Moment of inertia of 3 disc system

[Rosengrip]

Homework Statement

We have 3 discs, arranged as the sketch below shows. Find the moment of inertia of the whole system around the axis, passing horizontally through center of the bigger disk:

1. Two smaller disks are fixed and cannot rotate around their axes.

2. Two smaller disks can rotate freely around their axes.


R is the radius of big circle
r is the radius of 2 smaller disks
L is the distance between COM and axes of smaller disks.
M is the mass of bigger disk
m is the mass of 2 smaller disks

[PLAIN]http://www.shrani.si/f/3L/Os/9CLgTJj/rotacija.jpg

Homework Equations

I = (1/2)*m*R^2 ---> moment of inertia for disk
Parallel axis theorem

The Attempt at a Solution

For the 1st point, my solution would be:
I = (0.5)*M*R^2 + 2*((0.5)*m*r^2 + m*L^2)

However I am not sure about the 2nd one, any hints? Thanks in advcance

 

[tiny-tim]

Hi Rosengrip!

(try using the X² icon just above the Reply box )

We have 3 discs, arranged as the sketch below shows. Find the moment of inertia of the whole system around the axis, passing horizontally through center of the bigger disk:

2. Two smaller disks can rotate freely around their axes.
…
For the 1st point, my solution would be:
I = (0.5)*M*R^2 + 2*((0.5)*m*r^2 + m*L^2)

However I am not sure about the 2nd one, any hints? Thanks in advcance

Your 1 looks fine.

For 2, I think you're meant to assume that the small discs do not rotate on their own axes (ie they always face the same way).

 

[Rosengrip]

Hm, I don't have the right formula just yet but I somehow think that system moment of inertia in 2nd scenario would be bigger than one in 1st. Is that the correct assumption?

From center of bigger disk POV, the smaller disk rotates in point 2., whereas in point 1. it stands still. The situation is reversed from outside POV (in point 2. vertical line drawn on the smaller disk would always point downwards).

What bothers me is that I can't really formulate the problem under 2nd point, since all we did were cases which fall under point 1 :(

 

[tiny-tim]

Hi Rosengrip!

Hm, I don't have the right formula just yet but I somehow think …

erm … physics is equations!

stop philosophising, and do the equations …

if necessary, go back to the definition and do an ∫ (do it for something easier, like a rod!)

 

[rwisz]

I would first clarify the question and make sure I understand the system and its properties correctly. It appears that the bigger disk is fixed and cannot rotate, while the two smaller disks have different conditions (fixed vs. free rotation) depending on the scenario. I would also confirm the units of measurement being used for the radii and distances.

For the first scenario, where the two smaller disks are fixed, the moment of inertia can be calculated by summing the moments of inertia of each individual disk around the given axis. This would result in:

I = (0.5)*M*R^2 + 2*((0.5)*m*r^2)

For the second scenario, where the two smaller disks can rotate freely, the moment of inertia would also take into account the rotation of these disks around their own axes. In this case, we can use the parallel axis theorem to calculate the additional moment of inertia due to the rotation around their axes. This would result in:

I = (0.5)*M*R^2 + 2*((0.5)*m*r^2 + m*L^2) + 2*((0.5)*m*r^2)

This additional term takes into account the moment of inertia of each smaller disk around its own axis, which is given by (0.5)*m*r^2, and the distance between the axis of rotation and the axis of the bigger disk, which is L. This gives the final moment of inertia for the system.

In summary, the moment of inertia for the 3 disc system depends on the conditions of the smaller disks, and can be calculated using the appropriate equations and the parallel axis theorem. It is important to carefully consider the properties and conditions of the system in order to arrive at an accurate solution.