# Computing net Moment of Intertia

1. Nov 12, 2013

### Nicolaus

1. The problem statement, all variables and given/known data
An object is made from 2 solid disks: m1 = 23 kg, m2 = 39kg, and r1 = 17cm, and r2 = 33cm. The disks are connected by a thin rod of M = 11kg, L = 67cm. The object is held in a horizontal position and then let go. The object is free to rotate about a pivot through centre of mass of of larger disk. Find the speed of the CM of smaller disk when it has swung 90degrees from original horizontal position.

2. Relevant equations
I = ML^2 / 3 (rod - axis running through edge perpendicular to bar) or ML^2/12 for CM axis?
I = mr^2 / 2 (disks)

3. The attempt at a solution
To compute the net rotational inertia:
I (total) = I(rod) + I(small disk) + I(large disk)
Do I also have to incorporate the rotational inertia of both the small disk and rod from the distance from CM of large disk (which acts as pivot)?
(ie. would I also include mr^2 (small disk, where r is length from CM of large disk), mr^2 (for rod))?

And for the height that the CM of small disk falls (which is on the far end), would it be the length of rod + radii of both small and large disk?

2. Nov 12, 2013

### haruspex

The geometry is unclear. What is the relationship between the axis of the rod and those of the disks? I'll assume the disks are parallel and the rod is at right angles, running through the axes of both disks.
That's for a disk rotating around its axis. Is that what's happening here?

3. Nov 12, 2013

### Nicolaus

( m1 )---rod----(m2)
They are attached like so. The rod is connected to the end of either disk, not to their respective CMs. Imagine the disks laying flat on a surface when looking at the above diagram.

4. Nov 12, 2013

### SteamKing

Staff Emeritus
Since the object is rotating about an axis thru the C.O.M. of the larger disk, the inertia must be computed about that axis. It's more than just adding up the inertias of the individual objects: the parallel axis theorem will come into play.

5. Nov 12, 2013

### Nicolaus

M1 = large mass, m2 = small mass, M = rod mass
I = M1r^2 / 2 + [m2r^2 / 2 + m2r^2] + [ML^2 / 3 + Mr^2]
r in m2r^2 is the length from the CM of M1 to CM m2.
r in Mr^2 (rod) is merely the radius of M1 because rod is connected to it and I computed its moment of inertia with respect to axis perpendicular to and running through edge of rod.
Does this look right?

6. Nov 12, 2013

### haruspex

That's the M.I. for a disk rotating on its axis. The disk here is rotating about a diameter.
You cannot add displacements like that. The parallel axis theorem is for adding a single displacement to the MI about the c.o.m. Take the MI of the rod about its centre, then use the displacement from there to the axis.

7. Nov 12, 2013

### Nicolaus

Isn't the larger, pivot, mass being rotated about its axis?
I (rod) = ML^2 / 12 + M(L/2 +r1)^2 -> where r1 is the radius of the pivot mass. ?
Don't we have to compute BOTH the M.I. of each mass with respect to their CMs AND their mass times distance from the CM of the pivot mass?
What are you getting for this?

8. Nov 12, 2013

### haruspex

It's not the axis of the disk in the usual sense (which would be perpendicular to the disk). See http://hyperphysics.phy-astr.gsu.edu/hbase/tdisc.html
Yes, that's better. Do you see that this is different from what you wrote before?