Question

A DVD is spinning at 592.309 rpm. What is the size of the angular momentum of the DVD if it has a circumference of 39.543 cm and a mass of 10.803 g? Enter your answer in kg m2/s to the thousandth's place.

Answer 1

Angular velocity of the DVD, $\omega$ = 592.309 rpm

Convert the unit of angular velocity in rad/s -

$\omega$ = (592.309 / 60)*(2*pi) rad/s = 62.01 rad/s.

Now, circumference of the DVD = 39.543 cm

Hence, its diameter, d = (39.543) / (pi) = (39.543) / 3.141 = 12.59 cm

Hence, its radius, r = d/2 = 12.59/2 = 6.29 cm = 0.0629 m

Mass of DVD, m = 10.803 g = 0.010803 kg

So, moment of inertia of DVD about its center point, I = (1/2)*m*r^2

= (1/2)*0.010803*0.0629^2 = 2.14 x 10^-5 kg*m^2

Therefore, angular momentum of the DVD, L = I*$\omega$

= 2.14 x 10^-5 kg*m^2 x 62.01 rad/s

= 1.327 x 10^-3 kg*m^2/s