Question

(II) 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 (rad/s) 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 0.50 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?

Answer 1

Answer:

Explanation:

Given:

Angular velocity, w = 7200 rpm

Radius, r = 3 cm

= 0.03 m

1 bit = 0.5 × 10^-6 m

A.

1 rev = 2pi rad

7200 rev/min × 2pi rad/1 rev × 1 min/60 sec

= 753.98 rad/s

B.

Linear velocity, v = r × w

= 0.03 × 753.98

= 22.62 m/s

C.

1 bit = 0.5 × 10^-6 m

Velocity = 22.62 m/s

1 bit/0.5 × 10^-6 m × 22.62 m/s

= 4.524 × 10^7 bits/s

Answer 2

Using the appropriate relation, the angular velocity, linear speed and speed in bits are calculated thus :

Given the Parameters :

• Radius, r = 3 cm = 0.03 m
• Angular velocity, w = 7200 rpm
• bit to meters = 0.5 × 10^-6 m

A.)

Angular velocity in rad/sec :

• 1 rev = 2pi rad
• 1 minute = 60 seconds

Using dimensional analysis :

7200 rev/min × 2pi rad/1 rev × 1 min/60 sec

= 753.98 rad/s

B.)

Linear speed of the platter :

• Linear speed , s = r × w

s = 0.03 × 753.98

s = 22.62 m/s

C.)

Number of bits per second required :

$1 bit = 0.5 times 10^{-6} m$

Linear speed, s = 22.62 m/s

Converting the speed in m/s to bits :

$\frac{1}{0.5 × 10^{-6}} \times 22.62 m/s$

$= 4.524 × 10^{7} bits/s$

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