Ratio of angular momentum = ?

[shadowice]

Homework Statement

Compact discs and long-playing records are made from similar materials. The former have a diameter of about 12 cm, and the latter, about 32 cm. When in use, records spin at 33.333 rev/min, and compact discs spin at, say, 405 rev/min. Ignoring the holes in both objects and assuming that a compact disc has half the thickness of a record and 0.90 of its density, what is the ratio of the angular momentum of a compact disc in use to that of a record?
mc= mass of cd
mR= mass of record
Wrec = 33 1/3 rev/min = 3.49 rad/s
Wcd = 405 rev/min = 42.4 rad/s
rcd= 6cm = .06m
rR = 32cm = .16m
Height R = 1?
Height cd = .5?
density R = 1?
density CD = .9?

Homework Equations

m= d*v
v= 2pi*r^2*H
rev/min = 2pi/60 rad/s
I = 1/2mr^2

The Attempt at a Solution

i have all the work written out I am just kind of stuck on what to give the height and density values. So ill show you guys the symbol math and hopefully you can help me figure out what numerical values to give them.
mR = p*2pi*rR^2*HR
mcd = .9p*2pi*rcd^2*Hcd

Lcd/LR = Icd*Wcd/IR*Wr
= [(1/2mcd*rcd^2)Wcd]/[(1/2mR*rR^2)WR]

 

[LowlyPion]

Since what they want is a ratio just develop the mass as a ratio to begin with.

For instance M_c/M_r = (6/16)²*(1/2)*(.9)

That should get you almost the whole way there, because otherwise

L_c/L_r = I_c * ω_c / I_r * ω_r

And ω/ω = 405/33.3 because here radians, revs work out the same.

 

[nlightner]

= [(.9p*2pi*.06^2*.5)(42.4)]/[(.9p*2pi*.16^2)(3.49)]
= [(.000408p)(42.4)]/[.001766p]
= .0931

The ratio of angular momentum of a compact disc to that of a record is approximately 0.0931. This means that the compact disc has significantly less angular momentum compared to the record, given the same materials and spinning at different speeds. This could be due to the smaller diameter and thickness of the compact disc, resulting in a smaller moment of inertia.