Question

A computer hard drive has a disc that rotates at 10,000 rpm. The reader head is positioned 0.0005 in. above the disc’s surface. Estimate the shear force on the reader head due to the air between the disc and head.

Answer 1

shearing force is
`3.40* 10^(-4) lb`

Step-by-step explanation:

we know that

force can be determined
`F = \tau * A`

Area can be determine as

`A  = (\pi)/(4) d^2 = (\pi)/(4) [(0.2)/(12)]^2 =   2.18* 10^(-4) ft^2`

linear velocity can be determines as

`\tau = \mu_(air) (U)/(b)`

dynamic viscosity of air
`\mu_(air) = 3.74* 10^(-7) lb-s/ft^2`

veolcity of disc

`U =\omega R`

`U = (2\pi N)/(60) * R = (2\pi 10,000)/(60) * (2)/(12)`

U = 174.5 ft/s

so

`\tau = 3.74* 10^(-7) * (174.5)/((0.0005)/(12))`

`\tau = 1.56 lb/ft^2`

`F = 1.56* 2.18* 10^(-4) = 3.40* 10^(-4) lb`

shearing force is
`3.40* 10^(-4) lb`