Moment of inertia of two coaxial disks

[Nexus99]

Homework Statement

A rigid body consists of two thin and coaxial disks having the same density and thickness. One disk has a radius $R$, while second has a radius of $2R$. The total mass is $m$. Calculate the total moment of inertia respect to an axis passing through the center of the rigid body and perpendicular to the disk surfaces

Relevant Equations

density, lenght of a circumference ecc.

The total moment of inertia is:
$I_{tot} = 2 M_1 R^2 + \frac{1}{2} M_2 R^2$
We have $M_1 = (4 \pi R^2) \sigma$ and $M_2 = (\pi R^2) \sigma$ , where $\sigma$ is the density of the disks.
We also know that:
$\sigma = \frac{m}{ \pi 5 R^2}$

this leads us to say that:
$I_{tot} = \frac{17}{10} m R^2$

Is it ok?

 

[Lnewqban]

Perhaps I don’t understand the problem, but, since each disk is revolving about its centre of mass and a common axis, should not the total moment of inertia of that rigid body be given by the sum of the moi of the individual disks?
If so, the initial step should be to calculate the mass of each disk, based on the radius.

 

[Nexus99]

Isn't what i did?

 

[Lnewqban]

Isn't what i did?

Yes, my result is the same:
$I_{total}=\frac {17} {10} mR^2$

I did not use density, but compared individual masses to total mass of the system.
$m_1=\frac {4m} {5}$

$m_2=\frac {m} {5}$

 

[Nexus99]

Ok thanks. Which metod did you use? My only idea was to use density

 

[haruspex]

Ok thanks. Which metod did you use? My only idea was to use density

The methods are the same. @Lnewqban used density to obtain the ratio of the masses.

 

[Lnewqban]

Ok thanks. Which metod did you use? My only idea was to use density

I meant that calculating the individual masses was not necessary, since the values of density and thickness of the disks were not given, but were said to be the same for both.
Since the equation for the total moment of inertia of the system only involves total mass (m) of the system and R, calculating the proportion of each mass respect to given m, based on radius of each disk, was sufficient.
Hence, the proportions of masses that I showed in post #4 above.

Mass of each disk equals the product of volume and density.
Your equations for $m_1$, $m_2$ and $\sigma$ in post #1 are including the circular area of the disks but are missing the thickness.

 

[Lnewqban]

Ok thanks. Which metod did you use? My only idea was to use density

I meant that calculating the individual masses was not necessary, since the values of density and thickness of the disks were not given, but were said to be the same for both.

Since the equation for the total moment of inertia of the system only involves total mass (m) of the system and R, calculating the proportion of each mass respect to given m, based on radius of each disk, was sufficient.
Hence, the proportions of masses that I showed in post #4 above.

Mass of each disk equals the product of volume and density.
Your equations for $m_1$, $m_2$ and $\sigma$ in post #1 are including the circular area of the disks but are missing the thickness.

Let’s $t$ be the thickness of each disk,

$m_1 = (4 \pi R^2t)\sigma$
$m_2 = (\pi R^2t)\sigma$
$\sigma = \frac{m}{ 5\pi R^2t}$

 

[Nexus99]

Ok thanks, i got it