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, 410 rev/min. Ignoring the holes in both objects and assuming that a compact disc has two thirds 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

two points before we start:

1) angular momentum (p) and moment of inertia (I) must be described in relation to a stated axis of rotation. I assume that in this question the axis is intended to be that passing through the centres of the discs, and normal to their plane.

2) CDs do not operate at constant speed, so I assume that the final sentence in your question should include the phrase "at the stated speeds".

The basic concept we should use is this formula for the moment of inertia of a uniform disc of mass m and radius r. The axis of rotation passes through the centre of the disc, and is normal to its plane:

I = m*r^2/2.

angular momentum is p = w*I = m*w*r^2/2. Here w is the angular velocity measured in rad/s. Since we are asked for a ratio of p we can express w in terms of rpm for simplicity.

For a uniform disc of thickness t, density d and radius r the mass is m = pi*r^2*d*t, giving p = w*d*t*r^4 when all of the constants are omitted (they will cancel when the ratio is found)

we can write p1/p2 = w1*d1*t1*r1^4/(w2*d2*t2*r2^4).

#I leave the arithmetic to you.....