The platter of the hard drive of a computer rotates at 7200 rpm (rpm = revolutions per minute = rev/min). (a) What is the angular velocity \(\left( {{{{\bf{rad}}} \mathord{\left/{\vphantom {{{\bf{rad}}} {\bf{s}}}} \right.} {\bf{s}}}} \right)\) of the platter? (b) If the reading head of the drive is located 3.00 cm from the rotation axis, what is the linear speed of the point on the platter just below it? (c) If a single bit requires \({\bf{0}}{\bf{.50}}\;{\bf{\mu m}}\) of length along the direction of motion, how many bits per second can the writing head write when it is 3.00 cm from the axis?

The expression that shows the relationship between the linear velocity and angular velocity is as follows:

\(v = r\omega \)

Here, v is the linear velocity, r is the radius of the rounded path, and\(\omega \)is the angular velocity.

Given data:

The angular speed of the platter in the hard disk drive is \(\omega =7200\;{\rm{rpm}}\).

The distance of the reading head from the rotation axis is \(r = 3.00\;{\rm{cm}}\).

The length of the one bit along the direction of motion is \(l = 0.50\;{\rm{\mu m}}\).

(a)

The angular velocity of the platter in terms of rad/s can be calculated as:

\(\begin{aligned}{l}\omega &= \left( {7200\;{\rm{rpm}}} \right)\left( {\frac{{2\pi\;{\rm{rad}}}}{{1\;{\rm{rev}}}}} \right)\left( {\frac{{1\;\min}}{{60\;{\rm{s}}}}} \right)\\\omega &= 753.6\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}}{\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Thus, the angular velocity of the platter in terms of rad/s is 753.6 rad/s.

(b)

The linear speed of the point on the platter can be calculated as:

\(\begin{aligned}{l}v &= r\omega \\v &= \left( {\left( {3.00\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)} \right)\left( {753.6\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}} \right)\\v &= 22.6\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Thus, the linear speed of the point on the platter is 22.6 m/s.

(c)

The number of bits that a track can accommodate can be calculated as:

\(\begin{aligned}{l}n &= \frac{v}{l}\\n &= \frac{{\left( {22.6\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}} \right)}}{{\left( {0.50\;{\rm{\mu m}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{\mu m}}}}} \right)}}\\n &= 4.51 \times {10^7}\;{{{\rm{bit}}} \mathord{\left/{\vphantom {{{\rm{bit}}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Thus, the number of bits that a track can accommodate is \(4.51 \times{10^7}\;{{{\rm{bit}}}\mathord{\left/{\vphantom{{{\rm{bit}}}{\rm{s}}}}\right.}{\rm{s}}}\).