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**Figure 1**

## Homework Statement

Three identical marbles are on a turntable disk of radius [tex]R[/tex]. The mass of the disk is [tex]{m}_{d}[/tex]. The mass of each marble is [tex]{m}_{b}[/tex]. What is the moment of inertia of the disk and the marbles around an axis of rotation at the edge of the disk (the dot in

*Figure 1*)?

- [tex]\left({m}_{d} + 4{m}_{b}\right){R}^{2}[/tex]
- [tex]\left(\frac{5}{2}{m}_{d} + 3{m}_{b}\right){R}^{2}[/tex]
- [tex]\left(\frac{3}{2}{m}_{d} + 5{m}_{b}\right){R}^{2}[/tex]
- [tex]\left(\frac{1}{2}{m}_{d} + 3{m}_{b}\right){R}^{2}[/tex]
- [tex]\left(\frac{1}{2}{m}_{d} + 2{m}_{b}\right){R}^{2}[/tex]

## Homework Equations

- Solid Disk Inertial Moment: [tex]{I}_{disk} = \frac{1}{2}m{r}^{2}[/tex]
- Solid Sphere Inertial Moment: [tex]{I}_{solid\ sphere} = \frac{2}{5}m{r}^{2}[/tex]
- Inertial Moment Additivity: [tex]{I}_{sys} = {I}_{1} + {I}_{2} + \ldots + {I}_{n}[/tex]
- Parallel Axis Theorem: [tex]I = {I}_{cm} + m{d}^{2}[/tex]

## The Attempt at a Solution

I initially looked at this problem and I was not sure what assumptions I could make. I guessed that the marbles were not attached, but I am not sure what that means. The only experiment I did with the inertial moment of a system was dropping a hollow cylinder onto a rotating disk. While not attached, the cylinder had enough mass that it did not rotate independent of the disk and thus the system could be treated as a whole such that the principle of inertial moment additivity could be used. However, what am I supposed to assume here? Well, since this was a test question, I was forced to go with my instinct, which was to treat it like the experiment I did earlier. But what about the radius of the sphere? As you'll see below, I accounted for it, but why is the problem not accounting for it?

So, with that in mind, I started by listing what moments of inertia made up the system in terms of the knowns:

Let [tex]d[/tex] be the disk and [tex]b1[/tex], [tex]b2[/tex], and [tex]b3[/tex] be the marbles. Then we know that:

[tex]{I}_{sys} = {I}_{d} + {I}_{b1} + {I}_{b2} + {I}_{b3}[/tex]

I then found the inertial moments of each object independent of each other in terms of the knowns (but here is where I had to account for the marble's radius):

For the moments of inertia for each object, we simply substitute our variables into the given equation for the object type:

[tex]{I}_{dcm} = \frac{1}{2}{m}_{d}{R}^{2}[/tex]

[tex]{I}_{b1cm} = \frac{2}{5}{m}_{b}{r}_{b}^{2}[/tex]

[tex]{I}_{b2cm} = \frac{2}{5}{m}_{b}{r}_{b}^{2}[/tex]

[tex]{I}_{b3cm} = \frac{2}{5}{m}_{b}{r}_{b}^{2}[/tex]

My next step was to apply the parallel axis theorem to find the axis-adjusted moments of inertia for each object:

Since the axis is moved to one end of the disk and all distances from the new axis can be expressed in terms of the disk's radius, the axis-adjusted moments of inertia are as follows:

[tex]{I}_{d} = {I}_{dcm} + {m}_{d}{R}^{2} = \frac{1}{2}{m}_{d}{R}^{2} + {m}_{d}{R}^{2} = \left(\frac{1}{2}{m}_{d} + \frac{2}{2}{m}_{d}\right){R}^{2} = \frac{3}{2}{m}_{d}{R}^{2}[/tex]

[tex]{I}_{b1} = {I}_{b1cm} + {m}_{b}{0}^{2} = \frac{2}{5}{m}_{b}{r}_{b}^{2}[/tex]

[tex]{I}_{b2} = {I}_{b2cm} + {m}_{b}{R}^{2} = \frac{2}{5}{m}_{b}{r}_{b}^{2} + {m}_{b}{R}^{2}[/tex]

[tex]{I}_{b3} = {I}_{b3cm} + {m}_{b}{\left(2R\right)}^{2} = \frac{2}{5}{m}_{b}{r}_{b}^{2} + 4{m}_{b}{R}^{2}[/tex]

With those equations in hand, it was easy to substitute back into the equation for the moment of inertia of the system and perform some algebraic simplification. The result was:

[tex]{I}_{sys} = \frac{3}{2}{m}_{d}{R}^{2} + \frac{2}{5}{m}_{b}{r}_{b}^{2} + \frac{2}{5}{m}_{b}{r}_{b}^{2} + {m}_{b}{R}^{2} + \frac{2}{5}{m}_{b}{r}_{b}^{2} + 4{m}_{b}{R}^{2}[/tex]

[tex]{I}_{sys} = \frac{3}{2}{m}_{d}{R}^{2} + 3\left(\frac{2}{5}{m}_{b}{r}_{b}^{2}\right) + 5{m}_{b}{R}^{2}[/tex]

[tex]{I}_{sys} = \frac{3}{2}{m}_{d}{R}^{2} + \frac{6}{5}{m}_{b}{r}_{b}^{2} + 5{m}_{b}{R}^{2}[/tex]

[tex]{I}_{sys} = \frac{3}{2}{m}_{d}{R}^{2} + 5{m}_{b}{R}^{2} + \frac{6}{5}{m}_{b}{r}_{b}^{2}[/tex]

[tex]{I}_{sys} = \left(\frac{3}{2}{m}_{d} + 5{m}_{b}\right){R}^{2} + \frac{6}{5}{m}_{b}{r}_{b}^{2}[/tex]

As you can see, I arrive at a solution which is close to one of the options presented in the multiple choice answer block. On the test, I was forced to select option

*c*, but that is not why am posting this. I'm just not satisfied with the answer. Am I making incorrect assumptions? Is the question itself making incorrect assumptions? Is it possible that I am procedurally incorrect? How can they just discount the size of the marbles? I understand that we are already in unreal terrirtory since marbles and turntables are not of uniform density and mass. Thanks in advance for helping to clear my mind of what has been a nagging issue for me since I took the test.