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An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s1045 rad/s ). If a particular disk is spun at 910.0 rad/s910.0 rad/s while it is being read, and then is allowed to come to rest over 0.167 seconds0.167 seconds , what is the magnitude of the average angular acceleration of the disk?

Respuesta :

To solve this problem we must keep in mind the concepts related to angular kinematic equations. For which the angular velocity is defined as

[tex]\omega_f =\omega_i-\alpha t[/tex]

Where

[tex]\omega_f =[/tex] Final angular velocity

[tex]\omega_i =[/tex] Initial angular velocity

[tex]\alpha =[/tex]Angular acceleration

t= time

In this case we do not have a final angular velocity, then

[tex]\omega_i = \alpha t[/tex]

Re-arrange for [tex]\alpha[/tex]

[tex]\alpha= \frac{\omega_i}{t}[/tex]

[tex]\alpha = \frac{910}{0.167}[/tex]

[tex]\alpha = 5449.1 rad\s^2[/tex]

Therefore the mangitude of the angular aceleration is 5449.1rad/s²