A compact disc of a CD player has a moment of inertia I = 2.5 *10^-5 kg · m² and rotates at 500 rev/min. Taking L = I$ \omega $ and using the quantization of angular momentum, the approximate value of L is ( = 1.055* 10^-34 J · s) A) 1.2* 10³² B) 2.0 * 10³⁰ C) 1.2* 10³¹ D) 130 E) 3.5 * 10¹⁵

Question

Quizplus · Accepted Answer

Using the given formula L = Iω, we can calculate L: 
L = (2.5×10^(-5) kg·m^2)×(500 rev/min)×(2π/60) = 2.09×10^(-2) kg·m^2/s
We are asked to choose the closest value to L that satisfies the quantization of angular momentum principle, which states that L must be an integer multiple of h/2π (where h is Planck's constant). Therefore: 
L = (n h)/2π 
Solving for n: 
n = (2πL)/h ≈ 1.94 
The closest integer to 1.94 is 2, and therefore the best choice is B, 2.0×10^30 kg·m^2/s.

A Compact Disc of a CD Player Has a Moment ω \omega ω

A Compact Disc of a CD Player Has a Moment $\omega$