# Moment of inertia of two coaxial disks

• Nexus99
In summary, the total moment of inertia for a system of two disks revolving about a common axis is given by ##I_{tot} = \frac{17}{10} m R^2##, where ##m## is the total mass of the system and ##R## is the radius of each disk. The individual masses of the disks can be calculated by using the thickness of each disk and the density of the disks, which are assumed to be the same.
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?

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.

Isn't what i did?

Okpluto said:
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

Okpluto said:
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.

Okpluto said:
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.

Okpluto said:
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

## 1. What is moment of inertia of two coaxial disks?

The moment of inertia of two coaxial disks refers to the measure of an object's resistance to changes in its rotational motion around its central axis. It is determined by the mass distribution of the two disks and their distances from the axis of rotation.

## 2. How is moment of inertia of two coaxial disks calculated?

The moment of inertia of two coaxial disks can be calculated by using the formula I = ½MR², where I is the moment of inertia, M is the mass of the disks, and R is the distance between the disks and the axis of rotation.

## 3. What factors affect the moment of inertia of two coaxial disks?

The moment of inertia of two coaxial disks is affected by the mass distribution of the disks, their distances from the axis of rotation, and the size and shape of the disks.

## 4. How does the moment of inertia of two coaxial disks relate to rotational kinetic energy?

The moment of inertia of two coaxial disks is directly related to the rotational kinetic energy of the system. A larger moment of inertia means that more energy is required to rotate the disks at a given angular velocity.

## 5. Can the moment of inertia of two coaxial disks be changed?

Yes, the moment of inertia of two coaxial disks can be changed by altering the mass distribution of the disks, their distances from the axis of rotation, or the size and shape of the disks. It can also be changed by adding or removing additional disks to the system.

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