# Finding mass from moment of inertia graph - inconsistency?

1. Feb 9, 2015

### marenubium

1. The problem statement, all variables and given/known data
I have a disk that can rotate about an axis at a radial distance h from the center of the disk. I am given a graph showing the the rotational inertia I of the disk as a function of the distance h from the center of the disk out to the edge of the disk. From the graph (see attached picture) I can see that the ends of the graph are as follows: when h is zero (i.e. the axis is right on the center of the disk) I = 0.03 kg*m^2 and when h = 0.2m, I = 0.63 kg*m^2.

I am asked to determine the mass of the disk from these data.

2. Relevant equations

$I_p = I_{cm} + Md^2$ (parallel axis theorem)
$I_{disk} = 1/2Mr^2$ (the moment of inertia of a disk about its CM)

3. The attempt at a solution

Let point A be when h = 0, i.e. the axis is right on the CM. Let point B be when h = 0.2, i.e. on the edge of the disk. Applying the parallel axis theorem to both locations and given the values from the problem:

$I_B = I_{disk} + M(0.2)^2 = 0.63$
$I_A = I_{disk} + M(0)^2 = 0.03$

Subtracting the second equation from the first gives

$I_B - I_A = 0.63 - 0.03 = 0.60 = M[(0.2)^2 - 0^2]$

and so M = 15kg, which is the correct answer. But if I actually try to use this to recalculate the moment of inertia of the disk I get

$I_{disk} = 1/2Mr^2 = 1/2(15)(0.2^2) = 0.3 \not= I_A = 0.03$

and similarly I get the "wrong" answer for I_B using the parallel axis theorem. Is the problem just written poorly, or am I missing something really obvious here?

Thanks for any help... first time using Tex... hope I did okay.

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2. Feb 9, 2015

### TSny

Hello, and welcome to PF!

You're thinking correctly. Note that the problem does not state that the disk is of uniform mass density. So, it might be a disk where Icm is not given by (1/2)Mr2.

3. Feb 9, 2015

### marenubium

Oh, of course! I was completely on autopilot and made that assumption without thinking.

Thanks so much!

4. Feb 9, 2015

### lightgrav

The problem was poorly written ... the disk's radius is only .067m, but they can spin it around an axis that's 3x that far from the center!
(expensive massless washer way bigger than the real disk?) . They should re-program the range of I_A values.

5. Feb 9, 2015

### marenubium

I'm admittedly not sure where you are getting 0.067 m from. The problem statement implies that the edge of the disk from the center and thus the radius is 0.2 m.

6. Feb 10, 2015

### lightgrav

oops, 0.0632 m ... from I_disk = 0.03 kgm^2 and mass 15kg.

7. Feb 10, 2015

### marenubium

Right, but as TSny pointed out it's probably not reasonable here to assume the disk is uniform.

8. Feb 10, 2015

### lightgrav

they called it a disk ... they're supposed to guide your model, not mis-guide it.