The Seagate Momentus 7200 laptop hard-disk drive (the one with the “G-Force Prot

Question

The Seagate Momentus 7200 laptop hard-disk drive (the one with the “G-Force Protection" specifications described in the discussion problem in Homework 2) spins at 7200 revolutions per minute (rpm). This is typical of hard disks, and is probably happening inside your laptop or desktop right now. The diameter of the disk is 2.5 inches. (a) What is the angular velocity of the disk? (b) What is the speed of the outer edge of the disk? Express your result in m/s and miles per hour. (Recall that 1 m/s = 2.2 miles/hour.) (c) What is the radial acceleration of a point at the outer edge of the disk? Express your result in terms of g, the acceleration due to gravity at the surface of the earth (i.e. divide your answer by 9.8 m/s2). Suppose the disk can spin down from 7200 rpm to zero angular speed in 0.3 seconds (the time it takes to lock the head, which we used in Problem Set 2). Assume constant angular deceleration. (d) What is the magnitude of the angular acceleration of the disk? (e) What is the tangential acceleration at the edge of the disk? Express your result in terms of g, i.e. divide your answer by 9.8 m/s2.

Explanation / Answer

a) 7200 rpm = 120 rotation /s

1 rotation = 2 pi rad

so , 120 rotation /s = 120* 2*pi rad /sec = 753.98 rad /s

b) v = w *r

r = 2.5 inch = 6.35 cm = 0.0635 m

so , v = 753.98 rad /s * 0.0635 m = 47.87 m/s

= 47.87 * 2.2 = 105.33 miles /hour

c) radial = v2/ r = (47.87 ) 2 / 0.0635

=36087.19 m/s2

in terms of g = 3678.61 g

d) alpha =( final angular speed - inital angular speed) / time

= (0- 753.98 rad /s ) / 0.3

= -2513 rad /s2

e) tangential acceleration = alpha * r

= -2513 rad /s2 * 0.0635 = 153.59 m/s 2

in terms of g it is 16.26 g

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