An optical disk drive in a computer can spin a disk at up to 10,000 rpm. If a particular disk is spun at 5050 rpm while it is being read, and then is allowed to come to rest over 0.435 s, what is the magnitude of the average angular acceleration of the disk?

α = 1215.71 rad/s²

Explanation:

The angular acceleration of an object is defined as the time rate of change of angular velocity of he object. The formula for the angular acceleration of an object is given as follows:

α = (ωf - ωi)/Δt

where,

α = angular acceleration of the disk = ?

ωf = Final Angular Velocity = 0 rad/s (Since disk finally stops)

ωi = Initial Angular Velocity = (5050 rpm)(2π rad/rev)(1 min/60 s)

ωi = 528.83 rad/s

Δt = time interval = 0.435 s

Therefore,

α = (0 rad/s - 528.83 rad/s)/(0.435 s)

α = - 1215.71 rad/s²

here, negative sign indicates that the direction of acceleration is opposite to the direction of angular velocity or the angular motion. So, the magnitude of acceleration will be:

α = 1215.71 rad/s²

it should be -8.6 cm hope this out