Question

How much work is done by the motor in a CD player to make a CD spin, starting from rest? The CD has a diameter of 12.0 cm and a mass of 15.8 g. The laser scans at a constant tangential velocity of 1.20 m/s. Assume that the music is first detected at a radius of 29.0 mm from the center of the disk. Use the tangential velocity of the the laser and the radius at which music is first detected to determine the angular velocity of the disc. Ignore the small circular hole at the CD’s center. J

Answer 1

Given data:

* The mass of the CD is m = 15.8 g.

* The diameter of the CD is D = 12 cm.

* The tangential velocity of the CD is v =1.2 m/s.

* The music is detected at the radius of r = 29 mm.

Solution:

The moment of inertia of CD is,

$\begin{gathered} I=\frac{1}{2}mR^2^{} \\ I=\frac{1}{2}m(\frac{D}{2})^2 \end{gathered}$

Substituting the known values,

$\begin{gathered} I=\frac{1}{2}\times15.8\times10^{-3}\times(\frac{12}{2}\times10^{-2})^2 \\ I=284.4\times10^{-3-4} \\ I=284.4\times10^{-7}kgm^2 \end{gathered}$

The initial kinetic energy of the CD is,

$K_1=\frac{1}{2}I\omega^2_1_{}$

where,

$\omega_1\text{ is the initial angular velocity of the CD}$

As the CD is initially at rest, thus, the initial kinetic energy of the CD is,

$K_1=0\text{ J}$

The final kinetic energy of the CD is,

$K_2=\frac{1}{2}I\omega^2_2$

where,

$\omega_2\text{ is the final angular velocity of CD}$

The angular velocity of the CD in terms of the translational velocity and radius of CD at which the music is detected,

$\begin{gathered} \omega_2=\frac{v}{r} \\ \omega_2=\frac{1.2}{29\times10^{-3}} \\ \omega_2=0.04138\times10^3 \\ \omega_2=41.38\text{ rad/s} \end{gathered}$

Thus, the final kinetic energy is,

$\begin{gathered} K_2=\frac{1}{2}\times284.4\times10^{-7}\times(41.38)^2 \\ K_2=243489.7\times10^{-7} \\ K_2=0.024\text{ J} \end{gathered}$

The work done on the CD is the change in its kinetic energy, thus, the work done is,

$\begin{gathered} W=K_2-K_1 \\ W=0.024-0 \\ W=0.024\text{ J} \end{gathered}$