 kg dog sits on the platform near the person 1.35 m from the center. find the moment of inertia of this system, consisting of the platform and its population, with respect to the axis. moment of inertia:

User Keshan
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Answer: To find the moment of inertia of the system consisting of the platform, person, and dog, we can use the formula:

I = I_platform + I_person + I_dog

where I_platform, I_person, and I_dog are the moments of inertia of the platform, person, and dog, respectively.

The moment of inertia of the platform can be found using the formula for the moment of inertia of a uniform disk:

I_platform = (1/2)MR^2

where M is the mass of the platform and R is its radius. Substituting the given values, we get:

I_platform = (1/2)(121 kg)(1.71 m)^2

I_platform = 182.34 kg·m^2

The moment of inertia of the person can be found using the parallel axis theorem, which states that the moment of inertia of a body about an axis parallel to its center of mass is given by:

I_person = I_cm + Md^2

where I_cm is the moment of inertia of the person about their center of mass, M is their mass, and d is the distance between the axis and their center of mass. We can assume that the person is a uniform rod, so the moment of inertia about their center of mass is:

I_cm = (1/12)ML^2

where L is their length. Substituting the given values, we get:

I_cm = (1/12)(66.9 kg)(2(1.19 m))^2

I_cm = 6.07 kg·m^2

Substituting this into the parallel axis theorem, we get:

I_person = 6.07 kg·m^2 + (66.9 kg)(1.19 m)^2

I_person = 83.18 kg·m^2

The moment of inertia of the dog can also be found using the parallel axis theorem, assuming that the dog is a uniform cylinder. The moment of inertia about the center of mass of a cylinder is (1/2)MR^2, so the moment of inertia about the axis passing through the center of mass is:

I_dog = (1/2)MR^2 + Md^2

where M is the mass of the dog, R is the radius of the dog, and d is the distance between the axis and the center of mass of the dog. Substituting the given values, we get:

I_dog = (1/2)(25.5 kg)(0.15 m)^2 + (25.5 kg)(1.35 m)^2

I_dog = 7.68 kg·m^2

Finally, we can substitute all the values into the formula for the total moment of inertia:

I = I_platform + I_person + I_dog

I = 182.34 kg·m^2 + 83.18 kg·m^2 + 7.68 kg·m^2

I = 273.20 kg·m^2

Therefore, the moment of inertia of the system consisting of the platform, person, and dog, with respect to the axis passing through the center, is 273.20 kg·m^2.

User Le Hibou
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