Edge of the CD

Problem

A disc of radius $R = 30\;\mathrm{cm}$ rotates at $3000\;\mathrm{rpm}$. Estimate the speed of the edge and discuss why, at these speeds, a cellulose or glass disc can break due to “centrifugal force”.

Solution

We convert revolutions per minute into angular velocity. $3000\;\mathrm{rpm}$ means $3000/60 = 50$ revolutions per second, so: $\omega = 2\pi \cdot 50 = 100\pi \approx 314\;\mathrm{rad/s}$ The speed of the edge (with $R = 0{,}30\;\mathrm{m}$): $v = \omega R = 314 \cdot 0{,}30 \approx 94\;\mathrm{m/s}$ In kilometres per hour: $v \approx 94 \cdot 3{,}6 \approx 340\;\mathrm{km/h}$ Why it breaks. Every element of material at the edge needs an enormous centripetal force to stay on the circular orbit. The centripetal acceleration at the edge is $a_c = \omega^2 R = 314^2 \cdot 0{,}30 \approx 3{,}0\cdot 10^4\;\mathrm{m/s^2} \approx 3000\,g$ This force must be transmitted by the internal cohesion of the material (tensile stress). In a rotating frame it appears as a “centrifugal force” that tends to fling the material outward. When the required tensile stress exceeds the strength of the material (low for cellulose or glass), the disc shatters. $\ev{v = \omega R \approx 94\;\mathrm{m/s} \approx 340\;\mathrm{km/h}}$

Links

Topics: Dinamica Concepts: Velocità angolare · Moto circolare uniforme Skills: Stime di Fermi Objects: Disco rotante