How Does Drilling a Hole Affect the Rotational Inertia of a Disk?

[gills]

Homework Statement

A disk of radius R has an initial mass M. Then a hole of radius (1/4)R is drilled, with its edge at the disk center (The center of mass of the cutout is in the x positive direction). Find the new rotational inertia about the central axis.

Hint: Find the rotational inertia of the missing piece, and subtract it from that of the whole disk. You'll need to determine what fraction of the missing mass is of the total M and use the parallet-axis theorem.

Homework Equations

Parallel-axis theorem:
I = I_cm + md^2

Rotational Inertia of solid disk:
I = (1/2)MR^2

The Attempt at a Solution

My attempt thus far is not very good. I having trouble getting the mass of the small disk. Any advice?

 

[mjsd]

assuming uniform distribution of mass, you need to work out the ratio between the two disks. note [tex]A=\pi r^2[/tex] and you are given two different r's. after that follow the hints and you should be right

 

[ogb p]

I would approach this problem by first breaking down the components involved. We have a disk with a radius R and a mass M, and a hole with a radius (1/4)R drilled at the center of the disk. To find the new rotational inertia, we need to determine the rotational inertia of the missing piece (the small disk) and subtract it from the rotational inertia of the whole disk.

To find the mass of the small disk, we can use the fact that the volume of a disk is given by V = πr^2h, where r is the radius and h is the height. In this case, the height of the small disk is equal to the radius of the whole disk, R. We can also use the fact that the volume of the missing piece is equal to the volume of the whole disk minus the volume of the small disk. So, we have:

V_small disk = V_whole disk - V_small disk
= πR^2h - π(1/4)R^2h
= (3/4)πR^2h

Now, we can use the density (ρ) of the material to find the mass (m) of the small disk:

m = ρV_small disk
= ρ(3/4)πR^2h

We also need to determine the distance (d) between the center of mass of the small disk and the central axis. This can be found by using the parallel-axis theorem:

d = (1/4)R

Finally, we can use the parallel-axis theorem to find the rotational inertia of the small disk:

I_small disk = (1/2)m(1/4)R^2 + md^2
= (1/8)mr^2 + (1/4)mR^2
= (1/8)(ρ(3/4)πR^2h)R^2 + (1/4)(ρ(3/4)πR^2h)(1/4)R^2
= (3/32)ρπR^4h + (3/64)ρπR^4h
= (9/64)ρπR^4h

Now, we can subtract this from the rotational inertia of the whole disk (I = (1/2)MR^2) to find the new rotational inertia:

I_new =