Question

A 137 kg horizontal platform is a uniform disk of radius 1.53 m and can rotate about the vertical axis through its center. A 68.7 kg person stands on the platform at a distance of 1.19 m from the center, and a 25.9 kg dog sits on the platform near the person 1.45 m from the center. Find the moment of inertia of this system, consisting of the platform and its population, with respect to the axis.

Answer 1

The moment of inertia is
`I= 312.09 \ kg \cdot m^2`

Step-by-step explanation:

From the question we are told that

The mass of the platform is m = 137 kg

The radius is r = 1.53 m

The mass of the person is
`m_p = 68.7 \ kg`

The distance of the person from the center is
`d_c =1.19 \ m`

The mass of the dog is
`m_d = 25.9 \ kg`

The distance of the dog from the person
`d_d = 1.45 \ m`

Generally the moment of inertia of the system is mathematically represented as

`I = I_1 + I_2 + I_3`

Where
`I_1` is the moment of inertia of the platform which mathematically represented as

`I_1 = (m * r^2)/(2)`

substituting values

`I_1 = ( 137 * (1.53)^2)/(2)`

`I_1 = 160.35 \ kg\cdot m^2`

Also
`I_2` is the moment of inertia of the person about the axis which is mathematically represented as

`I_2 = m_p * d_c^2`

substituting values

`I_2 = 68.7 * 1.19^2`

`I_2 = 97.29 \ kg \cdot m^2`

Also
`I_3` is the moment of inertia of the dog about the axis which is mathematically represented as

`I_3 = m_d * d_d^2`

substituting values

`I_3 = 25.9 * 1.45^2`

`I_3 = 54.45 \ kg \cdot m^2`

Thus

`I= 160.35 + 97.29 + 54.45`

`I= 312.09 \ kg \cdot m^2`