• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Values for mass

  • Thread starter shadowice
  • Start date
1. 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???


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

3. The Attempt at a Solution

i have all the work written out im 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

Homework Helper
3,079
4
Since what they want is a ratio just develop the mass as a ratio to begin with.

For instance Mc/Mr = (6/16)2*(1/2)*(.9)

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

Lc/Lr = Ic * ωc / Ir * ωr

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

Related Threads for: Values for mass

Replies
8
Views
2K
Replies
1
Views
649
Replies
1
Views
2K
Replies
18
Views
621

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top