# B Moment of Inertia of a System - Disk with seperate masses

1. May 25, 2016

### Bhollehday

Ive got the following disk I am attempting to find the torque required to turn it 1/4 turn in 1/4 second.

Am I correct when Im considering I(D)=1/2mr^2 as its a disk, and I(d)=mr^2? and The final moment of inertia to be I(D)+I(d)?

Here is the disk I am speaking of:
https://www.dropbox.com/s/ax2rwpffhl5d4m5/Capture (2).PNG?dl=0

2. May 25, 2016

### Staff: Mentor

Can you just post the diagram? (I cannot access the link.)

3. May 25, 2016

### jbriggs444

The drawing shows a top view of a flat disk which apparently spins in the horizontal plane. The disk appears to be of uniform density except for a small hub attached with four screws. [Think of the turntable on a phonograph player]

On the top of the disk are placed two additional flat disks. Their diameter is perhaps 40% of the diameter of the main disk. Each is secured to the original disk with twelve screws. The two disks are positioned diametrically opposite one another. Neither overlaps the hub. Neither hangs over the edge. They both fit evenly between hub and rim. Both are of equal size.

The main disk is labeled ID. One of the smaller disks is labelled Id. No measurement has been made of the distance between the center of the small disks and the center of the large disk. That measurement is, of course, key to computing the moment of inertia of the combined assembly.

Bhollehday, you may want to Google for the "parallel axis theorem". The small disks contribute their own moment of inertia to the total. But they also contribute based on their mass and their distance from the center of the big disk.

Edit:

In the formulas above you have $I_D$ as $\frac{1}{2}mr^2$ and $I_d$ as $mr^2$. That is an unfortunate use of notation because you are re-using both "m" and "r" with two different meanings in each case.

No, the final moment of inertia will not be $I_D + I_d$. It will have an additional component based on the parallel axis theorem as above.