Question

On a 12.0-cm-diameter audio compact disc (CD), digital bits of information are encoded sequentially along an outward spiraling path. The spiral starts at radius \({{\bf{R}}_{\bf{1}}}{\bf{ = 2}}{\bf{.5}}\;{\bf{cm}}\) and winds its way out to radius \({{\bf{R}}_{\bf{2}}}{\bf{ = 5}}{\bf{.8}}\;{\bf{cm}}\). To read the digital information, a CD player rotates the CD so that the player’s readout laser scans along the spiral’s sequence of bits at a constant linear speed of 1.25 m/s. Thus the player must accurately adjust the rotational frequency f of the CD as the laser moves outward. Determine the values for f (in units of rpm) when the laser is located at \({{\bf{R}}_{\bf{1}}}\) and when it is at \({{\bf{R}}_{\bf{2}}}\).

Short answer

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.

Step-by-step solution

Relation between linear and angular velocity

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.

Given information

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.

Determination of the relation between rotational frequency f and radius R

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.

Determination of the values of rotational frequency f

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.