College

A 137 kg horizontal platform is a uniform disk with a radius of 1.53 m and can rotate about a 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 occupants, with respect to the axis.

Answer :

Answer:

The moment of inertia is [tex]I= 312.09 \ kg \cdot m^2[/tex]

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 [tex]m_p = 68.7 \ kg[/tex]

The distance of the person from the center is [tex]d_c =1.19 \ m[/tex]

The mass of the dog is [tex]m_d = 25.9 \ kg[/tex]

The distance of the dog from the person [tex]d_d = 1.45 \ m[/tex]

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

[tex]I = I_1 + I_2 + I_3[/tex]

Where [tex]I_1[/tex] is the moment of inertia of the platform which mathematically represented as

[tex]I_1 = \frac{m * r^2}{2}[/tex]

substituting values

[tex]I_1 = \frac{ 137 * (1.53)^2}{2}[/tex]

[tex]I_1 = 160.35 \ kg\cdot m^2[/tex]

Also [tex]I_2[/tex] is the moment of inertia of the person about the axis which is mathematically represented as

[tex]I_2 = m_p * d_c^2[/tex]

substituting values

[tex]I_2 = 68.7 * 1.19^2[/tex]

[tex]I_2 = 97.29 \ kg \cdot m^2[/tex]

Also [tex]I_3[/tex] is the moment of inertia of the dog about the axis which is mathematically represented as

[tex]I_3 = m_d * d_d^2[/tex]

substituting values

[tex]I_3 = 25.9 * 1.45^2[/tex]

[tex]I_3 = 54.45 \ kg \cdot m^2[/tex]

Thus

[tex]I= 160.35 + 97.29 + 54.45[/tex]

[tex]I= 312.09 \ kg \cdot m^2[/tex]