91Ó°ÊÓ

If an object is rotating about a fixed axis, thelinear velocity (v) of any point on the object located at a distance r from the axis of rotation is related to the angular velocity\(\left( \omega \right)\)of the object by the following relation:

\(v = r\omega \)

In this problem,the linear velocity of the CD is related to its angular velocity in accordance with the above relation.

The diameter of the audio compact disc (CD) is\(D = 12.0\;{\rm{cm}} = 12.0 \times {10^{ - 2}}\;{\rm{m}}\).

The minimum radius of the smallest path is\({R_1} = 2.5\;{\rm{cm}} = 2.5 \times {10^{ - 2}}\;{\rm{m}}\).

The maximum radius of the spiral path is\({R_2} = 5.8\;{\rm{cm}} = 5.8 \times {10^{ - 2}}\;{\rm{m}}\).

The linear speed of a point in the CD is v= 1.25 m/s.

The rotational frequency f of the CD is related to the angular velocity \(\omega \) of the CD by the following relation:

\(\omega = 2\pi f\)

If the laser is located at radius R of the spiral path, then the linear velocity is given by:

\(\begin{aligned}{l}v = R\omega \\v = R\left( {2\pi f} \right)\\f = \frac{v}{{2\pi R}}\end{aligned}\)

This is the relation between rotational frequency and the radius of the spiral path.

When the laser is located at radius\({R_1}\), the rotational frequency of the CD is:

\(\begin{aligned}{c}{f_1} = \frac{v}{{2\pi {R_1}}}\\ = \frac{{1.25\;{\rm{m/s}}}}{{2 \times 3.14 \times \left( {2.5 \times {{10}^{ - 2}}\;{\rm{m}}} \right)}}\\ = 8.0\;{\rm{rps}}\\ = 8.0\; \times 60\;{\rm{rpm}}\\ = \;480\;{\rm{rpm}}\end{aligned}\)

When the laser is located at radius\({R_2}\), the rotational frequency of the CD is:

\(\begin{aligned}{c}{f_2} = \frac{v}{{2\pi {R_2}}}\\ = \frac{{1.25\;{\rm{m/s}}}}{{2 \times 3.14 \times \left( {5.8 \times {{10}^{ - 2}}\;{\rm{m}}} \right)}}\\ = 3.4\;{\rm{rps}}\\ = 3.4\; \times 60\;{\rm{rpm}}\\ = \;204\;{\rm{rpm}}\end{aligned}\)

Thus, the value of rotational frequency when the laser is located at radius \({R_1}\)is 480 rpm and at radius \({R_2}\)is 204 rpm.