High School

A 103 kg horizontal platform is a uniform disk with a radius of 1.51 m. It can rotate about the vertical axis through its center. A 64.1 kg person stands on the platform at a distance of 1.05 m from the center, and a 26.7 kg dog sits on the platform near the person, 1.39 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 :

The total moment of inertia for the system, which includes a horizontal platform and its population (a person and a dog), is calculated as 239.58682 kg * m^2 by summing the individual moments of inertia of the platform, person, and dog.

To find the moment of inertia of the system consisting of the platform and its 'population', we must consider the moment of inertia for each component and add them up. The formula for the moment of inertia of a uniform disc with respect to its center axis is I = 0.5 * M * R^2, where M is the mass of the disc and R is the radius.

First, we calculate the moment of inertia of the platform: I_platform = 0.5 * 103 kg * (1.51 m)^2.

Then we consider the person and the dog as point masses, so their moments of inertia are calculated using the formula I = m * r^2, where m is the mass and r is the distance from the rotation axis.

The moment of inertia for the person is I_person = 64.1 kg * (1.05 m)^2 and for the dog is I_dog = 26.7 kg*(1.39 m)^2.

Finally, we sum up all the moments of inertia to find the total moment of inertia for the system, I_total = I_platform + I_person + I_dog. Upon calculation, we get:

  • I_platform = 0.5 * 103 kg * (1.51 m)^2 = 117.4515 kg * m^2
  • I_person = 64.1 kg * (1.05 m)^2 = 70.50775 kg * m^2
  • I_dog = 26.7 kg * (1.39 m)^2 = 51.62757 kg * m^2

Adding these together, we have I_total = 117.4515 kg * m^2 + 70.50775 kg * m^2 + 51.62757 kg * m^2 = 239.58682 kg * m^2.

The moment of inertia of the system consisting of the platform and its occupants with respect to the axis is approximately [tex]\( 240.11 \, \text{kg} \times \text{m}^2 \)[/tex].

The moment of inertia [tex]\( I \)[/tex] of a uniform disk about its center is given by the formula:

[tex]I_{\text{disk}} = \frac{1}{2} m r^2[/tex]

where:

- [tex]\( m \)[/tex] is the mass of the disk, and

- [tex]\( r \)[/tex] is the radius of the disk.

The total moment of inertia [tex]\( I_{\text{total}} \)[/tex] of the system is the sum of the moment of inertia of the platform, the person, and the dog.

Let's calculate [tex]\( I_{\text{platform}} \)[/tex], [tex]\( I_{\text{person}} \)[/tex], and [tex]\( I_{\text{dog}} \)[/tex], and then sum them up to find [tex]\( I_{\text{total}} \)[/tex].

First, let's calculate the moment of inertia [tex]\( I_{\text{platform}}[/tex] of the platform:

[tex]\[ I_{\text{platform}} = \frac{1}{2} \times 103 \, \times (1.51 \, )^2 \][/tex]

[tex]\[ I_{\text{platform}} = \frac{1}{2} \times 103 \, \times 2.2801 \, \][/tex]

[tex]\[ I_{\text{platform}} = 117.73635 \, \text{kg} \times \text{m}^2 \][/tex]

Now, let's calculate the moment of inertia [tex]\( I_{\text{person}} \)[/tex] of the person:

[tex]\[ I_{\text{person}} = 64.1 \, \times (1.05 \, )^2 \][/tex]

[tex]\[ I_{\text{person}} = 64.1 \, \times 1.1025 \, \][/tex]

[tex]\[ I_{\text{person}} = 70.7645 \, \text{kg} \times \text{m}^2 \][/tex]

Next, let's calculate the moment of inertia [tex]\( I_{\text{dog}} \)[/tex] of the dog:

[tex]\[ I_{\text{dog}} = 26.7 \, \times (1.39 \, )^2 \][/tex]

[tex]\[ I_{\text{dog}} = 26.7 \, \times 1.9321 \, \][/tex]

[tex]\[ I_{\text{dog}} = 51.609 \, \text{kg} \times \text{m}^2 \][/tex]

[tex]\[ I_{\text{total}} = I_{\text{platform}} + I_{\text{person}} + I_{\text{dog}} \][/tex]

[tex]\[ I_{\text{total}} = 117.73635 \, \text{kg} \times \text{m}^2 + 70.7645 \, \text{kg} \times \text{m}^2 + 51.609 \, \text{kg} \times \text{m}^2 \][/tex]

[tex]\[ I_{\text{total}} = 240.10985 \, \text{kg} \times \text{m}^2 \][/tex]

So, moment of inertia [tex]\[ I_{\text{total}} = 240.10985 \, \text{kg} \times \text{m}^2 \][/tex].