Question

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, 395 rev/min. Ignoring the holes in both objects and assuming that a compact disc has three quarters 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?

Answer 1

Angular momentum L is a product of moment of inertia I and angularspeed ? such as

L = I?

Then L₁/L₂= I₁?₁ /I₂?₂   
where I for each cases is
I₁=0.5m₁R₁²
I₂=0.5m₂R₂²

L₁/L₂=0.5m₁R₁² ?₁ /0.5m₂R₂² ?₂
L₁/L₂= m₁R₁² ?₁ / m₂R₂² ?₂

L₁/L₂= m₁(12)²395 / m₂ (32)² 33.333
L₁/L₂= 1.67 m₁ / m₂

Since m₁= 0.9??R₁²h and m₂= ??R₂²(2h) we finally have
L₁/L₂= 1.67*9??R₁² h /  ??R₂² (2h)

L₁/L₂= 1.67x 0.9(12)²    / (32)² x 2
L₁/L₂= 0.1056