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 12.70 cm and a mass of 16.30 g. The laser scans at a constant tangential velocity of 1.150 m/s. Assume that the music is first detected at a radius of 20.90 mm from the center of the disk. Ignore the small circular hole at the CD's center.

Answer 1

Answer:

Explanation:

Given

Diameter of CD $d=12.70\ cm$

radius $r=6.35\ cm$

mass of CD $m=16.30\ gm$

Tangential velocity $v_t=1.150\ m/s$

music detected at $r'=20.90\ mm=2.090\ cm$

Moment of Inertia of disc $I=\frac{mr^2}{2}$

$I=\frac{16.30\times 10^{-3}\times (6.35\times 10^{-2})^2}{2}$

$I=6.572\times 10^{-5} kg-m^2$

Work done is equal to change in kinetic Energy of CD

$W=\Delta K$

$W=\frac{1}{2}I\omega_f^2-\frac{1}{2}I\omega_i^2$

where $\omega =$angular velocity

$v_t=\omega \times r$

$W=\frac{1}{2}\times 6.572\times 10^{-5}\left ( \left ( \frac{1.15}{0.0209}\right )^2-0\right )$

$W=0.0994\ J$